مسائل محلولة للفصل الاول الفيزياء المتسعات

1. ‫ﻣﺴﺎﺋﻞ‬‫ﻣﺤﻠﻮﻟﺔ‬
‫اﻷول‬ ‫اﻟﻔﺼﻞ‬ ‫ﺣﻮل‬
)‫اﻟﻤﺘﺴﻌﺎت‬(
‫اﻋﺪاد‬
‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﻴﺪ‬ ‫اﻟﻤﺪرس‬
email/abuhussen_72@yahoo.com
www.facebook.com/saeedmuhi
2015 – 2016 ( ‫اﻟﻄﺒﻌﺔ‬‫اﻟﺜﺎﻧﻴﺔ‬ )

2. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
1
‫اﻟﻔﺼﻞ‬‫اﻷول‬/‫اﻟﻤﺘﺴﻌﺎت‬
‫ﻣﺘﺴﻌﺔ‬‫ﻣﻨﻔﺮدة‬:
‫ﻣﺜﺎل‬1/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫رﺑﻄﺖ‬20µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫واﻟﺒﻌ‬0.3cm‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬12V
‫اﺣﺴﺐ‬:
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬.
3-‫اﻟﺼﻔﯿﺤﺘﯿ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫ﻦ‬.
‫اﻟﺤﻞ‬/
J10144)12(1020
2
1
)V.(C
2
1
PE3
m/V4000
103.0
12
d
V
E2,C2401220V.CQ1
5262
2
−−
−
×=××=∆=−
=
×
=
∆
=−µ=×=∆=−
‫ﻣﺜﺎل‬2/‫ﻰ‬‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺘﺴﻌﺘﺎن‬5µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫وﻓ‬12V‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺮى‬‫واﻻﺧ‬4µF‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫وﻓ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬6V‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﺣﺴﺐ‬.
‫اﻟﺤﻞ‬/
J107236102)6(104
2
1
)V.(C
2
1
PE
J10360144105
2
1
)12(105
2
1
)V.(C
2
1
PE
66262
222
66262
111
−−−
−−−
×=××=××=∆=
×=×××=××=∆=
‫ﻣﺜﺎل‬3/‫اﻟﻜﮭ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻣﻘﺪار‬ ‫ﻣﺎ‬‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬ ‫ﺮﺑﺎﺋﻲ‬20µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺸﺤﻨﺔ‬‫واﻟ‬
120µC‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫واﻟﺒﻌﺪ‬0.3cm
‫اﻟﺤﻞ‬/
V
Q
C
∆
= ⇒ V6
20
120
C
Q
V ===∆
m/V2000
103.0
6
d
V
E 2
=
×
=
∆
= −
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬4/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬2μF‫ﺤﻨﺘﮭﺎ‬‫وﺷ‬100μC‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫واﻟﺒﻌ‬‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬0.5cm‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﺪة‬‫ﺷ‬ ‫ﺎ‬‫ﻣ‬
‫؟‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬
‫اﻟﺤﻞ‬/
V50
2
100
C
Q
V ===∆
m/V10000
105
50
105.0
50
d
V
E 32
=
×
=
×
=
∆
= −−
‫ﻣﺜـﺎل‬5/‫ﺴﺎوي‬‫ﯾ‬ ‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﺎن‬‫ﻛ‬ ‫اذا‬5000V/m‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫واﻟﺒﻌ‬
0.4cm‫اﻟﻤﺘ‬ ‫ﺳﻌﺔ‬ ‫اﺣﺴﺐ‬‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ ‫ﺴﻌﺔ‬‫ﻓﻲ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬400µC‫؟‬
‫اﻟﺤﻞ‬/
d
V
E
∆
= ⇒ 5000= 2
104.0
V
−
×
∆
⇒ V20104.05000V 2
=××=∆ −

3. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
2
F20
20
400
V
Q
C µ==
∆
=
‫ﻣﺜﺎل‬6/‫ﻣﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻛﺎن‬ ‫اذا‬80µC‫وﻣﻘﺪار‬‫ﺳﻌﺘﮭﺎ‬5µF‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫واﻟﻤﺠﺎل‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬4000V/m‫ﺻﻔﯿﺤﺘﯿﮭﺎ؟‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
V
Q
C
∆
= ⇒ V16
5
80
C
Q
V ===∆
d
V
E
∆
= ⇒ cm4.0m104
4000
16
E
V
d 3
=×==
∆
= −
‫ﻣﺜــﺎل‬7/‫ﺴ‬ ‫ﻣﺘ‬‫ﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺴﺎﻓﺔ‬ ‫اﻟﻤ‬ ‫ﻮازﯾﯿﻦ‬ ‫ﻣﺘ‬ ‫ﻔﯿﺤﺘﯿﻦ‬ ‫ﺻ‬ ‫ذات‬ ‫ﻌﺔ‬)5mm(‫ﺎ‬ ‫ﻣﻨﮭﻤ‬ ‫ﻞ‬ ‫ﻛ‬ ‫ﺴﺎﺣﺔ‬ ‫وﻣ‬)1m2
(‫ﺎذا‬ ‫ﻓ‬
‫ﺑﯿﻨﮭﻤﺎ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫أﺻﺒﺢ‬ ‫ﺣﺘﻰ‬ ‫وﺷﺤﻨﺘﺎ‬ ‫ﺑﺎﻟﻔﺮاغ‬ ‫اﻟﺼﻔﯿﺤﺘﺎن‬ ‫وﺿﻌﺖ‬)2×10 4
V(‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﺷﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
pF1770F1077.1
105
11085.8
d
A
C 9
3
12
=×=
×
××
=
ε
= −
−
−
ο
1-
2- Q = C ∆V = 1.77×10- 9
× 2×10 4
= 3.54×10- 5
C
‫ﻣﺜﺎل‬8/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫اﺣﺴﺐ‬20µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬4×10-3
J
‫اﻟﺒ‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫ﻌﺪ‬0.2cm.
‫اﻟﺤﻞ‬/
2
)V.(C
2
1
PE ∆= ⇒ 263
)V(1020
2
1
104 ∆××=× −−
263
)V(1010104 ∆×=× −−
⇒ 400
1010
104
)V( 6
3
2
=
×
×
=∆ −
−
⇒ V20V =∆
m/V10000
102.0
20
d
V
E 2
=
×
=
∆
= −
‫ﻣﺜﺎل‬9/‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﺎن‬‫ﻛ‬ ‫اذا‬6000V/m‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫واﻟﺒﻌ‬0.1cm
‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬72×10-5
J‫اﻟﻤﺘﺴﻌﺔ؟‬ ‫ﺳﻌﺔ‬ ‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬
‫اﻟﺤﻞ‬/
d
V
E
∆
= ⇒ 2
101.0
V
6000 −
×
∆
= ⇒ V6106000V 3
=×=∆ −
2
)V.(C
2
1
PE ∆= ⇒ 25
)6.(C
2
1
1072 =× −
⇒ 36C
2
1
1072 5
×=× −
F40F104
18
1072
C 5
5
µ=×=
×
= −
−

4. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
3
‫ﻣﺜﺎل‬10/‫ﺎ‬‫ﻣﻨﮭﻤ‬ ‫ﻞ‬‫ﻛ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺒﻌ‬ ‫ﻣﺘﺴﻌﺘﺎن‬0.2cm‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻌﺔ‬‫ﺳ‬ ‫ﺖ‬‫وﻛﺎﻧ‬20µF‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻌﺔ‬‫وﺳ‬
‫اﻟﺜﺎﻧﯿﺔ‬5µF‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫اﻻوﻟﻰ‬ ‫رﺑﻄﺖ‬6V‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫ورﺑﻄﺖ‬
12V‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫اﻟﻄﺎﻗ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫وﻣ‬ ‫؟‬ ‫ﺎ‬‫ﻣﻨﮭﻤ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫ﺻﻔﯿ‬‫؟‬ ‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺤﺘﻲ‬
‫اﻟﺤﻞ‬/
J1036)12(105
2
1
)V.(C
2
1
PE
J1036)6(1020
2
1
)V.(C
2
1
PE
m/V6000
102
12
d
V
E,m/V3000
102
6
d
V
E
m102102.0cm2.0d
5262
222
5262
11
3
2
23
1
1
32
−−
−−
−−
−−
×=××=∆=
×=×××=∆=
=
×
=
∆
==
×
=
∆
=
×=×==
‫ﻣﺘﺴﻌﺔ‬ ‫ﻋﻠﻰ‬ ‫ﻋﺎزل‬ ‫إدﺧﺎل‬‫ﻣﻨﻔﺮدة‬:
‫ﻣﺜﺎل‬11/‫ﺳﻌﺘﮭﺎ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬16µF‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺖ‬‫وﺻ‬10V
‫اﺣﺴﺐ‬:
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﺷﺤﻨﺘﮭﺎ‬.
2-‫اﻟﺒﻄﺎر‬ ‫ﻦ‬‫ﻋ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫ﻟﻮ‬‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﮫ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﺖ‬‫وادﺧﻠ‬ ‫ﺔ‬‫ﯾ‬(k)‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺾ‬‫اﻧﺨﻔ‬
‫اﻟﻰ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬5V‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫وﻣﺎ‬ ‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫ﺑﻮﺟﻮد‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬.
‫اﻟﺤﻞ‬/
2
16
32
C
C
k,F32
5
160
V
Q
C2
J108)10(1016
2
1
)V.(C
2
1
PE,C1601016V.CQ1
k
k
k
4262
===µ==
∆
=−
×=××=∆=µ=×=∆=− −−
‫ﻣﺜﺎل‬12/‫ﯿﻦ‬‫ﺑ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫رﺑﻄﺖ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬12V‫ﻦ‬‫ﻋ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﺎذا‬‫ﻓ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫وادﺧﻞ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬3‫ﺑﻤﻘﺪار‬ ‫ﺳﻌﺘﮭﺎ‬ ‫ازدادت‬60µF‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬.
2-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J10720103604
2
1
Q.V
2
1
PE
J1021601036012
2
1
Q.V
2
1
PE3
V4
3
12
k
V
V2
C3601230V.CQ
F30C60C260CC360CkC60CC1
66
kk
66
k
k
−−
−−
×=×××=∆=
×=×××=∆=−
==
∆
=∆−
µ=×=∆=
µ=⇒=⇒+=⇒+=⇒+=−

5. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
4
‫ﻣﺜﺎل‬13/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬15µF‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺖ‬‫وﺻ‬24V
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫وادﺧﻞ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬)k(‫ﺑﻤﻘﺪار‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ھﺒﻂ‬16V‫اﺣﺴﺐ‬:
1-‫ﺻﻔ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﯿﺤﺘﻲ‬.
2-‫اﻟﻌﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
3-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k.(
4-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J101440103608
2
1
Q.V
2
1
PE
J1043201036024
2
1
Q.V
2
1
PE4
3
15
45
C
C
k3
F45
8
360
V
Q
C,V8162416VV2
C3602415V.CQ1
66
kk
66
k
k
kk
−−
−−
×=×××=∆=
×=×××=∆=−
===−
µ==
∆
==−=−∆=∆−
µ=×=∆=−
‫ﻣﺜﺎل‬14/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫رﺑﻄﺖ‬25µF‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬6V‫ﺎزل‬‫ﻋ‬ ‫ﻞ‬‫ادﺧ‬ ‫ﻢ‬‫ﺛ‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬)k(
‫ﺑﻤﻘﺪار‬ ‫ﺷﺤﻨﺘﮭﺎ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫وﻣﺎزاﻟﺖ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬300µC‫اﺣﺴﺐ‬:
1-‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫ﺑﻮﺟﻮد‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k. (
3-‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
4-‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺼﺎل‬ ‫اﻻﺗ‬ ‫ﻊ‬ ‫ﻗﻄ‬ ‫ﻮ‬‫ﻟ‬‫ﺔ‬ ‫واﻟﻄﺎﻗ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬ ‫ﻓﻤ‬ ‫ﺎزل‬ ‫اﻟﻌ‬ ‫ﺮج‬‫اﺧ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﻚ‬ ‫ذﻟ‬ ‫ﺪ‬ ‫ﺑﻌ‬ ‫ﺔ‬‫واﻟﺒﻄﺎرﯾ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬
‫؟‬ ‫اﻟﺤﺎﻟﺔ‬ ‫ھﺬه‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬
‫اﻟﺤﻞ‬/
J1040503241025
2
1
)18(1025
2
1
)V.(C
2
1
PE
V18
25
450
C
Q
V4
J101350)6(1075
2
1
)V.(C
2
1
PE3
3
25
75
C
C
k2
F75
6
450
V
Q
C
C450300150300QQ,C150625V.CQ1
66262
6262
kk
k
k
k
k
−−−
−−
×=×××=××=∆=
===∆−
×=×××=∆=−
===−
µ==
∆
=
µ=+=+=µ=×=∆=−

6. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
5
‫ﻣﺜﺎل‬15/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﻮازﯾﺘﯿﻦ‬ ‫ﺻﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬5μF‫ﺟﮭﺪھﺎ‬ ‫ﻓﺮق‬ ‫ﺑﺒﻄﺎرﯾﺔ‬ ‫وﺻﻠﺖ‬200V‫ﺟﺪ‬:
1-‫اﻟﻤﺘﺴ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻋﻠﻰ‬ ‫ﺗﺘﻜﻮن‬ ‫ﺳﻮف‬ ‫اﻟﺘﻲ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﻌﺔ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ادﺧﻞ‬ ‫ﻟﻮ‬)2(‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻛﻢ‬.
3-‫ﺮق‬‫ﻓ‬ ‫ﺼﺒﺢ‬‫ﯾ‬ ‫ﻢ‬‫وﻛ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺼﺒﺢ‬‫ﺗ‬ ‫ﻢ‬‫ﻛ‬ ‫ﺎزل‬‫اﻟﻌ‬ ‫ﺮج‬‫اﺧ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻚ‬‫ذﻟ‬ ‫ﺑﻌﺪ‬ ‫واﻟﻤﺼﺪر‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻻﺗﺼﺎل‬ ‫ﻗﻄﻊ‬ ‫ﻟﻮ‬
‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬.
‫اﻟﺤﻞ‬/
1- Q =C . ∆V =5 × 200 =1000µC
2- Qk = k Q =2 × 1000 =2000µC
3- Q = Qk =2000µC (‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫)ﻻن‬
V400
5
2000
C
Q
V ===∆
‫ﻣﺜﺎل‬16/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﻮازﯾﺘﯿﻦ‬‫اﻟﻤﺘ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬)10µF(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﺎطﺔ‬‫ﺑﻮﺳ‬ ‫ﺤﻨﺖ‬‫ﺷ‬
12V‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫وادﺧﻞ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬2‫ﺑ‬‫اﺣﺴﺐ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺤﯿﺰ‬ ‫ﯾﻤﻸ‬ ‫ﺤﯿﺚ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
3-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
1- Q =C.∆V =10×12=120µC
2- Ck =k C =2×10 =20µF
3- V6
2
12
k
V
Vk ==
∆
=∆
Or V6
20
120
C
Q
V
k
k
k
===∆
‫ﻣﺜﺎل‬17/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬5µF‫ﺪارھﺎ‬‫ﻣﻘ‬ ‫ﺸﺤﻨﺔ‬‫ﺑ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﺎطﺔ‬‫ﺑﻮﺳ‬ ‫ﺤﻨﺖ‬‫ﺷ‬60µC‫ﻞ‬‫وادﺧ‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﻦ‬‫ﻋ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﻢ‬‫ﺛ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬2‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- Ck=k C =2 × 5=10µF
V6
10
60
C
Q
V
k
k
k
===∆
or
V12
5
60
C
Q
V ===∆
∴ V6
2
12
k
V
Vk ==
∆
=∆
2- J103610360106012
2
1
Q.V
2
1
PE 566 −−−
×=×=×××=∆=
J10181018010606
2
1
Q.V
2
1
PE 566
kkk
−−−
×=×=×××=∆=

7. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
6
‫ﻣﺜـﺎل‬18/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬12µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫وﻣﻘ‬600µc‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺮ‬‫وﻏﯿ‬‫ﺎذا‬‫ﻓ‬ ‫ﺼﺪر‬‫ﺑﺎﻟﻤ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬ ‫ادﺧﻞ‬5‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫اﺻﺒﺢ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬10000V/m‫اﺣﺴﺐ‬:
1-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬.
‫اﻟﺤﻞ‬/
1- Ck=k C =5 × 12 =60µF
2- V10
60
600
C
Q
V
k
k
k
===∆
d
V
E k
k
∆
= ⇒
d
10
10000 = ⇒ d=0.001m=0.1cm
‫ﻣﺜﺎل‬19/‫ﺳﻌﺘﮭﺎ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬20µF‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﺑﻮﺳﺎطﺔ‬ ‫ﺷﺤﻨﺖ‬6V
‫ﻛﮭﺮﺑ‬ ‫ﻋﺎزل‬ ‫ادﺧﻞ‬ ‫ﺛﻢ‬‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﺎﺋﻲ‬‫ﺳﻌﺘﮭﺎ‬ ‫ﻓﺎﺻﺒﺤﺖ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬60µF‫ﻣﻘﺪا‬ ‫ﻣﺎ‬‫ر‬:
1-‫اﻟﻌﺎزل‬ ‫ﻟﻠﻮح‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬.
2-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻋﻞ‬ ‫اﻟﺸﺤﻨﺔ‬.
3-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
4-‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.5cm.
‫اﻟ‬‫ﺤﻞ‬/
1- 3
20
60
C
C
k k
===
2- Qk = Ck . ∆Vk = 60×6=360µC
or Q = C .∆V =20 × 6 =120µC
Qk = k Q = 3 × 120 = 360µC
3- J103610360101206
2
1
Q.V
2
1
PE 566 −−−
×=×=×××=∆=
J10108101080103606
2
1
Q.V
2
1
PE 566
kkk
−−−
×=×=×××=∆=
or J1010810363kPEPE 55
k
−−
×=××==
4- m/V1200
105.0
6
d
V
E 2
k
k =
×
=
∆
= −
‫ﻣﺜﺎل‬20/‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﺎطﺔ‬‫ﺑﻮﺳ‬ ‫ﺷﺤﻨﺖ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ذات‬ ‫ﻣﺘﺴﻌﺔ‬20V‫ﻞ‬‫ادﺧ‬ ‫ﺎذا‬‫ﻓ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬4‫واﻟﻤﺘﺴﻌﺔ‬‫ﻣﺎزاﻟﺖ‬‫ﺤﻨﺘﮭﺎ‬‫ﺷ‬ ‫ﺒﺤﺖ‬‫اﺻ‬ ‫ﺔ‬‫ﺑﺎﻟﺒﻄﺎرﯾ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬400µC‫ﺴﺐ‬‫اﺣ‬
‫ﻣﻘﺪار‬:
1-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﻗﺒﻞ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬.
3-‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺰﯾﺎدة‬ ‫ﻣﻘﺪار‬.
‫اﻟﺤﻞ‬/
1- F20
20
400
V
Q
C
k
k
k
µ==
∆
=
kCCk
= ⇒ F5
4
20
k
C
C k
µ===
2- Q = C.∆V =5 × 20 =100µC
Or Qk = k Q ⇒ 400 =4Q ⇒ Q =100µC

8. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
7
3- J101010020
2
1
Q.V
2
1
PE 36 −−
=×××=∆=
J1041040020
2
1
Q.V
2
1
PE 36
kkk
−−
×=×××=∆=
∴ J10310104PEPEPE 333
k
−−−
×=−×=−=∆
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬21/‫ﺪه‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫ﻲ‬‫طﺮﻓ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫رﺑﻄ‬20V‫ﮫ‬ ‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻞ‬‫ادﺧ‬ ‫ﻢ‬‫ﺛ‬4
‫اﻟﻄﺎ‬ ‫ﻓﺄﺻﺒﺤﺖ‬‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫ﻗﺔ‬32×10-4
J‫اﺣﺴﺐ‬:
1-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫ﺑﯿﻨﮭﻤﺎ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.1cm.
‫اﻟﺤﻞ‬/
1- 2
Kkk )V.(C
2
1
PE ∆= ⇒ 2
k
4
)20(C
2
1
1032 =× −
⇒ k
4
C4001064 =× −
F16F1016
400
1064
C 6
4
k µ=×=
×
= −
−
Ck=k C ⇒ 16 =4C ⇒ C=4µF
2- m/V200001020
101.0
20
d
V
E 3
2
k
k =×=
×
=
∆
= −
‫ﻣﺜﺎل‬22/‫ﺷﺤﻨﺘﮭﺎ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻣﺘﺴﻌﺔ‬300µc‫ﺛ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫ادﺧﻞ‬ ‫ﺑﺎﻟﻤﺼﺪر‬ ‫ﻣﺘﺼﻠﺔ‬ ‫وﻏﯿﺮ‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺎﺑ‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬5‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻓﺄﺻﺒﺤﺖ‬0.003J‫اﺣﺴﺐ‬:
1-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ ‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.2cm.
‫اﻟﺤﻞ‬/
1-
k
2
k
k
C
Q
2
1
PE = ⇒
k
26
C2
)10300(
003.0
−
×
= ⇒ 6×10-3
Ck=90000×10-12
F15F1015
106
1090000
C 6
3
12
k µ=×=
×
×
= −
−
−
Ck =k C ⇒ 15=5C ⇒ C=3µF
2- V20
15
300
C
Q
V
k
k
k
===∆
m/V10000
102.0
20
d
V
E 2
k
k =
×
=
∆
= −
‫ﻣﺜﺎل‬23/‫ﻣﺘ‬ ‫ﺳﻌﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬‫ﺑﻤﺼﺪر‬ ‫ﻣﺘﺼﻠﺔ‬ ‫ﺴﻌﺔ‬10µF‫ﻓﻲ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫وﻣﻘﺪار‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬400µC‫ﻞ‬‫ادﺧ‬ ‫ﻓﺎذا‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬2‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺘﺴﻌﺔ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬‫ﻣﻘﺪار‬ ‫ﻣﺎ‬ ‫ﺑﺎﻟﻤﺼﺪر‬ ‫ﻣﺘﺼﻠﺔ‬:
1-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- Ck = k C =2 × 10 =20µF
2- V40
10
400
C
Q
V ===∆
Qk =k Q =2 × 400 =800µC

9. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
8
J1081080001040040
2
1
Q.V
2
1
PE 366 −−−
×=×=×××=∆=
J101610160001080040
2
1
Q.V
2
1
PE 366
kkk
−−−
×=×=×××=∆=
or PEk=kPE =2×8×10-3
=16×10-3
J
‫ﻣﺜﺎل‬24/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬12µF‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﺑﻮﺳﺎطﺔ‬ ‫ﺷﺤﻨﺖ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫اﻟﮭﻮاء‬
10V‫اﻟﺒﻄ‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫وادﺧﻞ‬ ‫ﺎرﯾﺔ‬5‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫اﻟﻌﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.3-‫اﻟﻌﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬.
4-‫اﻟﺼﻔﯿ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﺤﺘﯿﻦ‬.
‫اﻟﺤﻞ‬/
1- Q =C . ∆V=12 × 10=120µC
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺘﺴﻌﺔ‬
Qk = Q =120µC
2- Ck =kC=5 × 12=60µF
3- V2
60
120
C
Q
V
k
k
k
===∆
4- J1061012010
2
1
Q.V
2
1
PE 46 −−
×=×××=∆=
J1012101202
2
1
Q.V
2
1
PE 56
kkk
−−
×=×××=∆=
‫ﻣ‬‫ﺜﺎل‬25/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬20µF‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎن‬‫وﻛ‬ ‫ﺼﺪر‬‫اﻟﻤ‬ ‫ﻦ‬‫ﻋ‬ ‫ﺼﻮﻟﺔ‬‫وﻣﻔ‬ ‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬
600µC‫ﺑﻤﻘﺪار‬ ‫ﺳﻌﺘﮭﺎ‬ ‫ﻓﺎزدادت‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫ادﺧﻞ‬ ،40µF‫اﺣﺴﺐ‬:
1-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬.
2-‫ﺑﯿﻨﮭﻤﺎ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.2cm.
‫اﻟﺤﻞ‬/
1- Ck=C + ∆C =20 + 40=60µF
3
20
60
C
C
k k
===
2- V10
60
600
C
Q
V
k
k
k
===∆
m/V5000
102.0
10
d
V
E 2
k
k =
×
=
∆
= −

10. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
9
‫ﻣﺜــﺎل‬26/‫ﺎ‬ ‫ﻣﻨﮭ‬ ‫ﻞ‬ ‫ﻛ‬ ‫ﻠﻊ‬ ‫ﺿ‬ ‫ﻮل‬ ‫ط‬ ‫ﺸﻜﻞ‬ ‫اﻟ‬ ‫ﺔ‬ ‫ﻣﺮﺑﻌ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺻ‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﻞ‬ ‫ﻛ‬ ‫ﻮازﯾﺘﯿﻦ‬ ‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬ ‫اﻟ‬ ‫ذات‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬5cm
‫وﯾﻔﺼﻞ‬‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫ﻓﺈذا‬ ‫اﻟﻔﺮاغ‬ ‫ﺑﯿﻨﮭﻤﺎ‬5pF‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬‫ﺻﻔﯿ‬‫ﺤﺘﯿﮭﺎ‬.
2-‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﻣﻘﺪاره‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺗﺴﻠﯿﻂ‬ ‫ﺑﻌﺪ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬12V‫اﻟﻔﺮاغ‬ ‫ﺳﻤﺎﺣﯿﺔ‬ ‫ان‬ ‫اﻟﻌﻠﻢ‬ ‫ﻣﻊ‬
‫اﻟﺤﻞ‬/
1- A=(5×10-2
) =25 ×10-4
m2
d
A
C ο
ε
= ⇒
d
10251085.8
105
412
12
−−
− ×××
=×
m25.44
105
10102585.8
d 12
412
=
×
×××
= −
−−
2- Q= C . ∆V =5 × 12 =60PC
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬27/‫ﺎ‬‫ﺑﯿﻨﮭﻤ‬ ‫ﺪ‬ ‫اﻟﺒﻌ‬ ‫ﻮازﯾﺘﯿﻦ‬‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ذات‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬0.5cm‫ﺼﻔﯿﺤﺔ‬‫اﻟ‬ ‫ﺴﺎﺣﺔ‬‫وﻣ‬25cm2
‫ﺎ‬ ‫ﺑﯿﻨﮭﻤ‬ ‫ﺼﻞ‬‫ﯾﻔ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬5‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﺟﮭﺪه‬ ‫ﻓﺮق‬ ‫ﻟﻤﺼﺪر‬ ‫رﺑﻄﺖ‬100V‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬2-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺷﺤﻨﺔ‬3-‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫اﻟﺤﻞ‬/
1- F1025.221
105.0
1025
1085.85
d
A
kC 13
2
4
12
k
−
−
−
−
ο ×=
×
×
×××=ε=
2- Qk =Ck . ∆V =221.25×10-13
× 100 =221.25×10-11
C
3- J105.110621025.221100
2
1
Q.V
2
1
PE 1111
kk
−−
×=×××=∆=
‫اﻟﺘﻮازي‬ ‫رﺑﻂ‬:
‫ﻣﺜﺎل‬28/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=1µF,C2=4µF(‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺤﻨﺖ‬‫ﺷ‬ ‫ﺈذا‬‫ﻓ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﺮق‬‫ﻓ‬ ‫ﺒﺢ‬‫أﺻ‬ ‫ﻰ‬‫ﺣﺘ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬120V‫اﺣﺴﺐ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬‫اﻟ‬‫ﺸﺤﻨﺔ‬‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬.2-‫ﻟﮭﻤﺎ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.
‫اﻟﺤﻞ‬/
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =120V
Q1 =C1 . ∆V1 =1 × 120=120µC , Q2 =C2 . ∆V2 =4 × 120 =480µC
Ceq = C1 + C2 =1 + 4 =5µF
‫ﻣﺜﺎل‬29/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=2µF , C2=5µF(‫ﺑـ‬ ‫ﻟﮭﻤﺎ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫وﺷﺤﻨﺖ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬)280µC. (
‫ﻣﻘﺪار‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬:
1-‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬.
2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- Ceq =C1 + C2 =2 + 5=7µF
V40
7
280
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT =∆V1 = ∆V2 =40V
Q1 =C1 . ∆V1 =2 × 40 = 80µC , Q2 =C2 . ∆V2 =5 × 40 =200µC
2- J1016108040
2
1
Q.V
2
1
PE 46
111
−−
×=×××=∆=
J1041020040
2
1
Q.V
2
1
PE 36
222
−−
×=×××=∆=

11. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
10
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬30/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=3µF,C2=5µF(‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺘﺎن‬‫ﻣﻮﺻ‬،‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬12V
‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬3-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
1- Ceq=C1 + C2 =3 + 5 =8µF
2- ∆V1 = ∆V2 = ∆VT =12V (‫)ﺗﻮازي‬
3- Q1 =C1 . ∆V1 =3 × 12 =36µC , Q2 =C2 . ∆V2 =5 × 12 =60µC
QT =Q1 + Q2 =36 + 60 =96µC
or
QT =Ceq . ∆VT =8 × 12 =96µC
‫ﻣﺜﺎل‬31/‫ﻣﺘﺴﻌﺎت‬ ‫ﺛﻼث‬)C1=2µF,C2=6µF,C3=12µF(‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻮﺻﻮﻟﺔ‬
‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﻜﺎﻧﺖ‬ ‫ﺑﻄﺎرﯾﺔ‬120µC‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬2-‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬3-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
C72612V.CQ
C3666V.CQ,C1262V.CQ3
V6
20
120
C
Q
V2
F201262CCCC1
33
2211
eq
T
321eq
µ=×=∆=
µ=×=∆=µ=×=∆=−
===∆−
=++=++=−
‫ﻣﺜﺎل‬32/‫ﻣﺘﺴﻌﺘﺎن‬)C1=8µF,C2=20µF(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬
6V‫اﺣﺴﺐ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬2-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J10504)6(1028
2
1
)V.(C
2
1
PE
J10360)6(1020
2
1
)V.(C
2
1
PE
J10144)6(108
2
1
)V.(C
2
1
PE2
C120620V.CQ
C4868V.CQ,C168628V.CQ
F28208CCC1
6262
eqT
6262
22
6262
11
22
11eqT
21eq
−−
−−
−−
×=××=∆=
×=××=∆=
×=××=∆=−
µ=×=∆=
µ=×=∆=µ=×=∆=
=+=+=−
‫ﻣﺜـﺎل‬33/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=2µF , C2=5µF(‫ﻠﺘ‬‫ووﺻ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﺎ‬‫ﺖ‬‫ﻓﻜﺎﻧ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫ﻰ‬‫اﻟ‬
‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺸﺤﻨﺔ‬)96µC(‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬:
1-‫اﻟﺜﺎﻧﯿﺔ؟‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬
2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.4cm.
‫اﻟﺤﻞ‬/
1- V48
2
96
C
Q
V
1
1
1
===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆V2 = ∆V1 =48V

12. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
11
Q2=C2 . ∆V2 =5 × 48 =240µC
2- m/V12000
104.0
48
d
V
E 2
1
1 =
×
=
∆
= −
‫ﻣﺜﺎل‬34/‫ﻣﺘﺴﻌﺘﺎن‬)C1=4µF,C2=6µF(‫ﺔ‬‫ﻟﻠﻤﺠﻤﻮﻋ‬ ‫ﺔ‬‫اﻟﻜﻠﯿ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺖ‬‫وﻛﺎﻧ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬120µC
‫اﺣﺴﺐ‬:
1-‫ا‬ ‫ﻓﺮق‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫اﻟﻜﻠﻲ‬ ‫ﻟﺠﮭﺪ‬2-‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬.
3-‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬.
‫اﻟﺤﻞ‬/
1- Ceq =C1 + C2 =4 + 6=10µF
V12
10
120
C
Q
V
eq
T
T ===∆
∆V1 = ∆V2 = ∆VT =12V (‫)ﺗﻮازي‬
2- Q1 =C1 . ∆V1 =4 × 12 =48µC , Q2 =C2 . ∆V2 =6 × 12=72µC
3- J10288104812
2
1
Q.V
2
1
PE 66
111
−−
×=×××=∆=
J10432107212
2
1
Q.V
2
1
PE 66
222
−−
×=×××=∆=
J107201043210288PEPEPE 666
21T
−−−
×=×+×=+=
or
J107201012012
2
1
Q.V
2
1
PE 66
TTT
−−
×=×××=∆=
‫ﻣﺜـﺎل‬35/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=5µF,C2=10µF(‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﻢ‬‫ﺛ‬‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺖ‬‫وﻛﺎﻧ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬
‫اﻷوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﻋﻠﻰ‬200µC‫اﺣﺴﺐ‬:
1-‫اﻟﺸﺤﻨﺔ‬‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬.
2-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻷوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.5cm.
‫اﻟﺤﻞ‬/
1- Ceq =C1 + C2 =5 + 10 =15µF
V40
5
200
C
Q
V
1
1
1
===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆V1=∆V2=∆VT=40V
Q2 =C2 . ∆V2=10 × 40=400µC
QT =Q1 + Q2 =200 + 400 =600µC
2- m/V8000
105.0
40
d
V
E 2
1
1 =
×
=
∆
= −
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬36/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1,C2=20µF(‫اﻟﻤﻜ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺖ‬‫وﻛﺎﻧ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬‫ﺎ‬‫ﻟﮭﻤ‬ ‫ﺔ‬‫ﺎﻓﺌ‬30µF‫ﺎل‬ ‫واﻟﻤﺠ‬
‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬)C1(5000V/m‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫واﻟﺒﻌﺪ‬0.4cm‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬)C1.(2-‫اﻟﻜﻠﻲ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬3-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬
4-‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- Ceq =C1 + C2 ⇒ 20C30 1
+= ⇒ F102030C1
µ=−=
2-
d
V
E 1
1
∆
= ⇒ 2
1
104.0
V
5000 −
×
∆
= ⇒ V201045000V 3
1
=××=∆ −

13. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
12
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆V1=∆V2=∆VT=20V
3- Q1=C1 . ∆V1=10 × 20=200µC , Q2=C2 . ∆V2=20 × 20=400µC
QT =Q1 + Q2 =200 + 400 =600µC
4- J1061060020
2
1
Q.V
2
1
PE 36
TTT
−−
×=×××=∆=
‫ﻣﺜﺎل‬37/‫ﻣﺘﺴﻌﺘﺎن‬)C1=6µF,C2(‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫اﻟﻄﺎﻗ‬ ‫ﺖ‬‫ﻓﻜﺎﻧ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫وﺻﻠﺘﺎ‬ ، ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻮﺻﻮﻟﺘﺎن‬‫ﻲ‬‫ﻓ‬
‫ا‬‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻟﻤﺠﺎل‬‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬72×10-6
J‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫واﻟﺴﻌﺔ‬10µF‫ﻣﻘﺪار‬ ‫ﺟﺪ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬C2.2-‫اﻟﻜﻠﻲ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬3-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺸﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
1- Ceq=C1 + C2 ⇒ 10 =6 + C2 ⇒ C2=10 – 6 =4µF
2- 2
222 )V.(C
2
1
PE ∆= ⇒ 2
2
66
)V(104
2
1
1072 ∆××=× −−
36
2
72
)V( 2
2 ==∆ ⇒ ∆V2 =6V
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆V2 = ∆V1 = ∆VT =6V
3- Q1 =C1 . ∆V1 =6 × 6 =36µC , Q2 =C2 . ∆V2 =4 × 6=24µC
QT = Q1 + QT =36 + 24 =60µC
‫ﻣﺜﺎل‬38/‫ﺴﻌﺎت‬‫ﻣﺘ‬ ‫ﻊ‬‫ارﺑ‬)C1=4µF,C2=12µF,C3=8µF,C4=6µF(‫ﺔ‬‫اﻟﻄﺎﻗ‬ ‫ﺖ‬‫وﻛﺎﻧ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬
‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬256×10-6
J‫ا‬‫ﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.2-‫اﻟﻜﻠﻲ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬3-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬
4-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻷوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.4cm.
‫اﻟﺤﻞ‬/
1- Ceq=C1 + C2 + C3 + C4 =4 + 12 + 8 + 6=30µF
2- 2
333 )V.(C
2
1
PE ∆= ⇒ 2
3
66
)V(108
2
1
10256 ∆×××=× −−
64
104
10256
)V( 6
6
2
3 =
×
×
=∆ −
−
⇒ V8V3
=∆
3- ∆V1=∆V2=∆V3=∆V4=∆V4=8V
4- Q1 =C1 .∆V1 =4 × 8=32µC , Q2 =C2 . ∆V2 =12 × 8=96µC
Q3 =C3 . ∆V3 =8 × 8=64µC , Q4 =C4 . ∆V4 =6 × 8=48µC
QT =Q1 + Q2 + Q3 + Q4 =32 + 96 + 64 + 48 =240µC
‫ﻣﺜﺎل‬39/‫ﻣﻨﮭﺎ‬ ‫واﺣﺪة‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺘﺴﻌﺎت‬ ‫ارﺑﻊ‬)C(‫ﺮق‬‫ﻓ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺖ‬
‫ﻗﻄﺒﯿﺔ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬30V‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫ﻓﺎذا‬240µC‫وﺷﺤﻨﺘﮭﺎ؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
F8
30
240
V
Q
C
T
T
eq
µ==
∆
=
Ceq =nC ⇒ 8=4C ⇒ C=2µF
Q1=C1.∆V1=2 × 30=60µC , Q2=C2.∆V2=2 × 30 =60µC
Q3=C3.∆V3=2 × 30=60µC , Q4=C4.∆V4=2 × 30 =60µC

14. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
13
‫اﻟﺘﻮاﻟﻲ‬ ‫رﺑﻂ‬:
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬40/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=3µF,C2=6µF(‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬‫ﻢ‬ ‫ﺛ‬‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬ ‫إﻟ‬ ‫ﻠﺘﺎ‬ ‫وﺻ‬
‫ﻗﻄﺒﯿﮭﺎ‬60V‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬3-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
1-
21eq
C
1
C
1
C
1
+= ⇒
6
1
3
1
C
1
eq
+= ⇒
2
1
6
3
6
12
C
1
eq
==
+
=
∴ Ceq=2µF
2- QT =Ceq . ∆VT =2 × 60=120µC
QT = Q1 = Q2 =120µC (‫)ﺗﻮاﻟﻲ‬
3- V40
3
120
C
Q
V
1
1
1
===∆ , V20
6
120
C
Q
V
2
2
2
===∆
‫ﻣﺜﺎل‬41/‫ﻣﺘﺴﻌﺘﺎن‬)C1=12µF,C2=6µF(‫رﺑﻄﺘﺎ‬ ، ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺤﻨﺖ‬‫وﺷ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫إﻟ‬
‫ﻣﻘﺪارھﺎ‬ ‫ﻛﻠﯿﺔ‬ ‫ﺑﺸﺤﻨﺔ‬60µC‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.2-‫اﻟﻜﻠﻲ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
3-‫واﻟﻄ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫اﻟﻜﻠﯿﺔ‬ ‫ﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1-
21eq
C
1
C
1
C
1
+= ⇒
6
1
12
1
C
1
eq
+= ⇒
4
1
12
3
12
21
C
1
eq
==
+
=
∴ Ceq=4µF
2- V5
12
60
C
Q
V
1
1
1
===∆ , V10
6
60
C
Q
V
2
2
2
===∆
∆VT =∆V1 + ∆V2 =5 + 10=15V
3- J1015010605
2
1
Q.V
2
1
PE 66
111
−−
×=×××=∆=
J10300106010
2
1
Q.V
2
1
PE 66
222
−−
×=×××=∆=
PET =PE1 + PE2 =150×10-6
+ 300×10-6
=450×10-6
J
‫ﻣﺜﺎل‬42/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=6µF,C2=3µF(‫ﺑﺸﺤﻨﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺷﺤﻨﺖ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬200μC‫اﺣﺴﺐ‬:
1-‫ﻓﺮق‬‫ﺟﮭﺪ‬‫اﻟﻤﺼﺪر‬‫اﻟﺸﺎﺣﻦ‬2-‫ﺷﺤ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﻨﺔ‬
3-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- F2
9
18
36
36
CC
C.C
C
21
21
eq
µ==
+
×
=
+
=
V100
2
200
C
Q
V
eq
total
total ===∆
‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬ ‫ان‬ ‫ﺑﻤﺎ‬
2- Q1 = Q2 = Qtotal =200μC
V
3
100
6
200
C
Q
V
1
1
===∆ , V
3
200
C
Q
V
2
2
==∆

15. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
14
3- ( ) J10
3
1
3
100
10200
2
1
VQ
2
1
PE 26
11
−−
×=×××=∆×=
( ) J10
3
2
3
200
10200
2
1
VQ
2
1
PE 26
22
−−
×=×××=∆×=
J01.010010200
2
1
VQ
2
1
PE 6
totaltotal =×××=∆×= −
‫ﻣﺜﺎل‬43/‫ﺳﻌﺎﺗﮭﺎ‬ ‫ﻣﺘﺴﻌﺎت‬ ‫ﺛﻼث‬ ‫رﺑﻄﺖ‬)C1=2μF , C2=3μF , C3=6μF(‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﺟﮭﺪھﺎ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫إﻟﻰ‬120V‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬2-‫ﺷﺤﻨ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺔ‬.3-‫اﻟﻤﺠﻤﻮﻋﺔ؟‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬
‫اﻟﺤﻞ‬/
1- F1
6
6
6
123
6
1
3
1
2
1
C
1
C
1
C
1
C
1
321eq
µ==
++
=++=++= ⇒ Ceq=1µF
2- QT =Ceq . ∆VT =1 × 120 = 120µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 = Q3 =120µC
V60
2
120
C
Q
V
1
1
===∆ , V40
3
120
C
Q
V
2
2
2
===∆ , V20
6
120
C
Q
V
3
3
3
===∆
J107210120120
2
1
Q.V
2
1
PE 46
TTT
−−
×=×××=∆=
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬44/‫ﻌﺎﺗﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺎت‬‫ﻣﺘ‬ ‫ﻼث‬‫ﺛ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=12μF , C2=36μF , C3=18μF(‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﺟﮭﺪھﺎ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫إﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬30V‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬2-‫ﻛ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻞ‬.
3-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.2cm‫؟‬
‫اﻟﺤﻞ‬/
1- F
6
1
36
6
36
213
18
1
36
1
12
1
C
1
C
1
C
1
C
1
321eq
µ==
++
=++=++= ⇒ Ceq=6µF
2- QT =Ceq . ∆VT =6 × 30 = 180µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 = Q3 =180µC
V15
12
180
C
Q
V
1
1
1
===∆ , V5
36
180
C
Q
V
2
2
2
===∆ , V10
18
180
C
Q
V
3
3
3
===∆
m/V5000
102.0
10
d
V
E 3
3
3 =
×
=
∆
= −
‫ﻣﺜﺎل‬45/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬)C1,C2=24µF(‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺖ‬‫وﻛﺎﻧ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫إﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﻟﮭﻤﺎ‬6µF‫اﻟﻤ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫واﻟﻤﺠﺎل‬‫اﻷوﻟﻰ‬ ‫ﺘﺴﻌﺔ‬5000V/m‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫واﻟﺒﻌﺪ‬0.3cm‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬C1.2-‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
1-
21eq
C
1
C
1
C
1
+= ⇒
24
1
C
1
6
1
1
+= ⇒
24
1
6
1
C
1
1
−=
8
1
24
3
24
14
C
1
1
==
−
= ⇒ C1=8µF

16. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
15
2-
d
V
E 1
1
∆
= ⇒ 2
1
103.0
V
5000 −
×
∆
= ⇒ V151035000V 3
1
=××=∆ −
Q1=C1 . ∆V1=8 × 15=120µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬Q1 = Q2 = QT =120µC
V5
24
120
C
Q
V
2
2
2
===∆
‫ﻣﺜــﺎل‬46/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=20µF,C2=30µF(‫ﻰ‬ ‫اﻷوﻟ‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬ ‫اﻟﻄﺎﻗ‬ ‫ﺖ‬ ‫وﻛﺎﻧ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬
36×10-5
J‫اﻟ‬ ‫ﺎن‬‫ﻛ‬ ‫اذا‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫واﻟﻤﺠ‬ ‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫ﺒﻌ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬0.1cm.
‫اﻟﺤﻞ‬/
21eq
C
1
C
1
C
1
+= ⇒
12
1
60
5
60
23
30
1
20
1
C
1
eq
==
+
+= ⇒ Ceq=12µF
2
111 )V.(C
2
1
PE ∆= ⇒ 2
1
65
)V(1020
2
1
1036 ∆×××=× −−
36)V( 2
1
=∆ ⇒ ∆V1 =6V
Q1=C1 . ∆V1=20 × 6=120µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬Q1 = Q2 = QT =120µC
V4
30
120
C
Q
V
2
2
2
===∆
m/V4000
101.0
4
d
V
E 2
2
2 =
×
=
∆
= −
‫اﻟﻌﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫ﺗﻮازي‬:
‫ﻣﺜــﺎل‬47/‫وازﯾﺗﯾن‬ ‫اﻟﻣﺗ‬ ‫ﺻﻔﯾﺣﺗﯾن‬ ‫اﻟ‬ ‫ذوات‬ ‫ن‬ ‫ﻣ‬ ‫ﺳﻌﺗﺎن‬ ‫ﻣﺗ‬)C1=26µF,C2=18µF(‫وازي‬ ‫اﻟﺗ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎن‬ ‫ﻣرﺑوطﺗ‬
‫ﻗطﺑﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﺑطﺎرﯾﺔ‬ ‫ﻗطﺑﻲ‬ ‫ﺑﯾن‬ ‫رﺑطت‬ ‫وﻣﺟﻣوﻋﺗﮭﻣﺎ‬)50V(‫ﺎ‬‫ﻋزﻟﮭ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﻣﺎدة‬ ‫ﻣن‬ ‫ﻟوح‬ ‫ادﺧل‬ ‫اذا‬ ،
)k(‫ﺔ‬ ‫ﻟﻠﻣﺟﻣوﻋ‬ ‫ﺔ‬ ‫اﻟﻛﻠﯾ‬ ‫ﺷﺣﻧﺔ‬ ‫اﻟ‬ ‫ت‬ ‫ﻓﻛﺎﻧ‬ ‫ﺔ‬ ‫ﺑﺎﻟﺑطﺎرﯾ‬ ‫ﺻﻠﺔ‬ ‫ﻣﺗ‬ ‫ﺔ‬ ‫اﻟﻣﺟﻣوﻋ‬ ‫ت‬ ‫وﻣﺎزاﻟ‬ ‫ﻰ‬ ‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬ ‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬ ‫ﺻ‬ ‫ﯾن‬ ‫ﺑ‬
)3500µC(‫؟‬ ‫ﻣﻘدار‬ ‫ﻣﺎ‬
1-‫اﻟﻌزل‬ ‫ﺛﺎﺑت‬)k.(2-‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻣﺎدة‬ ‫ادﺧﺎل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬.
‫اﻟﺤﻞ‬/
Q‫ﻟذﻟك‬ ‫ﺑﺎﻟﺑطﺎرﯾﺔ‬ ‫ﻣﺗﺻﻠﺔ‬ ‫ﻣﺎزاﻟت‬ ‫اﻟﻣﺟﻣوﻋﺔ‬∆VTk = ∆VT =50V1-
F70
50
3500
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk = C1k + C2 ⇒ 70 = C1k + 18 ⇒ C1k =70 – 18 =52µF
2
26
52
C
C
k
1
k1
===
Q‫ﻟذﻟك‬ ‫اﻟﺗوازي‬ ‫ﻋﻠﻰ‬ ‫ﻣرﺑوطﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬∆VTk = ∆V1k = ∆V2 =50V2-
Q1k = C1k . ∆V1k =52 × 50 = 2600µC , Q2 = C2 . ∆V2 =18 × 50 = 900µC

17. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
16
‫ﻣﺜﺎل‬48/‫اﻟﻤﺘﺴﻌﺘﺎن‬)C1=2µF , C2=4µF(‫ﻰ‬‫اﻟ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﮫ‬‫ﻗﻄﺒﯿ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﻟﻠﻔﻮﻟﻄﯿ‬ ‫ﻣﺼﺪر‬100V‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻊ‬‫وﺿ‬ ،)k(‫ﺪل‬‫ﺑ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﮭﻮاء‬)‫ﻣ‬‫ﺑﺎﻟﻤﺼﺪر‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺑﻘﺎء‬ ‫ﻊ‬(‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﺒﺤﺖ‬‫ﻓﺎﺻ‬)1600µC(‫ﺰل‬‫اﻟﻌ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬)k(
‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫وﻣﺎ‬.
‫اﻟﺤﻞ‬/
C4001004V.CQ,C120010012V.CQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C4001004V.CQ,C2001002V.CQ
‫ﻞ‬‫ﺒ‬‫ﻗ‬‫اﻟﻌﺎزل‬
6
2
12
C
C
k
F12416CCC
F16
100
1600
V
Q
C
22k1k1
2211
1
k1
2eqkk1
Tk
eqk
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
‫ﻣﺜﺎل‬49/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=5µF , C2=3µF(‫ر‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬‫ﯿﻦ‬‫ﺑ‬ ‫ﺖ‬‫ﺑﻄ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬15V.‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﺻﺒﺤﺖ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫زاﻟﺖ‬ ‫وﻣﺎ‬ ‫اﻻوﻟﻰ‬270µC‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k. (
2-‫ﻣﺘﺴ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫ادﺧﺎل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻌﺔ‬.
‫اﻟﺤﻞ‬/
C45153V.CQ,C2251515V.CQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C45153V.CQ,C75155V.CQ
‫ﻞ‬‫ﺒ‬‫ﻗ‬‫اﻟﻌﺎزل‬
)2
3
5
15
C
C
k
F15318CCC
F18
15
270
V
Q
C)1
22k1k1
2211
1
k1
2eqkk1
Tk
eqk
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=

18. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
17
‫ﻣﺜـﺎل‬50/‫ﺳﻌﺗﺎن‬‫اﻟﻣﺗ‬ ‫ت‬‫رﺑط‬)C1=2µF , C2=6µF(‫رق‬‫ﻓ‬ ‫ﺔ‬‫ﺑطﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ﻠت‬‫ووﺻ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﻗطﺑﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬40V.
1-‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬ ‫ﻣﻘدار‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻟﻛل‬ ‫اﺣﺳب‬.
2-‫ﻣن‬ ‫ﻟوح‬ ‫ادﺧل‬ ‫اذا‬‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬ ‫ﻛﮭرﺑﺎﺋﯾﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬)4(‫ﻰ‬‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﺑﯾن‬)‫ﺻﻠﺔ‬‫ﻣﺗ‬ ‫ﺔ‬‫اﻟﺑطﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫ﺑﺎﻟﻣﺟﻣوﻋﺔ‬(‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬ ‫ﻓﻣﺎ‬.
‫اﻟﺤﻞ‬/
C240406V.CQ,C320408V.CQ
F824CkC)2
C240406V.CQ,C80402V.CQ)1
22k1k1
1k1
2211
µ=×=∆=µ=×=∆=
µ=×==
µ=×=∆=µ=×=∆=
‫ﻣﺜﺎل‬51/‫اﻟﻣﺗﺳﻌﺗﺎن‬)C1=2µF , C2=8µF(‫ﻣﺟﻣ‬ ‫ووﺻﻠت‬ ‫اﻟﺗوازي‬ ‫ﻋﻠﻰ‬ ‫ﻣرﺑوطﺗﺎن‬‫رق‬‫ﻓ‬ ‫ﺻدر‬‫ﻣ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫وﻋﺗﮭﻣ‬
‫ﻗطﺑﯾﮫ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬90V‫ر‬‫اﺧ‬ ‫ﺎزل‬‫ﺑﻌ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﺑﯾن‬ ‫اﻟﻌﺎزل‬ ‫واﺑدل‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫اﻟﻣﺟﻣوﻋﺔ‬ ‫ﻓﺻﻠت‬ ‫ﻓﺎذا‬
‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬)2(‫اﻟﻌﺎزل‬ ‫اﺑدال‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫وﺷﺣﻧﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﻣﻘدار‬ ‫ﻓﻣﺎ‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬:
Ceq=C1 + C2 =2 + 8=10µF
QT =Ceq . ∆VT =10 × 90 =900µC
‫اﻟﻌﺎزل‬ ‫ﺑﻌد‬:
C1k =k C1 =2 × 2=4µF
Ceqk =C1k + C2 =4 + 8 =12µF
Q‫ﻟذﻟك‬ ‫اﻟﺷﺎﺣن‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫ﻣﻧﻔﺻﻠﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬QTk = QT =900µC
V75
12
900
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VTk = ∆V1k = ∆V2 = 75V
Q1k =C1k . ∆V1k =4 × 75 =300µC , Q2 =C2 . ∆V2 =8 × 75 =600µC
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬52/‫ﺳﻌﺗﺎن‬‫اﻟﻣﺗ‬ ‫ت‬‫رﺑط‬)C1=4µF , C2=2µF(‫ﺻدر‬‫ﻣ‬ ‫ﺎطﺔ‬‫ﺑوﺳ‬ ‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫ﺣﻧت‬‫ﺷ‬ ‫م‬‫ﺛ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫ﻓظﮭر‬ ‫ﻋﻧﮫ‬ ‫وﻓﺻﻠت‬ ‫اﻟﻣﺳﺗﻣرة‬ ‫ﻟﻠﻔوﻟطﯾﺔ‬)40V(‫ﻣﻛﮭﺎ‬‫ﺳ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ت‬‫ادﺧﻠ‬ ‫م‬‫ﺛ‬)0.2cm(
‫ﺑ‬‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫د‬‫ﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺑﺢ‬‫ﻓﺎﺻ‬ ‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺣﯾز‬ ‫ﺗﻣﻸ‬ ‫ﺑﺣﯾث‬ ‫اﻟﺛﺎﻧﯾﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﯾن‬)12V(‫زل‬‫ﻋ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻓﻣ‬
‫اﻟﺛﺎﻧﯾﺔ؟‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫ﯾﺻﺑﺢ‬ ‫وﻛم‬ ‫؟‬ ‫اﻟﻌﺎزل‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬:
Ceq =C1 + C2 =4 + 2=6µF
QT =Ceq . ∆VT =6 × 40 =240µC
‫اﻟﻌﺎزل‬ ‫ﺑﻌد‬:
Q‫ﻟذﻟك‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫ﻣﻧﻔﺻﻠﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬QTk = QT =240µC
F20
12
240
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1 + C2k ⇒ 20 = 4 + C2k ⇒ C2k =20 – 4 =16µF
C2k =k C2 ⇒ 16 = k × 2 ⇒ 8
2
16
k ==
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VTk = ∆V1=∆V2k =12V
m/V6000
102.0
12
d
V
E 2
k2
k2 =
×
=
∆
= −

19. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
18
‫ﻣﺜـﺎل‬53/‫ﻮازﯾﺘﯿﻦ‬‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬‫ﻣ‬ ‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=2µF , C2=8µF(‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﺑﺸﺤﻨﺔ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫ﺷﺤﻨﺖ‬ ‫ﻓﺎذا‬ ‫اﻟﺘﻮازي‬‫ﻛﻠﯿﺔ‬600µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫واﻟﻄﺎﻗ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻟﻜ‬ ‫ﺴﺐ‬‫اﺣ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬.
2-‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺒﻂ‬‫ﻓﮭ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬30V
‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬.
‫اﻟﺤﻞ‬/
C240308V.CQ,C3603012V.CQ
6
2
12
C
C
k
F12820CCC
F20
30
600
V
Q
C)2
J101441048060
2
1
Q.V
2
1
PE
J10361012060
2
1
Q.V
2
1
PE
C480608V.CQ,C120602V.CQ
V60
10
600
C
Q
V
F1082CCC)1
k22kk1k1
1
k1
2eqkk1
k
T
eqk
46
22
46
11
2211
eq
T
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
−−
−−
‫ﻣﺜﺎل‬54/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫رﺑﻄﺖ‬)C1=2µF , C2=6µF(‫ﺑـ‬ ‫ﺎ‬‫ﻟﮭﻤ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺤﻨﺖ‬‫وﺷ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﺔ‬‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬
‫ﺪارھﺎ‬ ‫ﻣﻘ‬)960µC(‫ﺮق‬ ‫ﻓ‬ ‫ﺪار‬ ‫ﻣﻘ‬ ‫ﺎ‬ ‫ﻓﻤ‬ ‫ﮫ‬ ‫ﻋﻨ‬ ‫ﺼﻠﺖ‬ ‫ﻓ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﺔ‬ ‫ﻟﻠﻔﻮﻟﻄﯿ‬ ‫ﺼﺪر‬ ‫ﻣ‬ ‫ﺎطﺔ‬ ‫ﺑﻮﺳ‬‫ﻲ‬ ‫طﺮﻓ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﯿﻈﮭﺮ‬ ‫ﺳ‬ ‫ﺬي‬ ‫اﻟ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬
‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻲ‬‫طﺮﻓ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫اﺻﺒﺢ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫وﺿﻊ‬ ‫واذا‬ ‫؟‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬)96V. (‫ﺎ‬‫ﻣ‬
‫اﻟﻌﺎزﻟﺔ؟‬ ‫اﻟﻤﺎدة‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬
‫اﻟﺤﻞ‬/
2
2
4
C
C
k
F4610CCC,F10
96
960
V
Q
C
V120
8
960
C
Q
V
F862CCC
1
k1
2eqkk1
k
T
eqk
eq
T
T
21eq
===
µ=−=−=µ==
∆
=
===∆
µ=+=+=

20. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
19
‫ﻣﺜـﺎل‬55/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=2µF, C2=8µF(‫ﺸﺤ‬‫ﺑ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺤﻨﺖ‬‫وﺷ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﻨﺔ‬‫ﺔ‬‫ﻛﻠﯿ‬‫ﺪارھﺎ‬‫ﻣﻘ‬
)1000μC(‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫؟‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﺎ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﺑﻤﺎدة‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﮭﻮاء‬ ‫اﺳﺘﺒﺪل‬ ‫اذا‬)6(‫اﻟﻤﺨ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺼﺒﺢ‬‫ﺗ‬ ‫ﻓﻜﻢ‬‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫ﺘﺰﻧ‬
‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬.
‫اﻟﺤﻞ‬/
C400508V.CQ,C6005012V.CQ
V50
20
1000
C
Q
V
F20812CCC
F1226CkC)2
J10410800100
2
1
Q.V
2
1
PE
J1010200100
2
1
Q.V
2
1
PE
C8001008V.CQ,C2001002V.CQ
V100
10
1000
C
Q
V
F1082CCC)1
k22kk1k1
eqk
T
k
2k1eqk
1k1
26
22
26
11
2211
eq
T
21eq
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
µ=×==
×=×××=∆=
=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
−−
−−

21. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
20
‫ﻣﺜﺎل‬56/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=3µF , C2=2µF(‫ـ‬‫ﺑ‬ ‫ﺎ‬‫ﻟﮭﻤ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺤﻨﺔ‬‫وﺷ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬)180µC(
‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻟﻜ‬ ‫ﺴﺐ‬‫اﺣ‬‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫واﻟﻄﺎﻗ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)6(‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬‫اﻟﻌﺎزل‬.
‫اﻟﺤﻞ‬/
J108641014412
2
1
Q.V
2
1
PE
J10216103612
2
1
Q.V
2
1
PE
C1441212V.CQ,C36123V.CQ
V12
15
180
C
Q
V
F15123CCC
F1226CkC)2
J101296107236
2
1
Q.V
2
1
PE
J1019441010836
2
1
Q.V
2
1
PE
C72362V.CQ,C108363V.CQ
V36
5
180
C
Q
V
F523CCC)1
66
k2kk2
66
1k1
kK2K2k11
eqk
T
k
k21eqk
2k2
66
22
66
11
2211
eq
T
21eq
−−
−−
−−
−−
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
µ=×==
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=

22. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
21
‫ﻣﺜــﺎل‬57/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=2µF , C2=3µF(‫ـ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻟﮭﻤ‬ ‫ﺔ‬ ‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬ ‫اﻟ‬ ‫ﺤﻨﺖ‬ ‫وﺷ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬
)400µC(‫ﮫ‬‫ﻋﻨ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﻢ‬‫ﺛ‬ ‫ﺔ‬‫ﻟﻠﻔﻮﻟﻄﯿ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.‫ﻦ‬‫ﻣ‬ ‫ﺪﻻ‬‫ﺑ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﺘﻌﻤﻠﺖ‬‫اﺳ‬
‫اﻟ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﮭﺒﻂ‬ ‫اﻟﮭﻮاء‬‫ﺪار‬‫ﺑﻤﻘ‬ ‫ﺔ‬‫ﻤﺠﻤﻮﻋ‬60V‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺎ‬‫وﻣ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺰل‬‫اﻟﻌ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬
‫اﻟﻌﺎزل؟‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬
‫اﻟﺤﻞ‬/
C3602018V.CQ,C40202V.CQ
6
3
18
C
C
k
F18220CCC
F20
20
400
V
Q
C
V20608060VV
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
V80
5
400
C
Q
V
F532CCC
kk2k2k11
2
k2
1eqkk2
k
T
eqk
k
eq
T
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
=−=−∆=∆
===∆
µ=+=+=
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬58/‫ﺴﻌﺘﺎن‬ ‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=12µF , C2=8µF(‫ﺪارھﺎ‬ ‫ﻣﻘ‬ ‫ﺸﺤﻨﺔ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺤﻨﺔ‬‫وﺷ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬
240µC‫ﺛﻢ‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﺑ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﺘﺒﺪل‬‫اﺳ‬ ‫اذا‬‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺮ‬‫اﺧ‬ ‫ﺎزل‬‫ﺑﻌ‬)k(‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺒﺢ‬‫اﺻ‬6V
‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬)k. (
‫اﻟﺤﻞ‬/
3
12
32
C
C
k
F32840CCC
F40
6
240
V
Q
C)2
C96128V.CQ,C1441212V.CQ
V12
20
240
C
Q
V
F20812CCC)1
1
k1
2eqkk1
k
T
eqk
2211
eq
T
21eq
===
µ=−=−=
µ==
∆
=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=

23. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
22
‫ﻣﺜﺎل‬59/‫ﻣﻧﮭﻣ‬ ‫دة‬‫واﺣ‬ ‫ل‬‫ﻛ‬ ‫ﻌﺔ‬‫ﺳ‬ ‫ﺳﻌﺗﺎن‬‫ﻣﺗ‬‫ﺎ‬2µF‫ﯾن‬‫ﺑ‬ ‫د‬‫اﻟﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺔ‬‫ﺑطﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺗﺎ‬‫وﺻ‬ ‫م‬‫ﺛ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺗﺎ‬‫وﺻ‬
‫ﺎ‬ ‫ﻗطﺑﯾﮭ‬50V‫ﺔ‬ ‫اﻟﺑطﺎرﯾ‬ ‫ن‬ ‫ﻋ‬ ‫ﺔ‬ ‫اﻟﻣﺟﻣوﻋ‬ ‫ﺻﻠت‬ ‫ﻓ‬ ‫ﺎذا‬ ‫ﻓ‬ ‫ﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﺻ‬ ‫ن‬ ‫ﻣ‬ ‫أي‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻣﺧﺗزﻧ‬ ‫ﺷﺣﻧﺔ‬ ‫اﻟ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺗ‬ ‫ل‬ ‫ﻟﻛ‬ ‫ﺳب‬ ‫اﺣ‬
‫ﮫ‬‫ﻋزﻟ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻊ‬‫ووﺿ‬)3(‫د‬‫ﺑﻌ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ل‬‫ﻛ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫د‬‫اﻟﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫دار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻣ‬ ‫ﺔ‬‫اﻟﺛﺎﻧﯾ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬
‫اﻟﻌﺎز‬ ‫ادﺧﺎل‬‫ل‬‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫واﻟﺷﺣﻧﺔ‬.
‫اﻟﺤﻞ‬/
C150256V.CQ,C50252V.CQ
VVV25
8
200
C
Q
V
F862CCC
F623CkC
QC200100100QQQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C100502V.CQ,C100502V.CQ
kk1k2k11
k21
eqk
Tk
k
k21eqk
2k2
Tk21T
1111
µ=×=∆=µ=×=∆=
∆=∆====∆
µ=+=+=
µ=×==
=µ=+=+=
µ=×=∆=µ=×=∆=
‫ﻣﺜﺎل‬60/‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺘﺴﻌﺘﺎن‬2µF‫ﺑـ‬ ‫ﻟﮭﻤﺎ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫ﺷﺤﻨﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬‫ﻣﻘﺪارھﺎ‬ ‫ﻛﻠﯿﺔ‬ ‫ﺸﺤﻨﺔ‬
)120µC(‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬‫ﻟ‬ ‫ﻞ‬‫ادﺧ‬ ‫ﺎذا‬‫ﻓ‬ ‫ﮫ‬‫ﻋﻨ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﻢ‬‫ﺛ‬ ‫ﺔ‬‫ﻟﻠﻔﻮﻟﻄﯿ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬)k(‫ﯿﻦ‬‫ﺑ‬
‫ﺑﻤﻘﺪار‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ھﺒﻂ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬18V‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﻌﺎزل؟‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬
‫اﻟﺤﻞ‬/
C96128V.CQ,C24122V.CQ
4
2
8
C
C
k
F8210CCC
F10
12
120
V
Q
C
V12183018VV
C120QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
V30
4
120
C
Q
V
F422CnC
kk22111
2
k2
1eqkk2
k
T
eqk
k
TTk
eq
T
eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
=−=−∆=∆
µ==
===∆
µ=×==

24. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
23
‫ﻣﺜﺎل‬61/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=8µF,C2=4µF(‫ﺎ‬‫ﻟﮭﻤ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺤﻨﺖ‬‫ﺷ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﺑ‬‫ﺔ‬‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬
‫ﻣﻘﺪارھﺎ‬240µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)2(‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬
‫اﻟﻌ‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬‫ﺎزل‬.
‫اﻟﺤﻞ‬/
C48124V.CQ,C1921216V.CQ
V12
20
240
C
Q
V
F20416CCC
F1682kCC2
C80204V.CQ,C160208V.CQ
V20
12
240
C
Q
V
F1248CCC1
22kk1k1
eqk
Tk
k
2k1eq
1k1
2111
eq
T
21eq
µ=×=∆=µ=×=∆=∴
===∆
µ=+=+=
µ=×==−
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−
‫ﻣﺜــﺎل‬62/‫ﺳﻌﺗﺎن‬ ‫اﻟﻣﺗ‬ ‫ت‬ ‫رﺑط‬)C1=2µF , C2=8µF(‫دارھﺎ‬ ‫ﻣﻘ‬ ‫ﺷﺣﻧﺔ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ﺣﻧﺔ‬ ‫وﺷ‬ ‫وازي‬ ‫اﻟﺗ‬ ‫ﻰ‬ ‫ﻋﻠ‬
100µC‫اﺣﺳب‬ ‫ﻋﻧﮫ‬ ‫ﻓﺻﻠت‬ ‫ﺛم‬ ‫اﻟﺟﮭد‬ ‫ﻟﻔرق‬ ‫ﻣﺻدر‬ ‫ﺑوﺳﺎطﺔ‬:
1-‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬.
2-‫ﮫ‬ ‫ﻋزﻟ‬ ‫ت‬ ‫ﺛﺎﺑ‬ ‫ﺔ‬ ‫ﻋﺎزﻟ‬ ‫ﺎدة‬ ‫ﻣ‬ ‫ن‬ ‫ﻣ‬ ‫وح‬ ‫ﻟ‬ ‫ﻊ‬ ‫وﺿ‬ ‫اذا‬)6(‫د‬ ‫ﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺎ‬ ‫ﻓﻣ‬ ‫ﻰ‬ ‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬ ‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬ ‫ﺑ‬‫ﺔ‬ ‫واﻟطﺎﻗ‬ ‫ﻲ‬ ‫اﻟﻛﻠ‬
‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬‫؟‬
‫اﻟﺤﻞ‬/
J1010025104)5(108
2
1
)V.(C
2
1
PE
J1015025106)5(1012
2
1
)V.(C
2
1
PE
V5
20
100
C
Q
V
F20812CCC
F1226CkC)2
C80108V.CQ,C20102V.CQ
V10
10
100
C
Q
V
F1082CCC)1
66262
k2k2
66262
k11
eq
k
k
2k1eqk
1k1
2211
eq
T
21eq
−−−
−−−
×=××=×××=∆=
×=××=×××=∆=
===∆
µ=+=+=
µ=×==
µ=×=∆=µ=×=∆=
===∆
µ=+=+=

25. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
24
‫ﻣﺜــﺎل‬63/‫ﺳﻌﺗﺎن‬ ‫اﻟﻣﺗ‬ ‫ت‬ ‫رﺑط‬)C1=2µF , C2=3µF(‫دارھﺎ‬ ‫ﻣﻘ‬ ‫ﺷﺣﻧﺔ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ﺣﻧﺔ‬ ‫وﺷ‬ ‫وازي‬ ‫اﻟﺗ‬ ‫ﻰ‬ ‫ﻋﻠ‬
60µC‫ﻋ‬ ‫ﻓﺻﻠت‬ ‫ﺛم‬ ‫اﻟﺟﮭد‬ ‫ﻟﻔرق‬ ‫ﻣﺻدر‬ ‫ﺑوﺳﺎطﺔ‬‫اﺣﺳب‬ ‫ﻧﮫ‬:
1-‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫واﻟطﺎﻗﺔ‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬.
2-‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣن‬ ‫ﻟوح‬ ‫وﺿﻊ‬ ‫اذا‬)6(‫ﺎزل‬‫اﻟﻌ‬ ‫د‬‫ﺑﻌ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ﻛل‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫ﻓﻣﺎ‬ ‫اﻟﺛﺎﻧﯾﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬
‫ﺻﻔﯾﺣﺗﯾﮭ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫واﻟطﺎﻗﺔ‬‫؟‬ ‫ﺎ‬
‫اﻟﺤﻞ‬/
J10225)5(1018
2
1
)V.(C
2
1
PE
J1025)5(102
2
1
)V.(C
2
1
PE
VVV5
20
100
C
Q
V
F20182CCC
F1836CkC)2
J10216)12(103
2
1
)V.(C
2
1
PE
J10144)12(102
2
1
)V.(C
2
1
PE
C36123V.CQ,C24122V.CQ
V12
5
60
C
Q
V
F532CCC)1
6262
k2k2
6262
k11
k21
eq
k
k
k21eqk
2k2
6262
22
6262
11
2211
eq
T
21eq
−−
−−
−−
−−
×=×××=∆=
×=×××=∆=
∆=∆====∆
µ=+=+=
µ=×==
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
‫ﻣﺜﺎل‬64/‫ﻣﺘﺴﻌﺘﺎن‬)C1=4µF , C2=6µF(‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﺎ‬ ‫ﻗﻄﺒﯿﮭ‬80V‫ﺪ‬ ‫ﺟﮭ‬ ‫ﺮق‬ ‫ﻓ‬ ‫ﺒﻂ‬ ‫ھ‬ ‫ﺔ‬ ‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺎزل‬ ‫ﻋ‬ ‫ﻞ‬ ‫وادﺧ‬ ‫ﺔ‬ ‫اﻟﺒﻄﺎرﯾ‬ ‫ﻦ‬ ‫ﻋ‬ ‫ﺔ‬ ‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺼﻠﺖ‬ ‫ﻓ‬ ‫ﺎذا‬ ‫ﻓ‬
‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬50V‫اﻟ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﺟﺪ‬‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫ﻤﺨﺘﺰﻧﺔ‬.
‫اﻟﺤﻞ‬/
C6005012V.CQ,C200504V.CQ
2
6
12
C
C
k
F12416CCC
F16
50
800
V
Q
C
C800QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C8008010V.CQ
F1064CCC
kk2k2k11
2
k2
1eqkk2
k
Tk
eqk
TTk
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
µ==
µ=×=∆=
µ=+=+=

26. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
25
‫ﻣﺜﺎل‬65/‫ﻣﺘﺴﻌﺘﺎن‬)C1=4µF , C2=6µF(‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﺎ‬ ‫ﻗﻄﺒﯿﮭ‬60V‫ﺮ‬ ‫ﻓ‬ ‫ﺒﻂ‬ ‫ھ‬ ‫ﺔ‬ ‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺎزل‬ ‫ﻋ‬ ‫ﻞ‬ ‫وادﺧ‬ ‫ﺔ‬ ‫اﻟﺒﻄﺎرﯾ‬ ‫ﻦ‬ ‫ﻋ‬ ‫ﺔ‬ ‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺼﻠﺖ‬ ‫ﻓ‬ ‫ﺎذا‬ ‫ﻓ‬‫ﺪ‬ ‫ﺟﮭ‬ ‫ق‬
‫ﺑﻤﻘﺪار‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬45V‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﺟﺪ‬.
‫اﻟﺤﻞ‬/
C5401536V.CQ,C60154V.CQ
6
6
36
C
C
k
F36440CCC
F40
15
600
V
Q
C
V15456045VV,C600QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C6006010V.CQ
F1064CCC
kk2k2k11
2
k2
1eqkk2
k
Tk
eqk
TTkTTk
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
=−=−∆=∆µ==
µ=×=∆=
µ=+=+=
‫ﻣﺜﺎل‬66/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=4µF , C2=2µF(‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﻗﻄﺒﯿﮫ‬100V‫اﻟﻤﺼﺪ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫وادﺧﻞ‬ ‫ر‬)k(‫ﺒﻂ‬‫ھ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫ﺪار‬‫ﺑﻤﻘ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬40V‫ﺎل‬‫ادﺧ‬ ‫ﺪ‬‫ﺑﻌ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫واﻟ‬ ‫ﺰل‬‫اﻟﻌ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺪ‬‫ﺟ‬
‫اﻟﻌﺎزل‬.
‫اﻟﺤﻞ‬/
C360606V.CQ,C240604V.CQ
3
2
6
C
C
k
F6410CCC
F10
60
600
V
Q
C
V604010040VV,C600QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C6001006V.CQ
F624CCC
kk2k2k11
2
k2
1eqkk2
k
Tk
eqk
TTkTTk
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
=−=−∆=∆µ==
µ=×=∆=
µ=+=+=

27. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
26
‫ﻣﺜﺎل‬67/‫ﻣﺘﺴﻌﺘﺎن‬)C1=15µF , C2=30µF(‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻮﺻﻮﻟﺘﺎن‬‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﻣﺼﺪر‬ ‫اﻟﻰ‬ ‫وﺻﻠﺘﺎ‬
‫ﻗﻄﺒﯿﮫ‬100V‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬ ‫وادﺧﻞ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬)k(‫ﺒﻂ‬‫ھ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬75V‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﺟﺪ‬.
‫اﻟﺤﻞ‬/
C22507530V.CQ,C22507530V.CQ
2
15
30
C
C
k
F303060CCC
F60
75
4500
V
Q
C
C4500QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C450010045V.CQ
F453015CCC
k22kK11
1
k1
2eqkk1
k
Tk
eqk
TTk
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
µ==
µ=×=∆=
µ=+=+=
‫ﻣ‬‫ﺜــﺎل‬68/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=5µF,C2=15µF(‫ﺔ‬ ‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬ ‫ﺑ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﺎطﺔ‬ ‫ﺑﻮﺳ‬ ‫ﺤﻨﺘﺎ‬ ‫ﺷ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬
‫ﻣﻘﺪارھﺎ‬1000µC‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺘﺎ‬ ‫ﺛﻢ‬.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﺣﺴﺐ‬.
2-‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺾ‬‫اﻧﺨﻔ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬
20V‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬)k(‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J109)20(1045
2
1
)V.(C
2
1
PE
J10)20(105
2
1
)V.(C
2
1
PE
3
15
45
C
C
k
F45550CCC
F50
20
1000
V
Q
C2
C7505015V.CQ,F250505V.CQ
V50
20
1000
C
Q
V
F20155CCC1`
3262
kk2k2
3262
k11
2
k2
1eqkk2
k
T
eqk
2211
eq
T
21eq
−−
−−
×=××=∆=
=××=∆=
===∴
µ=−=−=
µ==
∆
=−
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−

28. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
27
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬69/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=2µF,C2=8µF(‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬24V‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬‫ﻟ‬ ‫ﻞ‬‫ادﺧ‬ ‫ﺎذا‬‫ﻓ‬)k(‫ﺴ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﻌﺔ‬
‫ﺑﻤﻘﺪار‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬48µC‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬:
1-‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k.(
2-‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫إدﺧﺎل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
C192248V.CQ,C96244V.CQ
C192248V.CQ,C48242V.CQ1
2
2
4
C
C
k
F4812CCC
F12
24
288
V
Q
C
F2884824048QQ,C2402410V.CQ
F1082CCC1
22k1k1
2211
1
k1
2eqkk1
Tk
eqk
TTkeqT
21eq
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=−
===∴
µ=−=−=
µ==
∆
=
µ=+=+=µ=×=∆=
µ=+=+=−
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬70/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=8µF,C2=16µF(‫ا‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬ ‫ﺑﻌ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬‫ﺤﻨﺖ‬ ‫ﺷ‬ ‫ﺎذا‬ ‫ﻓ‬ ‫ﻮازي‬ ‫ﻟﺘ‬‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬
‫ﻣﻘﺪارھﺎ‬ ‫ﻛﻠﯿﺔ‬ ‫ﺑﺸﺤﻨﺔ‬480µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺎﺋﻲ‬ ‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬ ‫اﻟﻤﺠ‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬ ‫واﻟﻄﺎﻗ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺻ‬ ‫ﻦ‬ ‫ﻣ‬ ‫أي‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬ ‫اﻟ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬ ‫ﻞ‬ ‫ﻟﻜ‬ ‫ﺴﺐ‬ ‫اﺣ‬
‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬.
2-‫ﺛﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬‫ﻋﺰﻟﮭﺎ‬ ‫ﺑﺖ‬)2(‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬.
‫اﻟﺤﻞ‬/
J1023041038412
2
1
Q.V
2
1
PE
J576109612
2
1
Q.V
2
1
PE
F3841232V.CQ,C96128V.CQ
V12
40
480
C
Q
V,F40328CCC
F32162kCC2
J10321032020
2
1
Q.V
2
1
PE
J10161016020
2
1
Q.V
2
1
PE
C3202016V.CQ,C160208V.CQ
V20
24
480
C
Q
V
F24168CCC1
66
k2kk2
6
1k1
kk2k2k11
eqk
T
kk21eqk
2k2
46
22
46
11
2211
eq
T
21eq
−−
−
−−
−−
×=×××=∆=
=×××=∆=
µ=×=∆=µ=×=∆=
===∆µ=+=+=
µ=×==−
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−

29. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
28
‫ﻣﺜﺎل‬71/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=4µF,C2=8µF(‫ﺷ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﺪارھﺎ‬‫ﻣﻘ‬ ‫ﺔ‬‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬‫ﺑ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺤﻨﺖ‬
600µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺎﺋﻲ‬ ‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬ ‫اﻟﻤﺠ‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬ ‫واﻟﻄﺎﻗ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺻ‬ ‫ﻦ‬ ‫ﻣ‬ ‫أي‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬ ‫اﻟ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬ ‫ﻞ‬ ‫ﻟﻜ‬ ‫ﺴﺐ‬ ‫اﺣ‬
‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫اﻻوﻟ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬‫ﻰ‬‫اﻟ‬ ‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺾ‬‫اﻧﺨﻔ‬ ‫ﻰ‬30V‫ﺎ‬‫ﻓﻤ‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬)k(‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﺸﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
C240308V.CQ
3
4
12
C
C
k
F12820CCC
F20
30
600
V
Q
C2
J101040050
2
1
Q.V
2
1
PE
J1051020050
2
1
Q.V
2
1
PE
C400508V.CQ,C200504V.CQ
V50
12
600
C
Q
V
F1284CCC1
222
1
k1
2eqkk1
TK
Tk
eqk
26
22
36
11
2211
eq
T
T
21eq
µ=×=∆=
===
µ=−=−=
µ==
∆
=−
=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−
−−
−−
‫ﻣﺜــﺎل‬72/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=6µF,C2=12µF(‫ﺔ‬ ‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺤﻨﺖ‬ ‫ﺷ‬ ، ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬
‫ﻣﻘﺪارھﺎ‬540µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬
‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﻔﯿﺤﺘ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺾ‬‫اﻧﺨﻔ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬‫ﯿﮭﺎ‬
‫اﻟﻰ‬18V‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬ ‫وﻣﺎ‬ ‫؟‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬.
‫اﻟﺤﻞ‬/
J10541036030
2
1
Q.V
2
1
PE
J10271018030
2
1
Q.V
2
1
PE
C3603012V.CQ,C180306V.CQ
V30
18
540
C
Q
V
F18126CCC1
46
22
46
11
2211
eq
T
21eq
−−
−−
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−

30. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
29
C2161812V.CQ,C3241818V.CQ
3
6
18
C
C
k
F181230CCC
F30
18
540
V
Q
C
V18VV2
222k1k1k1
1
k1
2eqkk1
Tk
Tk
eqk
k1Tk
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=
=∆=∆−
‫ﻣﺜﺎل‬73/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫وﺻﻠﺖ‬)C1=2µF,C2=4µF(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬
50V.
1-‫اﺣﺴﺐ‬‫و‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ا‬‫ﻣ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫ﻟﺸﺤﻨﺔ‬‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻦ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫واﻟﻄﺎﻗ‬‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬.
2-‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬‫ﻟ‬ ‫ﻞ‬‫وادﺧ‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﻦ‬‫ﻋ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫اذا‬)k(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﻧﺨﻔﺾ‬30V‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫اﻟﻌﺎزل؟‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬ ‫وﻣﺎ‬ ‫؟‬
‫ا‬‫ﻟﺤﻞ‬/
C240308V.CQ,C60302V.CQ
2
4
8
C
C
k
F8210CCC
F10
30
300
V
Q
C2
C200504V.CQ,C100502V.CQ
C300506V.CQ
F642CCC1
k2k2k2111
2
k2
1eqkk2
Tk
Tk
eqk
2211
TeqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=−
µ=×=∆=µ=×=∆=
µ=×=∆=
µ=+=+=−
‫ﻣﺜــﺎل‬74/‫ﻮازﯾﺘﯿﻦ‬ ‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬ ‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=6µF,C2=2µF(‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬ ‫ﺑﻌ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺑﯿﻦ‬ ‫رﺑﻄﺖ‬ ‫وﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫اﻟﺘﻮازي‬)12V(‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬ ،:
1-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
2-‫ﻛﮭﺮﺑﺎﺋ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻲ‬)2(‫ﻰ‬‫اﻻوﻟ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬)‫ﻲ‬‫طﺮﻓ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬(‫اﻟﻜﻠﯿﺔ؟‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
C16824144QQQ
C24122V.CQ,C1441212V.CQ
F1262kCC
C962472QQQ
C24122V.CQ,C72126V.CQ1
2k1Tk
22k1k1
1k1
21T
2211
µ=+=+=
µ=×=∆=µ=×=∆=
µ=×==
µ=+=+=
µ=×=∆=µ=×=∆=−

31. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
30
‫ﻣﺜﺎل‬75/‫اﻟﻤﺘﻮا‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫ذوات‬ ‫ﻣﻦ‬ ‫ﻣﺘﺴﻌﺘﺎن‬‫ﺔ‬‫زﯾ‬)C1=4µF,C2=6µF(‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺑﯿﻦ‬ ‫رﺑﻄﺖ‬ ‫وﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬)20V(‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬ ، ‫ﻋﻨﮭﺎ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫اﻟﻜ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﮭﺒﻂ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬‫اﻟﻰ‬ ‫ﻠﻲ‬5V‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k(‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬‫؟‬
‫اﻟﺤﻞ‬/
C180536V.CQ,C2054V.CQ
6
6
36
C
C
k
F36440CCC
F40
5
200
V
Q
C2
C120206V.CQ,C80204V.CQ
C2002010V.CQ
C1064CCC1
k2k2k2111
2
k2
1eqkk2
Tk
Tk
eqk
2211
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=−
µ=×=∆=µ=×=∆=
µ=×=∆=
µ=+=+=−
‫ﻣﺜﺎل‬76/‫ﻣﺘﺴﻌﺘﺎن‬)C1=6µF,C2=12µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻲ‬‫ﻗﻄﺒ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬6V‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫ﻓﺎذا‬)k(‫ﺑ‬‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬
‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﻣﺎزاﻟﺖ‬180µC‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬:
1-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k.(
2-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
C72612V.CQ,C108618V.CQ
C72612V.CQ,C3666V.CQ2
3
6
18
C
C
k
F181230CCC
F30
6
180
V
Q
C1
22k1k1
2211
1
k1
2eqkk1
Tk
Tk
eqk
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=−
===
µ=−=−=
µ==
∆
=−

32. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
31
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬77/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=3µF,C2=2µF(‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎ‬ ‫رﺑﻄﺘ‬‫ﺔ‬ ‫ﻛﻠﯿ‬ ‫ﺸﺤﻨﺔ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻟﮭﻤ‬ ‫ﺔ‬ ‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬ ‫اﻟ‬ ‫ﺤﻨﺖ‬‫ﺷ‬ ‫ﻢ‬ ‫ﺛ‬
‫ﺪارھﺎ‬ ‫ﻣﻘ‬120µC‫ﺎ‬ ‫ﻋﻨﮭ‬ ‫ﺼﻠﺖ‬ ‫ﻓ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﺎطﺔ‬ ‫ﺑﻮﺳ‬‫و‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬ ‫ﺛﺎﺑ‬ ‫ﺔ‬ ‫ﻋﺎزﻟ‬ ‫ﺎدة‬ ‫ﻣ‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﻮح‬ ‫ﻟ‬ ‫ﻞ‬ ‫ادﺧ‬‫ﺎ‬)6(‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬
‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬‫ﻓﯿﮭﺎ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬.
‫اﻟﺤﻞ‬/
J1038410968
2
1
Q.V
2
1
PE
J109610248
2
1
Q.V
2
1
PE
C96812V.CQ,C2483V.CQ
V8
15
120
C
Q
V
F15123CCC
F1226kCC
66
k2kk2
66
1k1
kk2k2k11
eqk
Tk
k
k21eqk
2k2
−−
−−
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
µ=×==
‫ﻣﺜﺎل‬78/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫رﺑﻄﺖ‬)C1=14µF,C2=16µF(‫ﺮق‬‫ﻓ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﻗﻄﺒﯿﮫ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬5V.
1-‫اﻟﻤﺘﺴﻌﺎت‬ ‫ﻣﻦ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫ا‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫ﻟﺸﺤﻨﺔ‬390µC‫ﺪ‬‫ﺑﻌ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﺎ‬‫وﻣ‬ ‫؟‬ ‫ﺰل‬‫اﻟﻌ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﻌﺎزل؟‬
‫اﻟﺤﻞ‬/
C320564V.CQ,C70514V.CQ
4
16
64
C
C
k
F641478CCC
F78
5
390
V
Q
C2
C80516V.CQ,C70514V.CQ
C150530V.CQ
F301614CCC1
k2k211
2
k2
1eqkk2
Tk
eqk
2211
eqT
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=−
µ=×=∆=µ=×=∆=
µ=×=∆=
µ=+=+=−

33. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
32
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬79/‫ﺴﻌﺘﺎن‬ ‫اﻟﻤﺘ‬)C1=12µF,C2=3µF(‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﻜﺎﻧﺖ‬300µC.
1-‫ﺷﺤﻨﺔ‬ ‫اﺣﺴﺐ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬.
2-‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫وادﺧﻞ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫اذا‬)k(‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﻧﺨﻔﺾ‬ ‫اﻟﺜﺎﻧﯿﺔ‬10V‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫؟‬ ‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫؟‬
‫اﻟﺤﻞ‬/
C1801018V.CQ,C1201012V.CQ
6
3
18
C
C
k
F181230CCC
F30
10
300
V
Q
C2
C60203V.CQ,C2402012V.CQ
V20
15
300
C
Q
V
F15312CCC1
k2k211
2
k2
1eqkk2
Tk
eqk
2211
eq
T
21eq
µ=×=∆=µ=×=∆=
===
µ=−=−=
µ==
∆
=−
µ=×=∆=µ=×=∆=
===∆
µ=+=+=−
‫ﻣﺜﺎل‬80/‫ذ‬ ‫ﻣن‬ ‫ﻣﺗﺳﻌﺗﺎن‬‫وازﯾﺗﯾن‬‫اﻟﻣﺗ‬ ‫ﺻﻔﯾﺣﺗﯾن‬‫اﻟ‬ ‫وات‬(C1=6µF , C2=14µF)‫ﻰ‬‫ﻋﻠ‬ ‫ﺿﮭﻣﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣرﺑوطﺗ‬
‫ﻲ‬ ‫طرﻓ‬ ‫ﻰ‬ ‫اﻟ‬ ‫ت‬‫رﺑط‬ ‫ﺎ‬ ‫وﻣﺟﻣوﻋﺗﮭﻣ‬ ‫وازي‬ ‫اﻟﺗ‬‫ﺎ‬ ‫ﻗطﺑﯾﮭ‬ ‫ﯾن‬ ‫ﺑ‬ ‫د‬ ‫اﻟﺟﮭ‬ ‫رق‬ ‫ﻓ‬ ‫ﺔ‬ ‫ﺑطﺎرﯾ‬)30V(‫ﻓ‬ ،‫ﻓ‬ ‫ﺎذا‬‫ن‬ ‫ﻋ‬ ‫ﺔ‬ ‫اﻟﻣﺟﻣوﻋ‬ ‫ﺻﻠت‬
‫اد‬ ‫ﺛم‬ ‫اﻟﺑطﺎرﯾﺔ‬‫ﺎ‬‫ﻋزﻟﮭ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻣن‬ ‫ﻟوﺣﺎ‬ ‫ﺧل‬)k(‫ﻰ‬‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫د‬‫ﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺑط‬‫ھ‬
‫ﺑﻣﻘدار‬)18V(‫ﻣﻘدار‬ ‫ﻓﻣﺎ‬:
1-‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻌزل‬ ‫ﺛﺎﺑت‬2-‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬
Ceq =C1 + C2 =6 + 14 =20µF
QT =Ceq . ∆VT =20 × 30 =600µC
‫اﻟﻌﺎزل‬ ‫ﺑﻌد‬
QTk = QT =600µC
∆Vk =∆V – 18 =30 – 18 =12V
F50
12
600
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1k + C2 ⇒ 50 =C1k + 14 ⇒ C1k =50 – 14 =36µF
C1k =kC1 ⇒ 36 =k × 6 ⇒ k =6
Q‫ﺗوا‬ ‫اﻟرﺑط‬‫ﻟذﻟك‬ ‫زي‬∆VTk =∆V1k = ∆V2 =12V
Q1k =C1k . ∆V1k =36 × 12 =432µC
Q2 =C2 . ∆V2 =14 × 12 =168µC

34. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
33
‫ﻣﺜﺎل‬81/‫اﻟﻤﺘﻮازﯾﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫ذوات‬ ‫ﻣﻦ‬ ‫ﻣﺘﺴﻌﺘﺎن‬)C1=4µF , C2=6µF(‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫ﻣﻮﺻﻮﻟﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫وﺻﻠﺖ‬ ‫ﺛﻢ‬40V.
1-‫ﻣﻘﺪ‬ ‫ﻣﺎ‬‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬ ‫ار‬.
2-‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫اذا‬‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬ ‫وﺿﻊ‬ ‫ﺛﻢ‬6‫ﺰ‬‫اﻟﺤﯿ‬ ‫ﻸ‬‫ﯾﻤ‬ ‫ﺚ‬‫ﺑﺤﯿ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬
‫ﻓﻜ‬ ‫ﺑﯿﻨﮭﻤﺎ‬‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﯾﺼﺒﺢ‬ ‫ﻢ‬‫ﻣﺘﺴﻌﺔ؟‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬
‫اﻟﺤﻞ‬/
1- ∆VT = ∆V1 = ∆V2 =40V (‫)ﺗﻮازي‬
Q1 =C1 . ∆V1 =4 × 40 = 160µC , Q2 =C2 . ∆V2 =6 × 40 =240µC
QT =Q1 + Q2 =160 + 240 =400µC
2- C2k =k C2 =6 × 6 =36µF
Ceqk =C1 + C2k =4 + 36 =40µC
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬QTk = QT =400µC
V10
40
400
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk = ∆V1 = ∆V2k =10V
Q1 =C1 .∆V1 =4 ×10=40V , QTk =C2k . ∆V2k =36 × 10 =360µC
‫ﻣﺜـﺎل‬82/‫ﻮازﯾﺘﯿﻦ‬‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬‫ﻣ‬ ‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=2µF , C2=8µF(‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﺑ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫ﺷﺤﻨﺖ‬ ‫ﻓﺎذا‬ ‫اﻟﺘﻮازي‬‫ﻛﻠﯿﺔ‬ ‫ﺸﺤﻨﺔ‬600µC‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬‫ﻟ‬ ‫ﻞ‬‫وادﺧ‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﻦ‬‫ﻋ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﺼﻠﺖ‬ ‫اذا‬)k(‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫ﺑﻤﻘﺪار‬ ‫ﺷﺤﻨﺘﮭﺎ‬ ‫ازدادت‬240µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬(k).
‫اﻟﺤﻞ‬/
1- Ceq =C1 + C2 =2 + 8 =10µF
21
eq
T
T VVV60
10
600
C
Q
V ∆=∆====∆
J1036)60(102
2
1
)V.(C
2
1
PE 4262
111
−−
×=×××=∆=
J10144)60(108
2
1
)V.(C
2
1
PE 4262
222
−−
×=×××=∆=
2-
‫اﻟﻌﺎزل‬ ‫ﻗﺒﻞ‬:
Q1 = C1 . ∆V1 =2 × 60 =120µC , Q2 =C2 . ∆V2 =8 × 60=480µC
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =600µC
Q1k=Q1 + 240 =120 + 240 =360µC , Q2 =QTk – Q1 =600 – 360 =240µC
V30
8
240
C
Q
V
2
2
2
===∆ =∆V1k
F12
30
360
V
Q
C
k1
k1
k1
µ==
∆
=
C1k = k C1 ⇒ 12 =k × 2 ⇒ 6
2
12
k ==

35. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
34
‫ﻣﺜﺎل‬83/‫اﻟﻤﺘﺴﻌﺘﺎن‬)C1=2µF , C2=4µF(‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻊ‬‫وﺿ‬ ، ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻮﺻﻮﻟﺘﺎن‬
)6(‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫ﻓﻜﺎﻧﺖ‬ ‫ﺑﻤﺼﺪر‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺻﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﮭﻮاء‬ ‫ﺑﺪل‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬)1600µC(‫ﺎ‬‫ﻓﻤ‬
‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻤﺼﺪر‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬
‫اﻟﺤﻞ‬/
‫ا‬ ‫ﺑﻌﺪ‬‫ﻟﻌﺎزل‬:
C1k =k C1 =6 × 2=12µF
Ceqk =C1k + C2 =12 + 4 =16µF
V100
16
1600
C
Q
V
eqk
Tk
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1k =∆V2 =100V
Q1k =C1k . ∆V1k =12 × 100 =1200µC , Q2 =C2 . ∆V2 =4 × 100 =400µC
‫ﻣﺜﺎل‬84/‫اﻟﻤﺘﺴﻌﺔ‬)2µF(‫ﺑ‬ ‫ﯾﻔﺼﻞ‬‫ﻠﺖ‬‫وﺻ‬ ‫ﺛﻢ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﮭﻮاء‬ ‫ﺑﺪل‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫وﺿﻌﺖ‬ ‫اﻟﮭﻮاء‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﯿﻦ‬
‫ﺴﻌﺔ‬ ‫ﺑﺎﻟﻤﺘ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬)3µF(‫ﺔ‬ ‫اﻟﻜﻠﯿ‬ ‫ﺸﺤﻨﺔ‬ ‫اﻟ‬ ‫ﺖ‬‫ﻓﻜﺎﻧ‬ ‫ﺔ‬ ‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺤﻨﺖ‬ ‫ﺷ‬ ‫ﻢ‬ ‫ﺛ‬)1800µC(‫ﻲ‬ ‫طﺮﻓ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬ ‫ﺮق‬‫وﻓ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬)120V. (‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬.2-‫ﺻﻔﯿﺤﺘ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻲ‬.
‫اﻟﺤﻞ‬/
1-
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻨﻔﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =1800µC
F15
120
1800
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1k + C2 ⇒ 15 =C1k + 3 ⇒ C1k =15 – 3 =12µF
C1k =k C1 ⇒ 12 =k × 2 ⇒ 6
2
12
k ==
2-
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk = ∆V1k = ∆V2 =120V
Q1k=C1k . ∆V1k =12 × 120 =1440µC
Q2 =C2 . ∆V2 =3 × 120 =360µC
‫ﻣﺜﺎل‬85/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬)2µF(‫واﻟﺒﻌﺪ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬)0.1cm(‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫رﺑﻄﺖ‬)3µF(
‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﮭﻮاء‬ ‫ﯾﻔﺼﻞ‬.‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﻦ‬‫ﻣ‬ ‫ﺪﻻ‬‫ﺑ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫اﺳﺘﻌﻤﻠﺖ‬
‫ﺟﮭﺪه‬ ‫ﻓﺮق‬ ‫ﺑﻤﺼﺪر‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬)20V(‫اﻟﺜﺎﻧﯿﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫ﻓﺎﺻﺒﺤﺖ‬)360µC(‫ﻓﻤﺎ‬:
1-‫ا‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬‫؟‬ ‫ﻻوﻟﻰ‬2-‫اﻻوﻟﻰ؟‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬
‫اﻟﺤﻞ‬/
1-
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk = ∆V1 = ∆V2k =20V
F18
20
360
V
Q
C
k2
k2
k2
µ==
∆
=
C2k = k C2 ⇒ 18 =k × 3 ⇒ 6
3
18
k ==
Q1 =C1 . ∆V1 =2 × 20 =40µC
2- m/V20000
101.0
20
d
V
E 2
1
1 =
×
=
∆
= −

36. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
35
‫ﻣﺜــﺎل‬86/‫ﺴﻌﺎت‬ ‫ﻣﺘ‬ ‫ﻊ‬ ‫ارﺑ‬(C1=4µF,C2=2µF,C3=8µF,C4=6µF)‫ﺤﻨﺖ‬ ‫ﺷ‬ ‫ﺎذا‬ ‫ﻓ‬ ، ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻠﺖ‬ ‫وﺻ‬
‫ﻛﻠﯿﺔ‬ ‫ﺑﺸﺤﻨﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬)600µC(‫ﻋﻨﮫ‬ ‫ﻓﺼﻠﺖ‬ ‫ﺛﻢ‬ ‫اﻟﻤﺴﺘﻤﺮة‬ ‫ﻟﻠﻔﻮﻟﻄﯿﺔ‬ ‫ﻣﺼﺪر‬ ‫ﺑﻮﺳﺎطﺔ‬.
1-‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻟﻜ‬ ‫ﺴﺐ‬‫اﺣ‬‫ﯿﻦ‬‫ﺑ‬ ‫ﺎﺋﻲ‬‫اﻟﻜﮭﺮﺑ‬ ‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫واﻟﻄﺎﻗ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬
‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)6(‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬.
‫اﻟﺤﻞ‬/
J103106010
2
1
Q.V
2
1
PE
J10241048010
2
1
Q.V
2
1
PE
J10102010
2
1
Q.V
2
1
PE
J102104010
2
1
Q.V
2
1
PE
C60106V.CQ,C4801048V.CQ
C20102V.CQ,C40104V.CQ
V10
60
600
C
Q
V
F6064824CCCCC
F4886kCC2
J10271018030
2
1
Q.V
2
1
PE
J10361024030
2
1
Q.V
2
1
PE
J109106030
2
1
Q.V
2
1
PE
J10181012030
2
1
Q.V
2
1
PE
C180306V.CQ,C240308V.CQ
C60302V.CQ,C120304V.CQ
V30
20
600
C
Q
V
F206824CCCCC
46
4k4
46
k3kk3
46
2k2
46
1k1
k44kk3k3
k22k11
eqk
Tk
k
4k321eqk
3k3
46
44
46
33
46
22
46
11
4433
2211
eq
4321eq
−−
−−
−−
−−
−−
−−
−−
−−
×=×××=∆=
×=×××=∆=
=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+++=+++=
µ=×==−
×=×××=∆=
×=×××=∆=
×=×××=∆=
×=×××=∆=
µ=×=∆=µ=×=∆=
µ=×=∆=µ=×=∆=
===∆
µ=+++=+++=

37. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
36
‫اﻟﻌﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫ﺗﻮاﻟﻲ‬:
‫ﻣﺜـﺎل‬87/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=3µF , C2=6µF(‫ﮫ‬‫ﻋﻨ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﻢ‬‫ﺛ‬ ‫ﺼﺪر‬‫ﺑﻤ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺤﻨﺖ‬‫وﺷ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫رﺑﻄﺘ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫طﺮﻓﻲ‬ ‫ﺑﯿﻦ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﻈﮭﺮ‬)90V. (
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
2-‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫اﺳﺘﻌﻤﻠﺖ‬ ‫واذا‬‫ﺎ‬‫ﻋﺰﻟﮭ‬(2)‫ﻤﻜﮭﺎ‬‫وﺳ‬)0.6cm(‫ﺮق‬‫ﻓ‬ ‫ﺼﺒﺢ‬‫ﯾ‬ ‫ﻢ‬‫ﻓﻜ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﺪل‬‫ﺑ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻲ‬‫ﻓ‬
‫؟‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻣﻘﺪار‬ ‫ھﻮ‬ ‫وﻛﻢ‬ ‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻋﺒﺮ‬ ‫اﻟﺠﮭﺪ‬
‫اﻟﺤﻞ‬/
1-
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq =2µF
QT =Ceq . ∆VT =2 × 90=180µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =180µC
V60
3
180
C
Q
V
1
1
1
===∆ , V30
6
180
C
Q
V
2
2
2
===∆
2- C1k =k C1 =2 × 3 = 6µF
3
1
6
2
6
11
6
1
6
1
C
1
C
1
C
1
2k1eqk
==
+
=+=+= ⇒ Ceqk=3µF
Q‫اﻟﺸ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬‫ﻟﺬﻟﻚ‬ ‫ﺎﺣﻦ‬QTk = QT =180µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QTk = Q1k = Q2 =180µC
V30
6
180
C
Q
V
k1
k1
k1
===∆ , V30
6
180
C
Q
V
2
2
2
===∆
m/V5000
106.0
30
d
V
E 2
k1
k1 =
×
=
∆
= −
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬88/‫ﺴﻌﺘﺎن‬ ‫اﻟﻤﺘ‬)C1=2µF,C2=3µF(‫ﻰ‬ ‫اﻟ‬ ‫ﻠﺘﺎ‬ ‫وﺻ‬ ، ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬
‫ﻗﻄﺒﯿﮭﺎ‬24V
1-‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)3(‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬)‫ﺔ‬‫ﺑﺎﻟﺒﻄﺎرﯾ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺠﻤﻮﻋﺔ‬(
‫اﻟﻌﺎزﻟﺔ؟‬ ‫اﻟﻤﺎدة‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
V16
3
48
C
Q
V,V8
6
48
C
Q
V
C48242V.CQ
F2
9
18
36
36
CC
C.C
C
F623kCC2
6.9
3
8.28
C
Q
V,V4.14
2
8.28
C
Q
V
C8.28242.1V.CQ
F2.1
5
6
32
32
CC
C.C
C1
2
k
2
k1
k
k1
Tkeqkk
2k1
2k1
eqk
1k1
2
2
1
1
Teq
21
21
eq
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=
µ=×==−
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−

38. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
37
‫ﻣﺜﺎل‬89/‫اﻟﻣﺗوازﯾﺗﯾن‬ ‫اﻟﺻﻔﯾﺣﺗﯾن‬ ‫ذوات‬ ‫ﻣن‬ ‫ﻣﺗﺳﻌﺗﺎن‬)C1=20µF , C2=30µF(‫ﻰ‬‫ﻋﻠ‬ ‫ﺿﮭﻣﺎ‬‫ﺑﻌ‬ ‫ﻣﻊ‬ ‫ﻣرﺑوطﺗﺎن‬
‫اﻟﺗواﻟﻲ‬.‫ﻗطﺑﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﺑطﺎرﯾﺔ‬ ‫طرﻓﻲ‬ ‫اﻟﻰ‬ ‫ﻣﺟﻣوﻋﺗﮭﻣﺎ‬ ‫رﺑطت‬30V‫ل‬‫ﻛ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫ﺎزﻻ‬‫ﻋ‬ ‫واء‬‫اﻟﮭ‬ ‫ﺎن‬‫وﻛ‬
‫ﻣﺎ‬ ‫ﻣن‬ ‫ﻟوح‬ ‫ادﺧل‬ ، ‫ﻣﺗﺳﻌﺔ‬‫ﮫ‬‫ﻋزﻟ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭرﺑﺎﺋﯾ‬ ‫ﻋﺎزﻟﺔ‬ ‫دة‬3‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬C1)‫ﺻﻠﺔ‬‫ﻣﺗ‬ ‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫ﺑﺎﻟﺑطﺎرﯾﺔ‬(‫ﻓﯾﮭﺎ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫واﻟطﺎﻗﺔ‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫ﻣﻘدار‬ ‫اﺣﺳب‬.
1-‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬2-‫اﻟﻌﺎزل‬ ‫ﺑﻌد‬
‫اﻟﺤﻞ‬/
J1061060020
2
1
Q.V
2
1
PE
J1031060010
2
1
Q.V
2
1
PE
V20
30
600
C
Q
V,V10
60
600
C
Q
V
C6003020V.CQ
F20
90
1800
3060
3060
CC
C.C
C
F60203kCC2
J102161036012
2
1
Q.V
2
1
PE
J103241036018
2
1
Q.V
2
1
PE
V12
30
360
C
Q
V,V18
20
360
C
Q
V
C3603012V.CQ
F12
50
600
3020
3020
CC
C.C
C1
36
k22
36
kk1k1
2
k
2
k1
k
k1
Tkeqkk
2k1
2k1
eqk
1k1
56
22
56
11
2
2
1
1
Teq
21
21
eq
−−
−−
−−
−−
×=×××=∆=
×=×××=∆=
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=
µ=×==−
×=×××=∆=
×=×××=∆=
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬90/‫ﺴ‬ ‫ﻣﺘ‬‫ﺔ‬ ‫اﻟﻤﺘﻮازﯾ‬ ‫ﺼﻔﺎﺋﺢ‬ ‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﻌﺘﺎن‬)C1=9µF,C2=18µF(‫ﺖ‬ ‫ورﺑﻄ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻧﻀﯿﺪة‬ ‫اﻟﻰ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬12V.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬ ‫اﺣﺴﺐ‬.
2-‫اﻷوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﺑﻘﺎء‬ ‫ﻣﻊ‬‫ﺑﺎﻟﻤﺠﻤﻮﻋﺔ‬(‫ﻓﺄﺻﺒﺤﺖ‬
‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬144µC‫اﻟﻌ‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬‫ﻣﺘﺴﻌﺔ؟‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﺰل‬
‫اﻟﺤﻞ‬/
1-
6
1
18
3
18
12
18
1
9
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq=6µF
QT =Ceq . ∆VT =6 × 12 =72µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =72µC
V8
9
72
C
Q
V
1
1
1
===∆ , V4
18
72
C
Q
V
2
2
2
===∆
2-
Q‫ﻟﺬﻟﻚ‬ ‫ﺑﺎﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬∆VTk = ∆VT =12V

39. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
38
V12
12
144
V
Q
C
Tk
Tk
eqk
==
∆
=
2k1eqk
C
1
C
1
C
1
+= ⇒
18
1
C
1
12
1
k1
+= ⇒
36
1
36
23
18
1
12
1
C
1
k1
=
−
=−=
∴ C1k=36µF
C1k =k C1 ⇒ 36 =k × 9 ⇒ 4
9
36
k ==
V8
18
144
C
Q
V,V4
36
144
C
Q
V
2
2
2
k1
k1
k1 ===∆===∆
‫ﻣﺜــﺎل‬91/‫ﻮازﯾﺘﯿﻦ‬ ‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬ ‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=6µF,C2=3µF(‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬ ‫ﺑﻌ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬
‫رﺑﻄﺖ‬ ‫وﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫اﻟﺘﻮاﻟﻲ‬‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺑﯿﻦ‬)12V(‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬ ،:
1-‫ﻓ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﺟﮭﺪ‬ ‫ﺮق‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬)‫ﻲ‬‫طﺮﻓ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬(‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬48µC‫اﻟﻌﺰل؟‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬‫اﻟﻌﺎزل؟‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬
‫اﻟﺤﻞ‬/
V4
12
48
C
Q
V,V8
6
48
C
Q
V
4
3
12
C
C
k
F12C
12
1
12
23
6
1
4
1
C
1
C
1
C
1
F4
12
48
V
Q
C2
V8
3
24
C
Q
V,V4
6
24
C
Q
V
C24122V.CQ
F2
9
18
63
63
CC
C.C
C1
k2
k
k2
1
k
1
2
k2
k2
1eqkk2
Tk
k
eqk
2
2
1
1
Teq
21
21
eq
===∆===∆
===
µ=⇒=
−
=−=−=
µ==
∆
=−
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−

40. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
39
‫ﻣﺜﺎل‬92/‫اﻟﻤﺘﻮازﯾﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫ذوات‬ ‫ﻣﻦ‬ ‫ﻣﺘﺴﻌﺘﺎن‬)C1=12µF,C2=6µF(‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺑﯿﻦ‬ ‫رﺑﻄﺖ‬ ‫وﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬)60V(‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬.
2-‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎﺋﻲ‬‫ﻛﮭﺮﺑ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫ﺑﺎﻟﺒﻄﺎرﯾ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬480µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫اﻟﻌﺎزل؟‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬
‫اﻟﺤﻞ‬/
V20
24
480
C
Q
V,V40
12
480
C
Q
V
4
6
24
C
C
k
F24C
24
1
24
23
12
1
8
1
C
1
C
1
C
1
F8
60
480
V
Q
C2
V40
6
240
C
Q
V,V20
12
240
C
Q
V
C240604V.CQ
F4
)12(6
612
612
612
CC
C.C
C1
k2
k
k2
1
k
1
2
k2
k2
1eqkk2
Tk
k
eqk
2
2
1
1
Teq
21
21
eq
===∆===∆
===
µ=⇒=
−
=−=−=
µ==
∆
=−
===∆===∆
µ=×=∆=
µ=
+
×
=
+
×
=
+
=−
‫ﻣﺜﺎل‬93/‫ﻣﺘﺴﻌﺘﺎن‬)C1=24µF,C2=12µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻲ‬‫ﻗﻄﺒ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬6V‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬‫ﻟ‬ ‫ادﺧﻞ‬ ‫ﻓﺎذا‬)k(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﻣﺎزاﻟﺖ‬96µC‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬:
1-‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬)k.(
2-‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
V2
48
96
C
Q
V,V4
24
96
C
Q
V
V4
12
48
C
Q
V,V2
24
48
C
Q
V
C4868V.CQ,F8
)12(12
1224
1224
1224
CC
C.C
C2
4
12
48
C
C
k
F48C
48
1
48
23
24
1
16
1
C
1
C
1
C
1
F16
6
96
V
Q
C1
k2
k
k2
1
k
1
2
2
1
1
Teq
21
21
eq
2
k2
k2
1eqkk2
Tk
k
eqk
===∆===∆
===∆===∆
µ=×=∆=µ=
+
×
=
+
×
=
+
=−
===
µ=⇒=
−
=−=−=
µ==
∆
=−

41. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
40
‫ﻣﺜﺎل‬94/‫ﻣﺘﺴﻌﺘﺎن‬)C1=12µF,C2=4µF(‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫رﺑﻄﺘﺎ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬24V.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬‫ﺎ‬)k(‫ﻔ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﯿﺤﺘﻲ‬‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬‫ﺔ‬‫ﺑﺎﻟﺒﻄﺎرﯾ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬
‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬192µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫ﻣﺘﺴﻌﺔ؟‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫وﻣﺎ‬
‫اﻟﺤﻞ‬/
V8
24
192
C
Q
V,V16
12
192
C
Q
V
6
4
24
C
C
k
F24C
24
1
24
23
12
1
8
1
C
1
C
1
C
1
F8
24
192
V
Q
C2
V18
4
72
C
Q
V,V6
12
72
C
Q
V
C72243V.CQ
F3
16
48
412
412
CC
C.C
C1
k2
k
k2
1
k
1
2
k2
k2
1eqkk2
Tk
k
eqk
2
2
1
1
Teq
21
21
eq
===∆===∆
===
µ=⇒=
−
=−=−=
µ==
∆
=−
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−
‫ﻣﺜﺎل‬95/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=9µF,C2=18µF(‫اﻟ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫رﺑﻄﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺼﺪر‬ ‫ﻰ‬
‫ﻗﻄﺒﯿﮫ‬ ‫ﺑﯿﻦ‬6V.
1-‫اﻟﻤﺘﺴﻌﺎت‬ ‫ﻣﻦ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫واﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬72µC‫ﺪ‬‫ﺑﻌ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫وﻣﺎ‬ ‫؟‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﻌﺎزل؟‬
‫اﻟﺤﻞ‬/
V4
18
72
C
Q
V,V2
36
72
C
Q
V,4
9
36
C
C
k
F36C
36
1
36
23
18
1
12
1
C
1
C
1
C
1
F12
6
72
V
Q
C2
V2
18
36
C
Q
V,V4
9
36
C
Q
V
C3666V.CQ
F6
)21(9
189
189
189
CC
C.C
C1
2
k
2
k1
k
k1
1
k1
k1
2eqkk1
Tk
k
eqk
2
2
1
1
Teq
21
21
eq
===∆===∆===
µ=⇒=
−
=−=−=
µ==
∆
=−
===∆===∆
µ=×=∆=
µ=
+
×
=
+
×
=
+
=−

42. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
41
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬96/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬)C1=36µF,C2=18µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻮﻟﺘﺎن‬‫ﻣﻮﺻ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬30V.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
2-‫ﻟ‬ ‫ادﺧﻞ‬‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻮح‬)k(‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
‫اﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬720µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫؟‬ ‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫؟‬
‫اﻟﺤﻞ‬/
V10
72
720
C
Q
V,V20
36
720
C
Q
V
4
18
72
C
C
k
F72C
72
1
72
23
36
1
24
1
C
1
C
1
C
1
F24
30
720
V
Q
C2
V20
18
360
C
Q
V,V10
36
360
C
Q
V
C3603012V.CQ
F12
)12(18
1836
1836
1836
CC
C.C
C1
2
k
k2
1
k
1
2
k2
k2
1eqkk2
Tk
k
eqk
2
2
1
1
Teq
21
21
eq
===∆===∆
===
µ=⇒=
−
=−=−=
µ==
∆
=−
===∆===∆
µ=×=∆=
µ=
+
×
=
+
×
=
+
=−
‫ﻣﺜـﺎل‬97/‫ﺖ‬‫رﺑﻄ‬‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬)C1=20µF , C2=5µF(‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻠﺖ‬‫ووﺻ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬20V.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬)‫ﻊ‬‫ﻣ‬‫ﻲ‬‫طﺮﻓ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬(‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫ﻓﺎﺻﺒﺤﺖ‬240µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬)k(‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬.
‫اﻟﺤﻞ‬/
J1064108016
2
1
Q.V
2
1
PE
J101610804
2
1
Q.V
2
1
PE
V16
5
80
C
Q
V,V4
20
80
C
Q
V
C80204V.CQ
F4
25
100
520
520
CC
C.C
C1
56
22
56
11
2
2
1
1
Teq
21
21
eq
−−
−−
×=×××=∆=
×=×××=∆=
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−

43. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
42
V8
30
240
C
Q
V,V12
20
240
C
Q
V
6
5
30
C
C
k
F30C
30
1
60
35
20
1
12
1
C
1
C
1
C
1
F12
20
240
V
Q
C2
k2
k
k2
1
k
1
2
k2
k2
1eqkk2
Tk
k
eqk
===∆===∆
===
µ=⇒=
−
=−=−=
µ==
∆
=−
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬98/‫ﺴﻌﺘﺎن‬ ‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=15µF,C2=30µF(‫ﺮق‬‫ﻓ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬ ‫اﻟ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬24V.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)4(‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬)‫ﺔ‬‫ﺑﺎﻟﻤﺠﻤﻮﻋ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﻣﺎزاﻟﺖ‬ ‫واﻟﺒﻄﺎرﯾﺔ‬(
‫؟‬ ‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
V16
30
480
C
Q
V,V8
60
480
C
Q
V
C4802420V.CQ2
F20
90
1800
3060
3060
CC
C.C
C
F60154kCC2
V16
15
240
C
Q
V,V8
30
240
C
Q
V
C2402410V.CQ
F10
45
450
3015
3015
CC
C.C
C1
2
k
2
k1
k
k1
Tkeqkk
2k1
2k1
eqk
1k1
2
2
1
1
Teq
21
21
eq
===∆===∆
µ=×=∆=−
µ==
+
×
=
+
=
µ=×==−
===∆===∆
µ=×=∆=
µ==
+
×
=
+
=−

44. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
43
‫ﻣﺜـﺎل‬99/‫ر‬‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺖ‬‫ﺑﻄ‬)C1=20µF , C2=5µF(‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻠﺖ‬‫ووﺻ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬20V.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬)‫ﻲ‬‫طﺮﻓ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬(‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫ﻓﺎﺻﺒﺤﺖ‬240µC‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫اﺣﺴﺐ‬)k(‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬.
‫اﻟﺤﻞ‬/
V8
30
240
C
Q
V,V12
20
240
C
Q
V
6
5
30
C
C
k
F30C
30
1
60
2
60
35
C
1
20
1
12
1
C
1
C
1
20
1
12
1
C
1
C
1
C
1
F12
20
240
V
Q
C)2
J10640108016
2
1
Q.V
2
1
PE
J1016010804
2
1
Q.V
2
1
PE
V16
5
80
C
Q
V,V4
20
80
C
Q
V
C80204V.CQ
F4
)14(5
520
520
520
CC
C.C
C)1
k2
k
k2
1
k
1
2
k2
k2
k2
k2k2k21eqk
Tk
k
eqk
66
22
66
11
2
2
1
1
Teq
21
21
eq
===∆===∆
===
µ=⇒==
−
=
−=⇒+=⇒+=
µ==
∆
=
×=×××=∆=
×=×××=∆=
===∆===∆
µ=×=∆=
µ=
+
×
=
+
×
=
+
=
−−
−−
‫ﻣﺜﺎل‬100/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=80µF , C2=20µF(‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ووﺻﻠﺖ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬
‫ﻗ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬‫ﻄﺒﯿﮭﺎ‬10V.
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﺎزل‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)6(‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬)‫ﻲ‬‫طﺮﻓ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻣﺮﺑﻮط‬ ‫ﺔ‬‫اﻟﺒﻄﺎرﯾ‬ ‫ﺎء‬‫ﺑﻘ‬ ‫ﻊ‬‫ﻣ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬(‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬.
‫اﻟﺤﻞ‬/
V8
20
160
C
Q
V,V2
80
160
C
Q
V
C1601016V.CQ
F16
5
80
)14(20
2080
2080
2080
CC
C.C
C)1
2
2
1
1
Teq
21
21
eq
===∆===∆
µ=×=∆=
µ==
+
×
=
+
×
=
+
=

45. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
44
V4
120
480
C
Q
V,V6
80
480
C
Q
V
C4801048V.CQ
F48
5
240
)32(40
12080
12080
12080
CC
C.C
C
C120206CkC)2
k2
k
k2
1
k
1
Teqkk
k21
k21
eqk
2k2
===∆===∆
µ=×=∆=
µ==
+
×
=
+
×
=
+
=
µ=×==
‫ﻣﺜﺎل‬101/‫اﻟﻤﺘﺴﻌﺘﺎن‬)C1=20µF , C2=60µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻣﻮﺻﻮﻟﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬20V.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
2-‫ﮫ‬ ‫ﻋﺰﻟ‬ ‫ﺖ‬ ‫ﺛﺎﺑ‬ ‫ﻰ‬ ‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺎ‬ ‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺎزل‬ ‫ﻋ‬ ‫ﻮح‬ ‫ﻟ‬ ‫ﻞ‬ ‫ادﺧ‬ ‫اذا‬)k(‫ﻰ‬ ‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬
‫ﻣﺘﺼﻠﺔ‬ ‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺠﻤﻮﻋﺔ‬‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺷﺤﻨﺔ‬ ‫اﺻﺒﺤﺖ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬600µC‫اﻟﻜﮭﺮﺑﺎﺋﻲ؟‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
3
20
60
C
C
k
F60C
60
1
60
12
60
1
30
1
C
1
C
1
C
1
F30
20
600
V
Q
C
V20VV)2
V5
60
300
C
Q
V,V15
20
300
C
Q
V
C3002015V.CQ
F15
4
60
)31(20
6020
6020
6020
CC
C.C
C)1
1
k1
k1
2eqkk1
T
k
eqk
TTk
2
2
1
1
Teq
21
21
eq
===∴
µ=⇒=
−
=−=−=
µ==
∆
=
=∆=∆
===∆===∆
µ=×=∆=
µ==
+
×
=
+
×
=
+
=
‫ﻣﺜــﺎل‬102/‫ﻰ‬ ‫اﻻوﻟ‬ ‫ﻌﺔ‬ ‫ﺳ‬ ‫ﺔ‬ ‫اﻟﻤﺘﻮازﯾ‬ ‫ﺼﻔﺎﺋﺢ‬ ‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﺎن‬ ‫ﻣﺘﻮاﻟﯿﺘ‬ ‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬6μF‫ﺔ‬ ‫اﻟﺜﺎﻧﯿ‬ ‫ﻌﺔ‬ ‫وﺳ‬3μF‫ﺖ‬ ‫رﺑﻄ‬
‫ﺪه‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺼﺪر‬‫ﻣ‬ ‫إﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬24V‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎزﻻ‬‫ﻋ‬ ‫ﺮاغ‬‫اﻟﻔ‬ ‫ﺎن‬‫وﻛ‬‫ﺴﻌﺔ‬.‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫ﻢ‬‫ﺛ‬‫ﻦ‬‫ﻋ‬
‫اﻟﻤﺼﺪر‬‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬ ‫وادﺧﻞ‬2‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬:
1-‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﻗﺒﻞ‬2-‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌﺪ‬
‫اﻟﺤﻞ‬/
1-‫اﻟﻌﺎزل‬ ‫ﻗﺒﻞ‬:
2
1
6
3
6
21
3
1
6
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq=2µF
QT =Ceq . ∆VT =2 × 24 =48µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =48µC
V8
6
48
C
Q
V
1
1
1
===∆ , V16
3
48
C
Q
V
2
2
2
===∆
2-‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
C1k =k C1 =2 × 6=12µF , C2k=k C2 =2 × 3=6µF

46. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
45
4
1
12
3
12
21
6
1
12
1
C
1
C
1
C
1
k2k1eqk
==
+
=+=+= ⇒ Ceqk=4µF
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =48µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QTk = Q1k = Q2k =48µC
V4
12
48
C
Q
V
k1
k1
k1
===∆ , V8
6
48
C
Q
V
k2
k2
k2
===∆
‫ﻣﺜﺎل‬103/‫اﻟﻣﺗوازﯾﺗﯾن‬ ‫اﻟﺻﻔﯾﺣﺗﯾن‬ ‫ذوات‬ ‫ﻣن‬ ‫ﻣﺗﺳﻌﺗﺎن‬(C1=9µF , C2=18µF)‫ﻣرﺑوطﺗﺎ‬‫ﻰ‬‫ﻋﻠ‬ ‫ﺑﻌﺿﮭﻣﺎ‬ ‫ﻣﻊ‬ ‫ن‬
‫ﻗطﺑﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﺑطﺎرﯾﺔ‬ ‫ﻗطﺑﻲ‬ ‫ﺑﯾن‬ ‫رﺑطت‬ ‫وﻣﺟﻣوﻋﺗﮭﻣﺎ‬ ‫اﻟﺗواﻟﻲ‬)6V(‫ﺎ‬‫ﻛﮭرﺑﺎﺋﯾ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ن‬‫ﻣ‬ ‫وح‬‫ﻟ‬ ‫ل‬‫ادﺧ‬ ،
‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬ ‫ﻣﻧﮭﻣﺎ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬2‫ﻣﺎزاﻟت‬ ‫واﻟﻣﺟﻣوﻋﺔ‬‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫ﻣﻘدر‬ ‫ﻓﻣﺎ‬ ‫ﺑﺎﻟﺑطﺎرﯾﺔ‬ ‫ﻣﺗﺻﻠﺔ‬:
1-‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬2-‫اﻟﻌﺎزل‬ ‫ﺑﻌد‬.
‫اﻟﺤﻞ‬/
‫ﻗﺑل‬‫اﻟﻌﺎزل‬:
6
1
18
3
18
12
18
1
9
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq=6µF
QT =Ceq . ∆VT =6 × 6=36µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =36µC
V4
9
36
C
Q
V
1
1
1
===∆ , V2
18
36
C
Q
V
2
2
2
===∆
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
C1k=kC1 =2 × 9=18µF , C2k=kC2 =2 × 18 = 36µF
12
1
36
3
36
12
36
1
18
1
C
1
C
1
C
1
k2k1eqk
==
+
=+=+= ⇒ Ceqk=12µF
Q‫ﻟﺬﻟﻚ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬∆VTk=∆VT =6V
QTk =Ceqk . ∆VTk =12 × 6 =72µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QTk = Q1k = Q2k =72µC
V4
18
72
C
Q
V
k1
k1
k1
===∆ , V2
36
72
C
Q
V
k2
k2
k2
===∆
‫ﻣﺜﺎل‬104/‫ﻮازﯾﺘﯿﻦ‬‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬‫ﻣ‬ ‫ﻣﺘﺴﻌﺘﺎن‬)C1=6µF,C2=12µF(‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻗﻄﺒﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤﺎ‬ ‫رﺑﻄﺖ‬ ، ‫اﻟﺘﻮاﻟﻲ‬)60V(‫ﻞ‬‫ﻛ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزﻻ‬ ‫اﻟﻔﺮاغ‬ ‫وﻛﺎن‬
‫ﺎ‬ ‫ﻣﻨﮭﻤ‬.‫ا‬‫ﮫ‬ ‫ﻋﺰﻟ‬ ‫ﺖ‬ ‫ﺛﺎﺑ‬ ‫ﺔ‬ ‫ﻋﺎزﻟ‬ ‫ﺎدة‬ ‫ﻣ‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﻟﻮﺣ‬ ‫ﺎ‬ ‫ﻣﻨﮭﻤ‬ ‫ﻞ‬ ‫ﻛ‬ ‫ﻔﯿﺤﺘﻲ‬ ‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﻞ‬ ‫دﺧ‬)2(‫ﺎ‬ ‫ﺑﯿﻨﮭﻤ‬ ‫ﺰ‬ ‫اﻟﺤﯿ‬ ‫ﻶ‬ ‫ﯾﻤ‬
)‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﻣﺎزاﻟﺖ‬(‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
)1(‫ﻗﺒﻞ‬‫اﻟﻌﺎزل‬)2(‫ﺑﻌﺪ‬‫اﻟﻌﺎزل‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬:
4
1
12
3
12
12
12
1
6
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq=4µF
QT =Ceq . ∆VT =4 × 60=240µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =240µC
V40
6
240
C
Q
V
1
1
1
===∆ , V20
12
240
C
Q
V
2
2
2
===∆

47. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
46
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
C1k=kC1 =2 × 6=12µF , C2k=kC2 =2 × 12 = 24µF
8
1
24
3
24
12
24
1
12
1
C
1
C
1
C
1
k2k1eqk
==
+
=+=+= ⇒ Ceqk=8µF
Q‫ﻟﺬﻟﻚ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬∆VTk=∆VT =60V
QTk =Ceqk . ∆VTk =8 × 60 =480µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QTk = Q1k = Q2k =480µC
V40
12
480
C
Q
V
k1
k1
k1
===∆ , V20
24
480
C
Q
V
k2
k2
k2
===∆
‫ﻣﺜﺎل‬105/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬)C1=10µF , C2=40µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬15V.‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻛﮭﺮﺑﺎﺋﯿﺎ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)6(‫ﻊ‬‫ﻣ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬
‫اﻟﺠ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺑﻘﺎء‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﮭﺪ‬
1-‫اﻟﻌﺎزل‬ ‫ﻗﺒﻞ‬.2-‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬.
‫اﻟﺤﻞ‬/
V9
40
360
C
Q
V,V6
60
360
C
Q
V
C3601524V.CQ
F24
5
120
)23(20
4060
4060
4060
CC
C.C
C
C60106CkC)2
V3
40
120
C
Q
V,V12
10
120
C
Q
V
C120158V.CQ
F8
5
40
)41(10
4010
4010
4010
CC
C.C
C)1
2
k
k2
k1
k
1
Teqkk
2k1
2k1
eqk
1k1
2
2
1
1
Teq
21
21
eq
===∆===∆
µ=×=∆=
µ==
+
×
=
+
×
=
+
=
µ=×==
===∆===∆
µ=×=∆=
µ==
+
×
=
+
×
=
+
=

48. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
47
‫ــﺎل‬‫ـ‬‫ﻣﺜ‬106/‫ﻌﺎﺗﮭﺎ‬ ‫ﺳ‬ ‫ﻮازﯾﺘﯿﻦ‬ ‫اﻟﻤﺘ‬ ‫ﺼﻔﯿﺤﺘﯿﻦ‬ ‫اﻟ‬ ‫ذوات‬ ‫ﻦ‬ ‫ﻣ‬ ‫ﺴﻌﺎت‬ ‫ﻣﺘ‬ ‫ﻼث‬ ‫ﺛ‬)C1=6µF,C2=9µF,C3=18µF(
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫رﺑﻄﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫ﺑﻌﻀﮭﺎ‬ ‫ﻣﻊ‬ ‫رﺑﻄﺖ‬12V.
1-‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫ﺑﺎﻟﺒﻄﺎرﯾ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﺖ‬‫ﻣﺎزاﻟ‬ ‫ﺔ‬‫واﻟﻤﺠﻤﻮﻋ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬
‫اﻟﻰ‬ ‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎزدادت‬48µC‫ا‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬‫ﻟﻜﮭﺮﺑﺎﺋﻲ‬)k.(
‫اﻟﺤﻞ‬/
2
6
12
C
C
k
F12C
12
1
36
3
36
249
18
1
9
1
4
1
C
1
C
1
C
1
C
1
F4
12
48
V
Q
C2
J103610362
2
1
Q.V
2
1
PE
J107210364
2
1
Q.V
2
1
PE
J1010810366
2
1
Q.V
2
1
PE
V2
18
36
C
Q
V,V4
9
36
C
Q
V,V6
6
36
C
Q
V
F36123V.CQ
F3C
3
1
18
6
18
123
18
1
9
1
6
1
C
1
C
1
C
1
C
1
1
1
k1
k1
32eqkk1
Tk
k
eqk
66
33
66
22
66
11
32
2
1
1
Teq
eq
321eq
===
µ=⇒==
−−
=−−=−−=
µ==
∆
=−
×=×××=∆=
×=×××=∆=
×=×××=∆=
===∆===∆===∆
µ=×=∆=
µ=⇒==
++
=++=++=−
−−
−−
−−
‫ﻣﺜﺎل‬107/‫اﻻوﻟﻰ‬ ‫ﻣﺗﺳﻌﺗﺎن‬C1‫واﻟﺛﺎﻧﯾﺔ‬18µF‫د‬‫اﻟﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺔ‬‫ﺑطﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫وﻟﺗﺎن‬‫وﻣوﺻ‬ ‫واﻟﻲ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎن‬‫ﻣرﺑوطﺗ‬
‫ﻗطﺑﯾﮭﺎ‬ ‫ﺑﯾن‬12V‫ﺑﻣﻘدار‬ ‫ﺳﻌﺗﮭﺎ‬ ‫ﻓﺎزدادت‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬ ‫ﻋﺎزل‬ ‫وﺿﻊ‬27µF‫ﺔ‬‫اﻟﻛﻠﯾ‬ ‫ﺷﺣﻧﺔ‬‫اﻟ‬ ‫واﺻﺑﺣت‬
‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌد‬144µC‫ﺟد‬:
1-‫اﻻوﻟﻰ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺳﻌﺔ‬)C1(‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬2-‫اﻟﻌزل‬ ‫ﺛﺎﺑت‬k3-‫ﺟﮭد‬ ‫ﻓرق‬‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬.
‫اﻟﺤﻞ‬/
Q‫ﻟذﻟك‬ ‫ﺑﺎﻟﺑطﺎرﯾﺔ‬ ‫ﻣﺗﺻﻠﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬∆VTk=∆VT=12V
F12
12
144
V
Q
C
Tk
Tk
eqk
µ==
∆
=
2k1eqk
C
1
C
1
C
1
+= ⇒
18
1
C
1
12
1
k1
+= ⇒
36
1
36
23
18
1
12
1
C
1
k1
=
−
=−=
∴ C1k=36µF
C1k = C1 + ∆C ⇒ 36 =C1 + 27 ⇒ C1 =36 – 27 =9µF
C1k=kC1 ⇒ 36=k × 9 ⇒ 4
9
36
k ==
V8
18
144
C
Q
V,V4
36
144
C
Q
V
2
k2
2
k1
k1
k1 ===∆===∆

49. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
48
‫ﻣﺜﺎل‬108/‫اﻻوﻟﻰ‬ ‫ﻣﺘﺴﻌﺘﺎن‬6µF‫واﻟﺜﺎﻧﯿﺔ‬C2‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫وﻣﻮﺻﻮﻟﺘﺎن‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫ﻣﺮﺑﻮطﺘﺎن‬
‫ﻗﻄﺒﯿﮭﺎ‬24V‫ﺛﻢ‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزل‬ ‫وﺿﻊ‬)k(‫ﺪار‬‫ﺑﻤﻘ‬ ‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺎزدادت‬‫ﻓ‬9µF‫ﺒﺤﺖ‬‫واﺻ‬
‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌﺪ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬96µC‫ﺟﺪ‬:
1-‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬C2‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﻗﺒﻞ‬.
2-‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬)k(
3-‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
C
Q
V
C216249V.CQ
F936CCC
‫ﻞ‬‫ﺒ‬‫ﻗ‬‫اﻟﻌﺎزل‬)3
F4
3
12
C
C
k)2
F3912C9C129CC
F12C24C224C4C6
C424C6
C6
C6
4
CC
C.C
C
F4
24
96
V
Q
C)1
1
Teq
21eq
2
k2
222k2
k2k2k2k2
k2k2
k2
k2
k21
k21
eqk
T
k
eq
=∆
µ=×=∆=
µ=+=+=
µ===
µ=−=⇒+=⇒+=
µ=⇒=⇒=−
+=⇒
+
=⇒
+
=
µ==
∆
=
‫ﻣﺴﺒﻘ‬ ‫ﻣﺸﺤﻮﻧﺘﯿﻦ‬ ‫او‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫وﻏﯿﺮ‬ ‫ﻣﺸﺤﻮﻧﺔ‬‫ﺎ‬:
‫ﻣﺜﺎل‬109/‫ﺳﻌﺗﮭﺎ‬ ‫ﻣﺗﺳﻌﺔ‬)4µF(‫وﺷﺣﻧﺗﮭﺎ‬)300µC(‫ﺷﺣوﻧﺔ‬‫ﻣ‬ ‫ر‬‫ﻏﯾ‬ ‫رى‬‫اﺧ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ﻊ‬‫ﻣ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠت‬‫وﺻ‬
‫ﺳﻌﺗﮭﺎ‬)2µF. (‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫وﺷﺣﻧﺔ‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫اﺣﺳب‬.
‫اﻟﺤﻞ‬/
QT =Q1 + Q2 =300 + 0 =300µC
Ceq =C1 + C2 =4 + 2 =6µF
V50
6
300
C
Q
V
eq
T
T ===∆
Q‫اﻟرﺑط‬‫ﻟذﻟك‬ ‫ﺗوازي‬∆VT = ∆V1 = ∆V2 =50V
Q1 =C1 . ∆V1 =4 × 50 =200µC , Q2 =C2 . ∆V2 =2 × 50 =100µC
‫ـﺎ‬‫ـ‬‫ﻣﺜ‬‫ل‬110/‫ﻌﺘﮭﺎ‬ ‫ﺳ‬ ‫ﺸﺤﻮﻧﺔ‬ ‫ﻣ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬6μF‫ﺪھﺎ‬ ‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬30V‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﺄﺻﺒﺢ‬ ‫ﻣﺸﺤﻮﻧﺔ‬20V‫وﻣﺎ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺎ‬‫؟‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1 =C1 . ∆V1 =6 × 30 =180µC , Q2 =0 (‫ﻣﺸﺤﻮﻧﺔ‬ ‫)ﻏﯿﺮ‬
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =180 + 0 =180µC
F9
20
180
V
Q
C
T
T
eq
µ==
∆
=

50. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
49
Ceq =C1 + C2 ⇒ 9 =6 + C2 ⇒ C2 =9 – 6 =3µF
Q1 =C1 . ∆V1 =6 × 20 =120µC , Q2 =C2 × ∆V2 =3 × 20=60µC
‫ﻣﺜﺎل‬111/‫ﺳﻌﺗﮭﺎ‬ ‫ﻣﺗﺳﻌﺔ‬)4µF(‫دھﺎ‬‫ﺟﮭ‬ ‫وﻓرق‬)200V(‫ﺷﺣوﻧﺔ‬‫ﻣ‬ ‫ر‬‫ﻏﯾ‬ ‫رى‬‫اﺧ‬ ‫ﺳﻌﺔ‬‫ﺑﻣﺗ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠت‬‫وﺻ‬
‫اﻟﻣﺟﻣوﻋﺔ‬ ‫طرﻓﻲ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﻓﺎﺻﺑﺢ‬)80V(‫ﻣﺗﺳ‬ ‫ﻛل‬ ‫ﺷﺣﻧﺔ‬ ‫وﻣﺎ‬ ‫؟‬ ‫اﻟﻣﺟﮭوﻟﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻓﻣﺎ‬‫وﺻﻠﮭﻣﺎ؟‬ ‫ﺑﻌد‬ ‫ﻌﺔ‬
‫اﻟﺤﻞ‬/
‫اﻟﺗوﺻﯾل‬ ‫ﻗﺑل‬:
Q1 =C1 . ∆V1 =4 × 200 =800µC
Q2 =0 (‫ﻣﺷﺣوﻧﺔ‬ ‫)ﻏﯾر‬
‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
QT =Q1 + Q2 =800 + 0 = 800µC
F10
80
800
V
Q
C
T
T
eq
µ==
∆
=
Ceq =C1 + C2 ⇒ 10 =4 + C2 ⇒ C2 =10 – 4 =6µF
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VT = ∆V1 = ∆V2 = 80V
Q1 =C1 . ∆V1 =4 × 80 =320µC , Q2 =C2 . ∆V2 =6 × 80 = 480µC
‫ﻣﺜﺎل‬112/‫اﻻوﻟﻰ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺗﺳﻌﺗﺎن‬)3µF(‫دھﺎ‬‫ﺟﮭ‬ ‫رق‬‫وﻓ‬)60V(‫ﻌﺗﮭﺎ‬‫ﺳ‬ ‫ﺔ‬‫واﻟﺛﺎﻧﯾ‬)5µF(‫دھﺎ‬‫ﺟﮭ‬ ‫رق‬‫وﻓ‬)28V(
‫و‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫ﻓﻣﺎ‬ ‫اﻟﺗوازي‬ ‫ﻋﻠﻰ‬ ‫رﺑطﺗﺎ‬‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬ ‫ﺷﺣﻧﺔ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺗوﺻﯾل‬ ‫ﻗﺑل‬:
Q1 =C1 . ∆V1 =3 × 60 =180µC , Q2 =C2 . ∆V2 =5 × 28 =140µC
‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
QT =Q1 + Q2 =180 + 140 =320µC
Ceq =C1 + C2 =3 + 5=8µF
V40
8
320
C
Q
V
eq
T
T ===∆
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VT = ∆V1 = ∆V2 = 40V
Q1 =C1 . ∆V1 =3 × 40 =120µC , Q2 =C2 . ∆V2 =5 × 40 = 200µC
‫ﻣﺜﺎل‬113/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻣﺘﺴﻌﺔ‬100μF‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬50V‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬
‫ﯿﻞ‬ ‫اﻟﺘﻮﺻ‬ ‫ﺪ‬ ‫ﺑﻌ‬ ‫ﺔ‬ ‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬ ‫ﺟﮭ‬ ‫ﺮق‬ ‫ﻓ‬ ‫ﺒﺢ‬ ‫ﻓﺄﺻ‬ ‫ﺸﺤﻮﻧﺔ‬ ‫ﻣ‬20V.‫ﺎ‬ ‫ﻣﻨﮭﻤ‬ ‫ﻞ‬ ‫ﻛ‬ ‫ﺤﻨﺔ‬ ‫ﺷ‬ ‫ﺎ‬ ‫وﻣ‬ ‫ﺔ‬ ‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻌﺔ‬ ‫ﺳ‬ ‫ﺎ‬ ‫ﻣ‬‫ﺪ‬ ‫ﺑﻌ‬
‫اﻟﺘﻮﺻﯿﻞ؟‬
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1 =C1 . ∆V1 =100 × 50 =5000µC
Q2 = 0 (‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬ ‫)ﻻﻧﮭﺎ‬
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =5000 + 0 =5000µC
C250
20
5000
V
Q
C
T
T
eq
µ==
∆
=
Ceq =C1 + C2 ⇒ 250 =100 + C2 ⇒ C2 =250 – 100 =150µF
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =20V
Q1 = C1 . ∆V1 =100 × 20 =2000µC , Q2 = C2 . ∆V2 = 150 × 20 = 3000µC

51. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
50
‫ﻣﺜــﺎل‬114/‫ﻌﺘﮭﺎ‬ ‫ﺳ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬2μF‫و‬‫ﺮق‬ ‫ﻓ‬‫ﺪھﺎ‬ ‫ﺟﮭ‬30V‫ﻌﺘﮭﺎ‬ ‫ﺳ‬ ‫ﺮى‬ ‫وأﺧ‬3μF‫و‬‫ﺮق‬ ‫ﻓ‬‫ﺪھﺎ‬ ‫ﺟﮭ‬40V‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻠﺘﺎ‬ ‫وﺻ‬
‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬‫ﻓﻤﺎ‬ ‫ﺑﻌﻀﮭﻤﺎ‬‫ﻓﺮق‬‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1 =C1 . ∆V1 =2 × 30 =60µC , Q2 =C2 . ∆V2 =3 × 40 =120µC
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 = 60 + 120 =180µC
Ceq =C1 + C2 =2 + 3 =5µF
V36
5
180
C
Q
V
eq
T
T ===∆
Q1 =C1 . ∆V1 =2 × 36 =72µC , Q2 =C2 . ∆V2 =3 × 36 =108µC
‫ﻣﺜﺎل‬115/‫ﻣﺘﺴﻌﺔ‬‫ﺳﻌﺘﮭﺎ‬1µF‫ﻣﻘﺪارھﺎ‬ ‫ﺑﺸﺤﻨﺔ‬ ‫ﻣﺸﺤﻮﻧﺔ‬400µF‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫اﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫وﺻﻠﺖ‬
‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬4µF‫اﺣﺴﺐ‬:
1-‫اﻟﺸ‬ ‫ﻣﻘﺪار‬‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫ﺤﻨﺔ‬‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬.
2-‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬.
‫اﻟﺤﻞ‬/
1- QT =Q1 + Q2 =400 + 0=400µC
Ceq =C1 + C2 =1 + 4=5µF
V80
5
400
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =80V
Q1=C1 . ∆V1 =1 × 80 =80µC , Q2 =C2 . ∆V2 =4 × 80=320µC
2- J1032108080
2
1
Q.V
2
1
PE 46
111
−−
×=×××=∆=
J101281032080
2
1
Q.V
2
1
PE 46
221
−−
×=×××=∆=
‫ﻣﺜــﺎل‬116/‫ﻌﺘﮭﺎ‬ ‫ﺳ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬2µF‫ﺤﻨﺘﮭﺎ‬ ‫وﺷ‬800µC‫ﺎ‬ ‫ﻓﯿﮭ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬ ‫اﻟﻄﺎﻗ‬ ‫ﺪار‬ ‫ﻣﻘ‬ ‫ﺴﺐ‬ ‫اﺣ‬.‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻠﺖ‬ ‫وﺻ‬ ‫واذا‬
‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬ ‫اﺧﺮى‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬6µF‫ﺔ‬‫اﻟﻄﺎﻗ‬ ‫ﻲ‬‫ﻓ‬ ‫ﻨﻘﺺ‬‫اﻟ‬ ‫ﺪار‬‫وﻣﻘ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬
‫اﻟﻤﺨﺘﺰﻧﺔ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
J1016
104
10640000
1022
)10800(
C
Q
2
1
PE 2
6
12
6
26
1
2
1
1
−
−
−
−
−
×=
×
×
=
××
×
==
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =800 + 0=800µC
Ceq =C1 + C2 =2 + 6=8µF
V100
8
800
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT =∆V1 =∆V2=100V
Q1=C1 . ∆V1=2 × 100 =200µC , Q2=C2 . ∆V2 =6 × 100 =600µC

52. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
51
J10410800100
2
1
Q.V
2
1
PE 26
TTT
−−
×=×××=∆=
∆PE =PET – PE1=4×10-2
– 16×10-2
= F1012 2
µ×− −
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬117/‫ﺳﻌﺗﺎن‬ ‫ﻣﺗ‬)C1=2µF , C2=8µF(‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎ‬‫رﺑطﺗ‬ ‫ﺷﺣوﻧﺔ‬ ‫ﻣ‬ ‫ر‬‫ﻏﯾ‬ ‫ﺔ‬ ‫واﻟﺛﺎﻧﯾ‬ ‫ﺷﺣوﻧﺔ‬‫ﻣ‬ ‫ﻰ‬ ‫اﻻوﻟ‬ ‫ت‬‫وﻛﺎﻧ‬
‫اﻟﻣﺟﻣوﻋﺔ‬ ‫طرﻓﻲ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﻓﻛﺎن‬ ‫ﺑﻌﺿﮭﻣﺎ‬ ‫ﻣﻊ‬ ‫اﻟﺗوازي‬)20V. (‫ﻰ‬‫اﻻوﻟ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﺟﮭد‬ ‫وﻓرق‬ ‫ﺷﺣﻧﺔ‬ ‫ﻣﻘدار‬ ‫ﺟد‬
‫ﺗوﺻل‬ ‫ان‬ ‫ﻗﺑل‬‫اﻟﺛﺎﻧﯾﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﻣﻊ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
Ceq =C1 + C2 =2 + 8=10µF
QT =Ceq . ∆VT =10 × 20 =200µC
‫اﻟﺗوﺻﯾل‬ ‫ﻗﺑل‬:
QT =Q1 + Q2 ⇒ 200 =Q1 + 0 ⇒ Q1 =200µC
V100
2
200
C
Q
V
1
1
1
===∆
‫ﻣﺜـﺎل‬118/‫ﺳﻌﺗ‬‫اﻟﻣﺗ‬ ‫ت‬‫رﺑط‬‫ﺎن‬)C1=3µF,C2=6µF(‫ﻊ‬‫ﻣ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫ﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ت‬‫رﺑط‬ ‫م‬‫ﺛ‬ ‫واﻟﻲ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﺳﻌﺗﮭﺎ‬ ‫ﻣﻘدار‬ ‫ﻣﺷﺣوﻧﺔ‬ ‫ﺛﺎﻟﺛﺔ‬ ‫ﻣﺗﺳﻌﺔ‬4µF‫ﺷﺣﻧﺗﮭﺎ‬ ‫وﻣﻘدار‬180µC‫ل‬‫ﻛ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ن‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬ ‫اﺣﺳب‬
‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬ ‫ﻣﺗﺳﻌﺔ‬.
‫اﻟﺤﻞ‬/
C120304V.CQ,QQC60302V.CQ
V30
6
180
C
Q
V
F642CCC
F2
9
18
63
63
CC
C.C
C
C1800180QQQ
33212,12,1
eq
T
32,1eq
21
21
2,1
2,13T
µ=×=∆===µ=×=∆=
===∆
µ=+=+=
µ==
+
×
=
+
=
µ=+=+=
‫ﻋﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫وﻏﯿﺮ‬ ‫ﻣﺸﺤﻮﻧﺔ‬:
‫ﻣﺜﺎل‬119/‫ﻌ‬‫ﺳ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬‫ﺗﮭﺎ‬2µF‫ﻔﯾﺣﺗﯾﮭﺎ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫د‬‫اﻟﺟﮭ‬ ‫رق‬‫وﻓ‬100V‫ﻌﺗﮭﺎ‬‫ﺳ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ﻊ‬‫ﻣ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠت‬‫وﺻ‬
2µF‫ﻣﺷﺣوﻧﺔ‬ ‫ﻏﯾر‬.‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻊ‬‫وﺿ‬ ‫واذا‬ ‫؟‬ ‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﻣن‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻟﻛل‬ ‫اﺣﺳب‬
‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬ ‫اﻟﮭواء‬ ‫ﻣن‬ ‫ﺑدﻻ‬ ‫اﻟﺛﺎﻧﯾﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬3‫ﻣﺗﺳﻌ‬ ‫ﻛل‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫اﻟﺷﺣﻧﺔ‬ ‫ﺳﺗﺻﺑﺢ‬ ‫ﻓﻛم‬‫ﺔ؟‬
‫اﻟﺤﻞ‬/
‫اﻟﺗوﺻﯾل‬ ‫وﻗﺑل‬ ‫اﻟﻌﺎزل‬ ‫ﻗﺑل‬:
Q1 =C1 . ∆V1 =2 × 100=200µC
Q2 = 0 (‫ﻣﺷﺣوﻧﺔ‬ ‫)ﻏﯾر‬
‫اﻟﻌﺎزل‬ ‫وﻗﺑل‬ ‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
QT =Q1 + Q2 =200 + 0 =200µC
Ceq =C1 + C2 =2 + 2=4µF
V50
4
200
C
Q
V
eq
T
T ===∆
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VT = ∆V1 = ∆V2 =50V
Q1 =C1 . ∆V=2 × 50 =100µC , Q2 =C2 . ∆V2 =2 × 50 =100µC

53. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
52
‫اﻟﻌﺎزل‬ ‫وﺑﻌد‬ ‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬
C2k =k C2 =3 × 2=6µF
Ceqk =C1 + C2k =2 + 6=8µF
Q‫ﻟذﻟك‬ ‫اﻟﺷﺎﺣن‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫ﻣﻧﻔﺻﻠﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬QTk = QT =200µC
V25
8
200
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VTk = ∆V1 = ∆V2k =25V
Q1 =C1 . ∆V1 =2 × 25 =50µC , QTk =C2k . ∆V2k =6 × 25 = 150µC
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬120/‫ﻌﺗﮭﺎ‬‫ﺳ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬8µF‫ﻔﯾﺣﺗﯾﮭﺎ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫د‬‫اﻟﺟﮭ‬ ‫رق‬‫وﻓ‬30V‫رى‬‫اﺧ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ﻊ‬‫ﻣ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠت‬‫وﺻ‬
‫ﺳﻌﺗﮭﺎ‬4µF‫ﻣﺷﺣوﻧﺔ‬ ‫ﻏﯾر‬.‫ﯾل‬‫اﻟﺗوﺻ‬ ‫د‬‫ﺑﻌ‬ ‫ﺳﻌﺔ‬‫ﻣﺗ‬ ‫ل‬‫ﻛ‬ ‫ﺟﮭد‬ ‫ﻓرق‬ ‫اﺣﺳب‬‫ﻣ‬ ‫وح‬‫ﻟ‬ ‫ل‬‫ادﺧ‬ ‫واذا‬ ،‫ﺎ‬‫ﻛﮭرﺑﺎﺋﯾ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ن‬
‫ﻋزﻟﮫ‬ ‫ﺛﺎﺑت‬)k(‫ﻰ‬‫اﻟ‬ ‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫د‬‫ﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺑط‬‫ھ‬ ‫ﺔ‬‫اﻟﺛﺎﻧﯾ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﺑﯾن‬12V‫ﺎﺋﻲ‬‫اﻟﻛﮭرﺑ‬ ‫زل‬‫اﻟﻌ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺳب‬‫اﺣ‬
)k(‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﻣن‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻣﺧﺗزﻧﺔ‬ ‫واﻟﺷﺣﻧﺔ‬‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وﻗﺑل‬ ‫اﻟﺗوﺻﯾل‬ ‫ﻗﺑل‬:
Q1 =C1 . ∆V1 =8 × 30 =240µC , Q2 =0 (‫ﻣﺷﺣوﻧﺔ‬ ‫)ﻏﯾر‬
QT =Q1 + Q2 =240 + 0 =240µC
Ceq =C1 + C2 =8 + 4 =12µF
V20
12
240
C
Q
V
eq
T
T ===∆
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VT = ∆V1 = ∆V2 =20V
Q1 =C1 . ∆V1 =8 × 20 =160µC , Q2 =C2 . ∆V2 =4 × 20 =80µC
‫اﻟﻌﺎزل‬ ‫وﺑﻌد‬ ‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
Q‫ﻟذﻟك‬ ‫اﻟﺷﺎﺣن‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫ﻣﻔﺻوﻟﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬QTk = QT =240µC
F20
12
240
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1 + C2k ⇒ 20 =8 + C2k ⇒ C2k =20 – 8 =12µF
C2k = kC2 ⇒ 12 =k × 4 ⇒ k=3
Q‫ﻟذﻟك‬ ‫ﺗوازي‬ ‫اﻟرﺑط‬∆VTk = ∆V1 = ∆V2k =12V
Q1=C1 × ∆V1 =8 × 12 =96µC , Q2k =C2k × ∆V2k =12 × 12 =144µC
‫ﻣﺜﺎل‬121/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬2µF‫ﺟﮭﺪ‬ ‫ﻟﻔﺮق‬ ‫ﻣﺸﺤﻮﻧﺔ‬50V‫ﻮ‬‫اﻟﺘ‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺖ‬‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺮى‬‫اﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ازي‬8µF
‫ﯿﻞ‬‫اﻟﺘﻮﺻ‬ ‫ﺪ‬‫ﺑﻌ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫ﺤﻨﺔ‬‫وﺷ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻣﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬.‫ﺎدة‬‫ﻣ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﻊ‬‫وﺿ‬ ‫واذا‬
‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬6‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﻤﺎ‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وﻗﺒﻞ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1=C1 . ∆V1 =2 × 50 =100µC
Q2 = 0 ( ‫ﻣﺸﺤ‬ ‫ﻏﯿﺮ‬‫ﻮﻧﺔ‬ )
‫اﻟﻌﺎزل‬ ‫وﻗﺒﻞ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =100 + 0=100µC
Ceq =C1 + C2 =2 + 8 =10µF
V10
10
100
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =10V

54. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
53
Q1 =C1 . ∆V1 =2 × 10 =20µC , Q2 =C2 × ∆V2 =8 × 10 =80µC
‫وﻗ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬‫اﻟﻌﺎزل‬ ‫ﺒﻞ‬:
C1k =kC1 =6 × 2=12µF
Ceqk=C1k + C2 =12 + 8=20µF
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =100µC
V5
20
100
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk = ∆V1k = ∆V2=5V
Q1k =C1k . ∆V1k =12 × 5=60µC , Q2 =C2 . ∆V2 =8 × 5 =40µC
‫ﻣﺜﺎل‬122/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬2µF‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬30V‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺖ‬3µF‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬.
)1(‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫اﺣﺴﺐ‬.
)2(‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬6‫اﻟﮭﻮ‬ ‫ﺑﺪل‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬‫ﺤﻨﺔ‬‫وﺷ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬ ‫اء‬
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬.
‫اﻟﺤﻞ‬/
)1(‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1=C1 . ∆V1=2 × 30=60µC
Q2=0
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2=60 + 0=60µC
Ceq=C1 + C2 =2 + 3=5µF
F12
5
60
C
Q
V
eq
T
T µ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT =∆V1 =∆V2=12V
Q1=C1 . ∆V1=2 × 12=24µC , Q2=3 × 12=36µC
)2(‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
C2k=kC2 =6 × 3=18µF
Ceqk=C1 + C2k =2 + 18=20µF
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk =QT =60µC
V3
20
60
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk =∆V1 =∆V2k=3V
Q1=C1 . ∆V1=2 × 3=6µC , Q2k=C2k . ∆V2k=18 × 3=54µC
‫ﻣﺜﺎل‬123/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺮ‬‫ﻏﯿ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬2μF‫ﻰ‬‫ﻋﻠ‬ ‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﺪل‬‫ﺑ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎ‬‫ﻋﺎزﻟﮭ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻌﺖ‬‫وﺿ‬
‫ﺷﺤﻨﺘﮭﺎ‬ ‫أﺧﺮى‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬180μC‫ﻌﺘﮭﺎ‬‫وﺳ‬3μF‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺒﺢ‬‫ﻓﺄﺻ‬12V‫ﻞ‬‫ﻛ‬ ‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﺎ‬‫ﻓﻤ‬
‫ﻣﺘﺴﻌﺔ‬‫؟‬ ‫اﻷوﻟﻰ‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫وﻣﺎ‬ ‫؟‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QTk =Q1k + Q2 =0 + 180 =180µC
F15
12
180
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1k + C2 ⇒ 15 =C1k + 3 ⇒ C1k =15 – 3=12µF

55. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
54
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk = ∆V1k = ∆V2 =12V
Q1k =C1k . ∆V1k =12 × 12 =144µC , Q2 =C2 . ∆V2 =3 × 12=36µC
6
2
12
C
C
k
1
k1
===
‫ﻣﺜﺎل‬124/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬12μF‫ﻣﻘﺪاره‬ ‫ﺟﮭﺪ‬ ‫ﻟﻔﺮق‬ ‫ﻣﺸﺤﻮﻧﺔ‬60V‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬‫ﺑﻤﺘ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬
8μF‫اﺣﺴﺐ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬:
1-‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
2-‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزل‬ ‫وﺿﻊ‬ ‫ﻟﻮ‬)2(‫ﺑﺪﻻ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬‫اﻟﮭﻮاء‬ ‫ﻣﻦ‬‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﯾﺼﺒﺢ‬ ‫ﻓﻜﻢ‬
3-‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻣﻘﺪار‬.
4-‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫وﺑﻌﺪ‬ ‫ﻗﺒﻞ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1-
‫اﻟ‬ ‫ﻗﺒﻞ‬‫ﺘﻮﺻﯿﻞ‬:
Q1 =C1 . ∆V1 =12 × 60 =720µC , Q2 =0 (‫ﻣﺸﺤﻮﻧﺔ‬ ‫)ﻏﯿﺮ‬
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =720 + 0 =720µC
Ceq =C1 + C2 =12 + 8 =20µF
V36
20
720
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =36V
Q1 =C1 . ∆V1 =12 × 36 =432µC , Q2 =C2 . ∆V2 =8 × 36 =288µC
2- C2k =k C2 =2 × 8 =16µF
Ceqk =C1 + C2k =12 + 16 =28µF
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT = 720µC
V7.25
28
720
C
Q
V
eqk
Tk
Tk ===∆
3- J1094.3960)7.25(1012
2
1
)V.(C
2
1
PE 6262
111
−−
×=×××=∆=
J1092.5283)7.25(1016
2
1
)V.(C
2
1
PE 6262
2k2k2
−−
×=×××=∆=
4- J1012961072036
2
1
Q.V
2
1
PE 56
TTT
−−
×=×××=∆=
PETk =PE1 + PE2k =3960.94×10-6
+ 5283.92×10-6
=9244.86×10-6
J
‫ﻣﺜﺎل‬125/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬4µF‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬200V‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫اﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬
‫ﺳﻌﺘﮭﺎ‬6µF‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻊ‬‫وﺿ‬ ‫ﺪ‬‫وﻋﻨ‬ ، ‫ﯿﻞ‬‫اﻟﺘﻮﺻ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺟﺪ‬‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬
‫ﺷﺤﻨﺘﮭﺎ‬ ‫اﺻﺒﺤﺖ‬ ‫اﻟﮭﻮاء‬ ‫او‬ ‫اﻟﻔﺮاغ‬ ‫ﺑﺪل‬600µC‫اﻟﻌﺰل؟‬ ‫ﺛﺎﺑﺖ‬ ‫ﺟﺪ‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وﻗﺒﻞ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q1=C1×∆V1=4×200=800µF , Q2=0 (‫ﻣﺸﺤﻮﻧﺔ‬ ‫)ﻏﯿﺮ‬
‫اﻟﻌﺎزل‬ ‫وﻗﺒﻞ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT = Q1 + Q2 =800 + 0=800µF
Ceq =C1 + C2 =4+6=10µF

56. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
55
21
eq
T
T VVV80
10
800
C
Q
V ∆=∆====∆
Q1=C1.∆V1=4×80=320µC , Q2=C2.∆V2=6×80=480µC
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =800µC
QTk =Q1 + Q2k ⇒ 800 =Q1 +600 ⇒ Q1=200µC
K2
1
1
1
VV50
4
200
C
Q
V ∆====∆
F12
50
600
V
Q
C
K2
K2
K2
µ==
∆
= , 2
6
12
C
C
k
2
k2
===
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬126/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬4μF‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬200V‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺖ‬‫رﺑﻄ‬
‫ﺳﻌﺘﮭﺎ‬6μF‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬ ‫ﺟﺪ‬.‫ﺪل‬‫ﺑ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫وﺿﻊ‬ ‫وﻋﻨﺪ‬
‫اﻟﮭ‬‫أﺻﺒﺤﺖ‬ ‫ﺷﺤﻨﺘﮭﺎ‬ ‫ان‬ ‫وﺟﺪ‬ ‫ﻮاء‬600μC.‫ﻣﻘﺪ‬ ‫ﻣﺎ‬‫اﻟﻤﺴﺘﻌﻤﻠﺔ‬ ‫اﻟﻤﺎدة‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ار‬.
‫اﻟﺤﻞ‬/
Q1= C1∆V1 = 4×200 = 800μC
‫ﻏ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ان‬ ‫ﺑﻤﺎ‬‫ﻟﺬﻟﻚ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﯿﺮ‬)Q2=0(
∴ Qtotal = Q1+Q2 = 800 + 0 =800μC
Ceq = C1 + C2 = 4 + 6 =10μF , ∆Vtotal=
eq
total
C
Q
= V80
10
800
= =∆V1 =∆V2=∆V
∴ Q1 = C1∆V = 4×80 = 320μC , Q2 = C2∆V = 6×80 = 480μF
‫اﻟﻌﺎزﻟﺔ‬ ‫اﻟﻤﺎدة‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬
Qtotal = Q1 + Q2 ⇒ 800 = Q1 + 600 ⇒ Q1 = 200μC
∴ ∆V1 = V50
4
200
C
Q
1
1
== =∆V2 =∆V
F12
50
600
V
Q
C
2
2
k2
µ==
∆
=
2
6
12
C
C
K
2
k2
===
‫ﻣﺜﺎل‬127/‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺘﺴﻌﺘﺎن‬2μF‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫اﻷوﻟﻰ‬60V‫ﯿﻦ‬‫ﺑ‬ ‫ﻊ‬‫وﺿ‬ ‫ﺸﺤﻮﻧﺔ‬‫ﻣ‬ ‫ﺮ‬‫ﻏﯿ‬ ‫ﺔ‬‫واﻟﺜﺎﻧﯿ‬
‫ﺎزل‬‫ﻋ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﺪل‬‫ﺑ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬‫ﮫ‬‫ﻋﺰﻟ‬ ‫ﺖ‬‫ﺛﺎﺑ‬)4(‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺎ‬‫ﻣ‬ ، ‫ﻰ‬‫اﻷوﻟ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬
‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
Q1=C1∆V1= 2×60 =120μC , C2k= KC2= 4×2= 8μF
‫ﻓﺎن‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﺑﻌﺾ‬ ‫ﻣﻊ‬ ‫اﻟﻤﺘﺴﻌﺘﯿﻦ‬ ‫رﺑﻂ‬ ‫ﺑﻌﺪ‬
Qtotal = Q1 = 120μC , Ceq= C1 + C2k= 2 + 8= 10μF
∆Vtotal= V12
10
120
C
Q
eq
total
==
Q1=C1∆V=2×12=24μC , Q2 =C2×∆V =8×12= 96μC

57. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
56
‫ﻣﺜﺎل‬128/‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎزل‬‫اﻟﻌ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻣﺘﺴﻌﺔ‬6μF‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬100V‫ﯿﻦ‬‫ﺑ‬ ‫ﺎزل‬‫ﻋ‬ ‫ﻊ‬‫وﺿ‬
‫ﻋﺰﻟﮫ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬)2(‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺖ‬ ‫ﺛﻢ‬8μF‫ﺤﻨﺔ‬‫وﺷ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺪ‬‫ﺟ‬
‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وإدﺧﺎل‬ ‫اﻟﺮﺑﻂ‬ ‫ﻗﺒﻞ‬:
Q1=C1×∆V=6×100=600μC
‫اﻟﻌﺎزل‬ ‫ووﺿﻊ‬ ‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬:
Q1=Qtotal=600μF , CK1=KC1=2×6=12μF
Ceq=CK1 + C2 =12+8=20μF
V30
20
600
C
Q
V
eq
total
total ===∆
‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬ ‫ان‬ ‫ﺑﻤﺎ‬
∆V1=∆V2=∆Vtotal =30V
Q1=CK1×∆V=12×30=360μC , Q2=C2×∆V=8×30=240μC
‫ﻣﺜﺎل‬129/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬2µF‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬200V‫ﺳﻌﺘﮭﺎ‬ ‫اﺧﺮى‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺖ‬4µF‫وﻓﺮق‬
‫ﺟﮭﺪھﺎ‬50V‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬k‫ﺮ‬‫ﻓ‬ ‫ﺾ‬‫اﻧﺨﻔ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﺑﯿﻦ‬‫ﺪار‬‫ﺑﻤﻘ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ق‬
40V‫اﻟﻌﺎزل‬ ‫ادﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺎ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
1- Q1 =C1 . ∆V1 =2 × 200 =400µC , Q2 =C2 . ∆V2 =4 × 50 =200µC
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT = Q1 + Q2 =400 + 200 =600µC
Ceq =C1 + C2 =2 + 4 =6µF
V100
6
600
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =100V
Q1 =C1 . ∆V1 =2 × 100 =200µC , Q2 =C2 . ∆V2 =4 × 100 =400µC
2-
‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
∆VTk =∆VT – 40 =100 – 40 =60V
Q‫ﻣ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻨﻔﺼﻠﺔ‬QTk = QT =600µC
F10
60
600
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1k + C2 ⇒ 10 =C1k + 4 ⇒ C1k =10 – 4 =6µF
C1k =k C1 ⇒ 6 =k × 2 ⇒ 3
2
6
k ==

58. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
57
‫ﻣﺜﺎل‬130/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬)15μF(‫ﺪ‬‫ﺟﮭ‬ ‫ﺑﻔﺮق‬ ‫ﻣﺸﺤﻮﻧﺔ‬)300V(‫ﺮ‬‫ﻏﯿ‬ ‫ﺮى‬‫أﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺖ‬‫رﺑﻄ‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫طﺮﻓﻲ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﺄﺻﺒﺢ‬ ‫ﻣﺸﺤﻮﻧﺔ‬)100V(‫اﺣﺴﺐ‬:
1-‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬.
2-‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
3-‫ﺴﻌ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﻊ‬‫وﺿ‬ ‫إذا‬‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬ ‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺒﺢ‬‫أﺻ‬ ‫ﺔ‬ ‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻰ‬ ‫اﻷوﻟ‬ ‫ﺔ‬)75V(‫ﻚ‬ ‫ﺗﻠ‬ ‫ﺰل‬‫ﻋ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺪ‬ ‫ﺟ‬
‫اﻟﻤﺎدة؟‬
‫اﻟﺤﻞ‬/
1-
‫اﻟﺮﺑﻂ‬ ‫ﻗﺒﻞ‬:
Q1 =C1 . ∆V1 =15 × 300 =4500µC , Q2 =0 (‫ﻣﺸﺤﻮﻧﺔ‬ ‫)ﻏﯿﺮ‬
‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =4500 + 0 =4500µC
F45
100
4500
V
Q
C
T
T
eq
µ==
∆
=
Ceq =C1 + C2 ⇒ 45 =15 + C2 ⇒ C2 =45 – 15 =30µF
2- Q1 =C1 × ∆V1 =15 × 100 =1500µC , Q2 =C2 × ∆V2 =30 × 100 =3000µF
3-
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =4500µC
F60
75
4500
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1k + C2 ⇒ 60 =C1k + 30 ⇒ C1k =60 – 30 =30µF
2
15
30
C
C
k
1
k1
===
‫ﻣﺜﺎل‬131/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻣﺘﺴﻌﺔ‬2μF‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﻔﯿﺤﺘﯿﮭﺎ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻌﺖ‬‫وﺿ‬ ‫اﻟﮭﻮاء‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻌﺎزل‬
‫ﻋﺰﻟﮭﺎ‬2.5‫رﺑﻄﺖ‬ ‫ﺛﻢ‬‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬ ‫أﺧﺮى‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬3μF‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺖ‬‫ﻓﻜﺎﻧ‬
‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬45μC.‫اﻟﺮﺑﻂ‬ ‫ﻗﺒﻞ‬ ‫اﻷوﻟﻰ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬ ‫اﺣﺴﺐ‬.
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وإدﺧﺎل‬ ‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬:
CK =K C =2.5×2=5μF
V15
3
45
C
Q
V
2
2
2
===∆
‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬ ‫ان‬ ‫ﺑﻤﺎ‬
∆V2 = ∆V1 = ∆V
VCQ 1K1
∆×=∴ = 5×15=75μC
Qtotal = Q1 + Q2 = 75 + 45 =120μC
‫ﻓﺎن‬ ‫ﻟﺬﻟﻚ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﻓﺎن‬ ‫اﻟﻌﺎزل‬ ‫ووﺿﻊ‬ ‫اﻟﺮﺑﻂ‬ ‫ﻗﺒﻞ‬
Q1 = Qtotal =120μC , C60
2
120
C
Q
V
1
1
1
µ===∆

59. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
58
‫ﻣﺜﺎل‬132/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬)3µF(‫وﺷﺤﻨﺘﮭﺎ‬)180µC(‫وﺿ‬ ‫ﺑﺎﻟﻤﺼﺪر‬ ‫ﻣﺘﺼﻠﺔ‬ ‫ﻏﯿﺮ‬‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻌﺖ‬
‫طﺮﻓﯿﮭﺎ‬ ‫ﻋﻠﻰ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﻈﮭﺮ‬ ‫اﻟﮭﻮاء‬ ‫ﺑﺪل‬)30V(‫ﻓﻤﺎ‬:
1-‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬.
2-‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺮى‬‫اﺧ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫اﻟﻌﺎزل‬ ‫وﺿﻊ‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ھﺬه‬ ‫وﺻﻠﺖ‬ ‫اذا‬(4µF)‫ﺤﻨﺘﮭﺎ‬‫وﺷ‬(60µC)‫ﺎ‬‫ﻓﻤ‬
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
1- V60
3
180
C
Q
V ===∆ , 2
30
60
V
V
k
k
==
∆
∆
=
2- QTk =Q1k + Q2 =180 + 60 =240µC
C1k=k C1=2 × 3=6µF
Ceqk=C1k + C2 =6 + 4=10µF
V24
10
240
C
Q
V
eqk
Tk
Tk ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VTk =∆V1k = ∆V2 =24V
Q1k =C1k . ∆V1k =6 × 24 =144µC , Q2 =C2 .∆V2 =4 × 24 =96µC
‫ﻣﺜﺎل‬133/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬2μF‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﺛﻢ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫وﺿﻌﺖ‬ ‫اﻟﮭﻮاء‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻌﺎزل‬
‫ﺳﻌﺘﮭﺎ‬ ‫أﺧﺮى‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬6μF‫ﻰ‬‫اﻷوﻟ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﻓﻜﺎﻧﺖ‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫ﻣﻊ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ووﺻﻠﺖ‬320μC
‫واﻟﺜﺎﻧﯿﺔ‬240μC‫؟‬ ‫اﻟﻤﺎدة‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
V40
6
240
C
Q
V
2
2
2
===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆V1k =∆V2 =40V
F8
40
320
V
Q
C
k1
k1
k1
µ==
∆
=
4
2
8
C
C
k
1
k1
===
‫اﻟﻤﺨﺘﻠﻂ‬ ‫اﻟﺮﺑﻂ‬:
‫ﻣﺜﺎل‬134/‫ﻣﺘﺴﻌﺘﺎن‬)C1=6μF , C2=12μF(‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻠﺖ‬‫وﺻ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺘﺎن‬‫ﻣﻮﺻ‬
‫ﻣﺘﺴﻌﺔ‬‫ﺳﻌﺘﮭﺎ‬ ‫ﺛﺎﻟﺜﺔ‬16μF‫ﺟﮭﺪھﺎ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫إﻟﻰ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺻﻠﺖ‬ ‫ﺛﻢ‬20V‫اﺣﺴﺐ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1-
4
1
12
3
12
12
12
1
6
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=4µF
Ceq=C12 + C3 =4 + 16=20µF
∆VT = ∆V3 = ∆V12 =20V (‫)ﺗﻮازي‬
Q3 =C3 . ∆V3 =16 × 20 =320µC , Q12 =C12 . ∆V12 =4 × 20 =80µC = Q1 = Q2
2- J10321032020
2
1
Q.V
2
1
PE 46
333
−−
×=×××=∆=

60. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
59
‫ﻣﺜﺎل‬135/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=3µF,C2=6µF(‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫وﻣﻊ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬)C3=2µF(‫ﻰ‬‫ﻋﻠ‬
‫ﺟﮭﺪھﺎ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫إﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫رﺑﻄﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬12V‫اوﺟﺪ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
1-
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=2µF
Ceq =C12 + C3 =2 + 2=4µF
∆VT = ∆V3 = ∆V12 =12V (‫)ﺗﻮازي‬
Q3 =C3 . ∆V3 =2 × 12=24µC , Q12 =C12 . ∆V12=2 × 12=24µC = Q1 = Q2
V8
3
24
C
Q
V
1
1
1
===∆ , V4
6
24
C
Q
V
2
2
2
===∆
‫ﻣﺜﺎل‬136/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=3µF,C2=6µF(‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫وﻣﻊ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬)C3=4µF(‫ﻰ‬‫ﻋﻠ‬
‫ﺪارھﺎ‬‫ﻣﻘ‬ ‫ﺔ‬‫ﻛﻠﯿ‬ ‫ﺑﺸﺤﻨﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺷﺤﻨﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬270µC‫ﺎل‬‫اﻟﻤﺠ‬ ‫ﺎ‬‫وﻣ‬ ‫؟‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻞ‬‫ﻛ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫وﻓ‬ ‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬
‫ﺑﯿﻨﮭﻤﺎ‬ ‫اﻟﻤﺴﺎﻓﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫اذا‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬(0.2cm).
‫اﻟﺤﻞ‬/
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=2µF
Ceq=C12 + C3 =2 + 4=6µF
V45
6
270
C
Q
V
eq
T
T ===∆
∆VT =∆V3 =∆V12 =45V (‫)ﺗﻮازي‬
Q3 =C3 . ∆V3=4 × 45 =180µC , Q12=C12 . ∆V12=2 × 45=90µC
Q12 = Q1 = Q2 =90µC
V30
3
90
C
Q
V
1
1
1
===∆ , V15
6
90
C
Q
V
2
2
2
===∆
m/V15000
102.0
30
d
V
E 2
1
1 =
×
=
∆
= −
‫ﻣﺜﺎل‬137/‫اﻟﻣﺗﺳﻌﺗﺎن‬ ‫رﺑطت‬)C1=5µF , C2=7µF(‫ﺎ‬‫ﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ت‬‫رﺑط‬ ‫م‬‫ﺛ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﻋ‬‫ﻊ‬‫ﻣ‬ ‫واﻟﻲ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻠ‬
‫ﺛﺎﻟﺛﺔ‬ ‫ﻣﺗﺳﻌﺔ‬)C3=4µF(‫ﻗطﺑﯾﮫ‬ ‫ﺑﯾن‬ ‫اﻟﺟﮭد‬ ‫ﻓرق‬ ‫ﺑﻣﺻدر‬ ‫اﻟﻧﮭﺎﺋﯾﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬ ‫ﺷﺣﻧت‬ ‫ﺛم‬)60V(‫د‬‫ﺟﮭ‬ ‫رق‬‫وﻓ‬ ‫ﺣﻧﺔ‬‫ﺷ‬ ‫ﻓﻣﺎ‬
‫ﺻﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﺑﯾن‬ ‫اﻟﺑﻌد‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺛﺎﻟﺛﺔ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫وﻣﺎ‬ ‫ﻣﺗﺳﻌﺔ؟‬ ‫ﻛل‬)0.5cm(‫؟‬
‫اﻟﺤﻞ‬/
C12 =C1 + C2 =5 + 7 =12µF
3
1
12
4
12
31
4
1
12
1
C
1
C
1
C
1
312eq
==
+
=+=+= ⇒ Ceq=3µF
QT =Ceq . ∆VT =3 × 60 =180µC =Q12 = Q3
21
12
12
12
VVV15
12
180
C
Q
V ∆=∆====∆ , V45
4
180
C
Q
V
3
3
3
===∆
Q1 =C1 . ∆V1 =5 × 15 =75µC , Q2 =C2 . ∆V2 =7 × 15 =105µC
m/V9000
105.0
45
d
V
E 2
3
3 =
×
=
∆
= −

61. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
60
‫ﻣﺜﺎل‬138/‫ﺳﻌﺗﺎن‬‫اﻟﻣﺗ‬ ‫ت‬‫رﺑط‬)C1=3µF , C2=6µF(‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻊ‬‫وﻣ‬ ‫ﺿﮭﻣﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫واﻟﻲ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬)C3=4µF(
‫ﻔﯾﺣﺗﯾﮭﺎ‬ ‫ﺻ‬ ‫ﯾن‬ ‫ﺑ‬ ‫د‬ ‫اﻟﺟﮭ‬ ‫رق‬ ‫ﻓ‬ ‫ﺑﺢ‬ ‫اﺻ‬ ‫ﻰ‬ ‫ﺣﺗ‬ ‫ﺔ‬ ‫اﻟﻛﻠﯾ‬ ‫ﺔ‬ ‫اﻟﻣﺟﻣوﻋ‬ ‫ﺣﻧت‬ ‫ﺷ‬ ‫م‬ ‫ﺛ‬ ‫وازي‬ ‫اﻟﺗ‬ ‫ﻰ‬ ‫ﻋﻠ‬)45V(‫ﺣﻧﺔ‬ ‫ﺷ‬ ‫دار‬ ‫ﻣﻘ‬ ‫ﺎ‬ ‫ﻓﻣ‬
‫ﺑﯾﻧﮭﻣﺎ‬ ‫اﻟﺑﻌد‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻣﺗﺳﻌﺔ‬ ‫ﺻﻔﯾﺣﺗﻲ‬ ‫ﺑﯾن‬ ‫اﻟﻛﮭرﺑﺎﺋﻲ‬ ‫اﻟﻣﺟﺎل‬ ‫وﻣﺎ‬ ‫؟‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻛل‬)0.2cm(‫؟‬
‫اﻟﺤﻞ‬/
∆VT = ∆V3 = ∆V12 =45V
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=2µF
Q3 =C3 . ∆V3 =4 × 45 =180µC , Q12=C12 . ∆V12 = 2 × 45 =90µC = Q1 = Q2
V30
3
90
C
Q
V
1
1
1
===∆
m/V15000
102.0
30
d
V
E 2
1
1 =
×
=
∆
= −
‫ﻣﺜﺎل‬139/‫ﺳﻌﺗﺎن‬‫اﻟﻣﺗ‬ ‫ت‬‫رﺑط‬)C1=3µF , C2=6µF(‫واﻟﻲ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬‫ﻋ‬ ‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫ﻠت‬‫وﺻ‬ ‫م‬‫ﺛ‬ ‫ﺿﮭﻣﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬‫ﻰ‬‫ﻠ‬
‫ﻣﺷﺣوﻧﺔ‬ ‫ﺛﺎﻟﺛﺔ‬ ‫ﻣﺗﺳﻌﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺗوازي‬)C3=2µF(‫وﺷﺣﻧﺗﮭﺎ‬)300µC(‫اﻟﺗوﺻﯾل؟‬ ‫ﺑﻌد‬ ‫ﻣﻧﮭﻣﺎ‬ ‫ﻛل‬ ‫ﺟﮭد‬ ‫وﻓرق‬ ‫ﺷﺣﻧﺔ‬ ‫ﻓﻣﺎ‬
‫اﻟﺤﻞ‬/
Q1 = Q2 = Q12 = 0
QT =Q3 + Q12 =300 + 0 =300µC
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=2µF
Ceq =C12 + C3 =2 + 2 =4µF
V75
4
300
C
Q
V
eq
T
T ===∆
∆VT = ∆V3 = ∆V12 =75V
Q3 =C3 . ∆V3 =2 × 75 =150µC , Q12 =C12 . ∆V12 =2 × 75 =150µC = Q1 = Q2
V50
3
150
C
Q
V
1
1
1
===∆ , V25
6
150
C
Q
V
2
2
2
===∆
‫ﻣﺜﺎل‬140/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=2F,C2=4µF(‫ﺖ‬‫رﺑﻄ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬)C3=3µF(‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﺟﻌﻞ‬ ‫ﻣﺴﺘﻤﺮ‬ ‫ﻣﺼﺪر‬ ‫ﻣﻦ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫وﺷﺤﻨﺖ‬ ‫ﻣﻌﮭﻤﺎ‬)300μC(‫اﺣﺴﺐ‬:
1-‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.3-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
‫اﻟﺤﻞ‬/
C12 =C1 + C2 =2 + 4=6µF
2
1
6
3
6
21
3
1
6
1
C
1
C
1
C
1
312eq
==
+
=+=+= ⇒ Ceq =2µF
V150
2
300
C
Q
V
eq
T
T ===∆
QT = Q3 = Q12 =300µC
V50
6
300
C
Q
V
12
12
12
===∆
∆V12 = ∆V1 = ∆V2 =50V (‫)ﺗﻮازي‬ , V100
3
300
C
Q
V
3
3
3
===∆

62. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
61
Q1 =C1 . ∆V1 =2 × 50 =100µC , Q2 =C2 . ∆V2 =4 × 50 =200µC
‫ﻣﺜـﺎل‬141/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=20µF , C2=30µF(‫ﺖ‬‫رﺑﻄ‬ ‫ﺎ‬‫وﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺎن‬‫ﻣﺮﺑﻮطﺘ‬
‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬)C3=18µF(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻲ‬‫ﻗﻄﺒ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﻢ‬‫ﺛ‬12V.
‫ﻣﻘﺪار‬ ‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.2-‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬‫ﺔ‬.
3-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
4-‫ﺑﯿﻨﮭﻤﺎ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.4cm.
‫اﻟﺤﻞ‬/
1-
12
1
60
5
60
23
30
1
20
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12=12µF
Ceq =C12 + C3 =12 + 18 =30µF
2- QT =Ceq . ∆VT =30 × 12 =360µC
3- ∆VT = ∆V3 = ∆V12 =12V
Q3 =C3 . ∆V3 =18 × 12 =216µC , Q12 =QT – Q3 =360 – 216 =144µC
Q12 = Q1 = Q2 =144µC
4- V8.4
30
144
C
Q
V
2
2
2
===∆ , m/V1200
104.0
8.4
d
V
E 2
2
2 =
×
=
∆
= −
‫ﻣﺜﺎل‬142/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬ ‫ﻏﯿﺮ‬ ‫ﻣﺘﺴﻌﺎت‬ ‫ﺛﻼث‬)C1=3µF,C2=6µF,C3=18µF(‫رﺑ‬‫ﻋﻠﻰ‬ ‫واﻟﺜﺎﻧﯿﺔ‬ ‫اﻷوﻟﻰ‬ ‫ﻄﺖ‬
‫ﺪه‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﻣﺼﺪر‬ ‫ﻗﻄﺒﻲ‬ ‫ﻋﺒﺮ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ورﺑﻄﺖ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻌﮭﻤﺎ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫ورﺑﻄﺖ‬ ‫اﻟﺘﻮاﻟﻲ‬30V‫ﺮق‬‫وﻓ‬ ‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﺎ‬‫ﻣ‬
‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬
‫اﻟﺤﻞ‬/
C1,2= F2
9
18
63
63
CC
CC
21
21
µ==
+
×
=
+
, Ceq=C1,2 + C3= 2+18=20μF
Qtotal= Ceq ∆V=20×30=600μC , Q3=C3∆V=18×30=540μC ,
Q1,2=C1,2∆V=2×30=60μC
∴ Q1=Q2=60μC
∆V1= V20
3
60
C
Q
1
1
== , ∆V2= V10
6
60
C
Q
2
2
==
‫ﻣﺜــﺎل‬143/‫ﺸﺤﻮﻧﺔ‬ ‫ﻣ‬ ‫ﺮ‬ ‫ﻏﯿ‬ ‫ﺴﻌﺎت‬ ‫ﻣﺘ‬ ‫ﻼث‬ ‫ﺛ‬)C1=5µF,C2=7µF,C3=6µF(‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺔ‬ ‫واﻟﺜﺎﻧﯿ‬ ‫ﻰ‬ ‫اﻷوﻟ‬ ‫ﺖ‬ ‫رﺑﻄ‬
‫ﻣ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫ورﺑﻄﺖ‬ ‫اﻟﺘﻮازي‬‫ﺟﮭﺪ‬ ‫ﻓﺮق‬ ‫ﯾﻌﻄﻲ‬ ‫ﻣﺼﺪر‬ ‫ﻗﻄﺒﻲ‬ ‫ﻋﺒﺮ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ووﺻﻠﺖ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫ﻌﮭﻤﺎ‬60V.‫ﺤﻨﺔ‬‫ﺷ‬ ‫ﻣﺎ‬
‫؟‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬
‫اﻟﺤﻞ‬/
C1,2=C1 + C2=5+7=12μF
4
1
12
3
12
21
6
1
12
1
C
1
C
1
C
1
32,1eq
==
+
=+=+= ⇒ Ceq=4μF
Qtotal=Ceq∆V=4×60=240μC
∴ Q3=Qtotal=240μC , Q1,2=Qtotal=240μC
∆V1,2= V20
12
240
C
Q
2,1
2,1
== , ∆V1=∆V2=20V
Q1=C1∆V1=5×20=100μC , Q2=C2 ∆V=7×20=140V

63. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
62
∆V3= V40
6
240
C
Q
3
3
==
‫ﻣﺜﺎل‬144/‫ﺴﻌﺘﺎن‬‫اﻟﻤﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=20µF , C2=4µF(‫ﻊ‬‫ﻣ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬
‫ﺔ‬ ‫ﺛﺎﻟﺜ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬)C3=12µF(‫ﺎ‬ ‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﺪ‬ ‫اﻟﺠﮭ‬ ‫ﺮق‬ ‫ﻓ‬ ‫ﺔ‬ ‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬ ‫اﻟ‬ ‫ﻊ‬ ‫اﻟﺠﻤﯿ‬ ‫ﻂ‬ ‫رﺑ‬ ‫ﻢ‬ ‫ﺛ‬6V‫ﺸﺤﻨﺔ‬ ‫اﻟ‬ ‫ﺴﻌﺔ‬ ‫ﻣﺘ‬ ‫ﻞ‬ ‫ﻟﻜ‬ ‫ﺴﺐ‬ ‫اﺣ‬
‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬.
‫اﻟﺤﻞ‬/
V426VVV
VVV2
24
48
C
Q
V
QQC4868V.CQ
F8
3
24
)12(12
1224
1224
1224
CC
CC
C
F24420CCC
2,1T3
21
2,1
2,1
2,1
2,13TeqT
32,1
32,1
eq
212,1
=−=∆−∆=∆
∆=∆====∆
==µ=×=∆=
µ==
+
×
=
+
×
=
+
=
µ=+=+=
‫ﻣﺜﺎل‬145/‫ﻣﺘﺴﻌﺎت‬ ‫ﺛﻼث‬)C1=12µF,C2=6µF,C3=16µF(‫رﺑﻄﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫واﻟﺜﺎﻧﯿﺔ‬ ‫اﻻوﻟﻰ‬ ‫رﺑﻄﺖ‬
‫اﻟﺜﺎﻟ‬‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﻜﺎﻧﺖ‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺠﻤﯿﻊ‬ ‫ورﺑﻂ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﻣﻌﮭﻤﺎ‬ ‫ﺜﺔ‬300µC‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.
2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
3-‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J10181018001024015
2
1
Q.V
2
1
PE3
,C2401516V.CQ
QQC60154V.CQ
VVV15
20
300
C
Q
V2
F20164CCC
F4
4
12
)12(6
612
612
612
CC
C.C
C1
466
333
333
212,12,12,1
32,1
eq
T
T
32,1eq
21
21
2,1
−−−
×=×=×××=∆=−
µ=×=∆=
==µ=×=∆=
∆=∆====∆−
µ=+=+=
µ==
+
×
=
+
×
=
+
=−

64. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
63
‫ﻣﺜﺎل‬146/‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫رﺑﻄﺖ‬)C1=5µF,C2=7µF(‫ﺔ‬‫ﺛﺎﻟﺜ‬ ‫ﺴﻌﺔ‬‫ﻣﺘ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫ﻣﻌﮭﻤ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬
)C3=12µF(‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻓﻜﺎن‬ ‫ﺑﻄﺎرﯾﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺠﻤﯿﻊ‬ ‫ورﺑﻂ‬6V‫اﺣﺴﺐ‬:
1-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬.
2-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
3-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.2cm.
3-‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
J105.3110213
2
1
Q.V
2
1
PE3
m/V1500
2
3000
102.0
3
d
V
E,V3
12
36
C
Q
V3
C2137V.CQ,C1535V.CQ
VVV3
12
36
C
Q
V
QQC3666V.CQ2
F6
)11(12
1212
1212
1212
CC
C.C
C
F1275CCC1
66
222
2
3
3
3
3
3
222111
21
2,1
2,1
2,1
2,13TeqT
32,1
32,1
eq
212,1
−−
−
×=×××=∆=−
==
×
=
∆
====∆−
µ=×=∆=µ=×=∆=
∆=∆====∆
==µ=×=∆=−
µ=
+
×
=
+
×
=
+
=
µ=+=+=−
‫ﻋﺎزل‬ ‫ﺑﻮﺟﻮد‬ ‫اﻟﻤﺨﺘﻠﻂ‬ ‫اﻟﺮﺑﻂ‬:
‫ﻣﺜﺎل‬147/‫ﺳﻌﺗﺎن‬‫ﻣﺗ‬)C1=8µF,C224µF(‫واﻟ‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎن‬‫ﻣرﺑوطﺗ‬‫ﻊ‬‫ﻣ‬ ‫وازي‬‫اﻟﺗ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠت‬‫وﺻ‬ ‫ﺎ‬‫وﻣﺟﻣوﻋﺗﮭﻣ‬ ‫ﻲ‬
‫ﺳﻌﺗﮭﺎ‬ ‫ﻣﺷﺣوﻧﺔ‬ ‫ﻣﺗﺳﻌﺔ‬)C3=4µF(‫ﺟﮭدھﺎ‬ ‫وﻓرق‬100V‫اﺣﺳب‬:
1-‫اﻟﻣﺟﻣوﻋﺔ‬ ‫ﺟﮭد‬ ‫ﻓرق‬.
2-‫ﺔ‬‫اﻟﻣﺟﻣوﻋ‬ ‫د‬‫ﺟﮭ‬ ‫رق‬‫ﻓ‬ ‫ﺑﺢ‬‫اﺻ‬ ‫ﺔ‬‫اﻟﺛﺎﻟﺛ‬ ‫ﺳﻌﺔ‬‫اﻟﻣﺗ‬ ‫ﻔﯾﺣﺗﻲ‬‫ﺻ‬ ‫ﯾن‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ت‬‫ادﺧﻠ‬ ‫اذا‬25V‫ﺎدة‬‫اﻟﻣ‬ ‫زل‬‫ﻋ‬ ‫ت‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻓﻣ‬
‫اﻟﻌﺎزﻟﺔ؟‬
‫اﻟﺤﻞ‬/
‫اﻟﺗوﺻﯾل‬ ‫ﻗﺑل‬:
1- Q1 = Q2 = Q12 = 0
Q3 =C3 × ∆V3 =4 × 100 =400µC
‫اﻟﺗوﺻﯾل‬ ‫ﺑﻌد‬:
QT = Q3 + Q12 = 400 + 0 =400µC
6
1
24
4
24
13
24
1
8
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12 =6µF
Ceq =C12 + C3 =6 + 4 =10µF
V40
10
400
C
Q
V
eq
T
T ===∆
2-
Q‫اﻟﺷﺎﺣن‬ ‫اﻟﻣﺻدر‬ ‫ﻋن‬ ‫ﻣﻔﺻوﻟﺔ‬ ‫اﻟﻣﺟﻣوﻋﺔ‬‫ﻟذﻟك‬QTk = QT =400µC
F16
25
400
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C3k + C12 ⇒ 16 =C3k + 6 ⇒ C3k =16 – 6 =10µF

65. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
64
C3k =kC3 ⇒ 10 =k × 4 ⇒ 5.2
4
10
k ==
‫ﻣﺜــﺎل‬148/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=8µF,C2=24µF(‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎن‬ ‫ﻣﺮﺑﻮطﺘ‬،‫ﺴﻌﺔ‬ ‫ﻣﺘ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻠﺘﺎ‬ ‫وﺻ‬
‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺸﺤﻮﻧﺔ‬)C3=4µF(‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬100V‫اﺣﺴﺐ‬:
1-‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.
2-‫ﺷﺤﻨﺘﮭﺎ‬ ‫اﺻﺒﺤﺖ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ادﺧﻠﺖ‬ ‫اذا‬250µC‫اﻟﻌﺎ‬ ‫اﻟﻤﺎدة‬ ‫ﻋﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ھﻮ‬ ‫ﻓﻜﻢ‬‫زﻟﺔ؟‬
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﻗﺒﻞ‬:
Q3=C3 . ∆V3 =4 × 100 =400µC
Q1 = Q2 = Q12 =0
‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
QT =Q12 + Q3 =0 + 400 =400µC
6
1
24
4
24
13
24
1
8
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12 =6µF
Ceq=C12 + C3 =6 + 4 =10µF
V40
10
400
C
Q
V
eq
T
T ===∆
‫اﻟﻌﺎزل‬ ‫وﺑﻌﺪ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﺸﺎﺣﻦ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =400µC
QTk =Q1k + Q2 ⇒ 400 =250 + Q2 ⇒ Q2 =400 – 250 =150µC
V25
6
150
C
Q
V
12
12
12
===∆
∆V12 = ∆V3k =25V
F10
25
250
V
Q
C
k3
k3
k3
µ==
∆
= , C3k =kC3 ⇒ 10 =k × 4 ⇒ 5.2
4
10
k ==
‫ﻣﺜــﺎل‬149/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬ ‫ﺖ‬ ‫رﺑﻄ‬)C1=3µF,C2=6µF(‫رﺑﻄ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬‫ﺖ‬‫ﻮازي‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﮭﻤ‬ ‫ﻊ‬ ‫ﻣ‬
‫اﻟﻤﺘﺴﻌﺔ‬)C3=9µF(‫إﻟﻰ‬ ‫اﻟﺠﻤﯿﻊ‬ ‫ورﺑﻂ‬‫ﻣﺴﺘﻤﺮ‬ ‫ﻣﺼﺪر‬‫ﺟﮭﺪ‬ ‫ﻓﺮق‬‫ه‬100V‫ا‬‫ﺣﺴﺐ‬:
1-‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
2-‫ﺪ‬ ‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺼﺒﺢ‬ ‫ﯾ‬ ‫ﺔ‬ ‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﺑﻤ‬ ‫ﺔ‬ ‫اﻟﺜﺎﻟﺜ‬ ‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬ ‫ﺑ‬ ‫ﻮاء‬‫اﻟﮭ‬ ‫ﺪل‬ ‫وأﺑ‬ ‫ﺼﺪر‬ ‫اﻟﻤ‬ ‫ﻦ‬ ‫ﻋ‬ ‫ﺔ‬ ‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺼﻠﺖ‬‫ﻓ‬ ‫اذا‬
‫اﻟﻤﺠﻤﻮﻋﺔ‬55V‫؟‬ ‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﺟﺪ‬
‫اﻟﺤﻞ‬/
1-
2
1
6
3
6
12
6
1
3
1
C
1
C
1
C
1
2112
==
+
=+=+= ⇒ C12 =2µF
Ceq =C12 + C3 =2 + 9=11µF
QT =Ceq .∆VT =11 × 100=1100µC
∆VT = ∆V12 = ∆V3 =100V (‫)ﺗﻮازي‬
Q3 =C3 . ∆V3 =9 × 100 =900µC , Q12 =C12 . ∆V12 =2 × 100=200µC
Q12 = Q1 = Q2 =200µC (‫)ﺗﻮاﻟﻲ‬
V
3
200
C
Q
V
1
1
1
==∆ , V
3
100
6
200
C
Q
V
2
2
2
===∆
2-
Q‫ﻟﺬﻟﻚ‬ ‫اﻟﻤﺼﺪر‬ ‫ﻋﻦ‬ ‫ﻓﺼﻠﺖ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬QTk = QT =1100µC

66. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
65
F20
55
1100
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C12 + C3k ⇒ 20 =2 + C3k ⇒ C3k =20 – 2 =18µF
2
9
18
C
C
k
3
k3
===
‫ﻣﺜــﺎل‬150/‫ﺴﻌﺘﺎن‬ ‫اﻟﻤﺘ‬ ‫ﺖ‬ ‫رﺑﻄ‬)C1=4µF,C2=12µF(‫ﺴﻌﺔ‬ ‫اﻟﻤﺘ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬ ‫رﺑﻄ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬
)C3=7µF(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫اﻟﻤﺘﺴﻌﺎت‬ ‫ﻣﺠﻤﻮﻋﺔ‬ ‫رﺑﻄﺖ‬ ‫ﺛﻢ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬10V‫ﻮح‬‫ﻟ‬ ‫ﻞ‬‫ادﺧ‬ ‫ﺎذا‬‫ﻓ‬
‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬)k(‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﻓﺎزدادت‬ ‫ﺑﺎﻟﺒﻄﺎرﯾﺔ‬ ‫ﻣﺘﺼﻠﺔ‬ ‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻻوﻟﻰ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫اﻟﻜ‬‫ﺑﻤﻘﺪار‬ ‫ﻠﯿﺔ‬50µF‫اﻟﻌﺰل‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬)k(‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫وﺷﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
6
4
24
C
C
k
F24C
24
1
24
23
C
1
12
1
8
1
C
1
12
1
C
1
8
1
C
1
C
1
C
1
F8715C7C15CCC
F15
10
150
V
Q
C
)‫ﺔ‬‫ﻠ‬‫ﺼ‬‫ﺘ‬‫ﻣ‬(V10VV
C1505010050QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
C1001010V.CQ
F1073CCC
F3
4
12
)31(4
124
124
124
CC
C.C
C
1
k1
k1
k1
k1k13k1K2,1
K2,1K2,13K2,1eqk
T
Tk
eqk
TTk
TTk
TeqT
32,1eq
21
21
2,1
===
µ=⇒=
−
=
−=⇒+=⇒+=
µ=−=⇒+=⇒+=
µ==
∆
=
=∆=∆
µ=+=+=
µ=×=∆=
µ=+=+=
µ==
+
×
=
+
×
=
+
=

67. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
66
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬151/‫ﺴﻌﺎت‬‫ﻣﺘ‬ ‫ﻼث‬ ‫ﺛ‬)C1=5µF,C2=7µF,C3=6µF(‫ﺖ‬‫ورﺑﻄ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺔ‬‫واﻟﺜﺎﻧﯿ‬ ‫ﻰ‬‫اﻻوﻟ‬ ‫ﺖ‬‫رﺑﻄ‬
‫ﺟﮭﺪه‬ ‫ﻓﺮق‬ ‫ﻣﺼﺪر‬ ‫اﻟﻰ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ووﺻﻠﺖ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫ﻣﻌﮭﻤﺎ‬ ‫اﻟﺜﺎﻟﺜﺔ‬60V.
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﺣﺴﺐ‬.
2-‫ﻋﺰﻟﮭﺎ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﺎزﻟﺔ‬ ‫ﻣﺎدة‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬ ‫اذا‬)k(‫ﺼﺪر‬‫ﺑﺎﻟﻤ‬ ‫ﺼﻠﺔ‬‫ﻣﺘ‬ ‫ﻣﺎزاﻟﺖ‬ ‫واﻟﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﺜﺎﻟﺜﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬
‫ﺪار‬‫ﺑﻤﻘ‬ ‫ﺤﻨﺘﮭﺎ‬‫ﺷ‬ ‫ازدادت‬120µC‫ﺰل‬‫اﻟﻌ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺪار‬‫ﻣﻘ‬ ‫ﺎ‬‫ﻓﻤ‬)k(‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺎ‬‫وﻣ‬ ‫؟‬
‫اﻻوﻟﻰ‬‫اﻟﻌﺎزل‬ ‫إدﺧﺎل‬ ‫ﺑﻌﺪ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫واﻟﻤﺘﺴﻌﺔ‬.
‫اﻟﺤﻞ‬/
C210307V.CQ,C150305V.CQ
VVV30
12
360
C
Q
V
2
6
12
C
C
k
F12C
12
1
12
12
12
1
6
1
C
1
C
1
12
1
6
1
C
1
C
1
C
1
F6
60
360
V
Q
C
V60VV
C360120240120QQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﻌﺎزل‬
2
C140207V.CQ,C100205V.CQ
VVV20
12
240
C
Q
V
QQC240604V.CQ
F4
3
12
)12(6
612
612
612
CC
C.C
C
F1275CCC1
222111
21
2,1
K2,1
k2,1
3
k3
k3
k3
k3k32,1eqk
Tk
Tk
eqk
TTk
TTk
222111
21
2,1
2,1
2,1
2,13TeqT
32,1
32,1
eq
212,1
µ=×=∆=µ=×=∆=
∆=∆====∆
===
µ=⇒=
−
=−=
+=⇒+=
µ==
∆
=
=∆=∆
µ=+=+=
−
µ=×=∆=µ=×=∆=
∆=∆====∆
==µ=×=∆=
µ==
+
×
=
+
×
=
+
=
µ=+=+=−

68. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
67
‫أﻣﺜﻠﺔ‬‫اﻟﻤﺘﺴﻌﺎت‬ ‫ﻋﻠﻰ‬ ‫اﺧﺮى‬:
‫ﻣﺜﺎل‬152/‫ﻣﺘﺴﻌﺘﺎن‬)C1=8µF,C2=4µF(‫ا‬ ‫ﺖ‬‫ﻓﻜﺎﻧ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫اﻟ‬ ‫وﺻﻠﺘﺎ‬ ‫ﺛﻢ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫وﺻﻠﺘﺎ‬‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫ﻟﻄﺎﻗ‬
‫ﻓﻲ‬‫اﻷوﻟﻰ‬64×10-6
J‫اﻟﺜﺎﻧﯿﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬32×10-6
J‫اﺣ‬‫ﺴﺐ‬:
1-‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬.2-‫اﻟﻜﻠﯿﺔ‬ ‫واﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺷﺤﻨﺔ‬.
‫اﻟﺤﻞ‬/
1- 2
111 )V.(C
2
1
PE ∆= ⇒ 2
1
66
)V(108
2
1
1064 ∆×××=× −−
16
4
64
)V( 2
1 ==∆ ⇒ V4V1
=∆
2
222 )V.(C
2
1
PE ∆= ⇒ 2
2
66
)V(104
2
1
1032 ∆×××=× −−
16
2
32
)V( 2
2 ==∆ ⇒ V4V2
=∆
Q 21
VV ∆=∆
∴‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬
V4VVV 21T
=∆=∆=∆
2- Q1 = C1 . ∆V1 =8 × 4 =32µC , Q2 =C2 . ∆V2 =4 × 4=16µC
QT =Q1 + Q2 =32 + 16 =48µC
‫ﻣﺜــﺎل‬153/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=3µF , C2=6µF(‫ﻰ‬ ‫اﻷوﻟ‬ ‫ﻲ‬ ‫ﻓ‬ ‫ﺔ‬ ‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬ ‫اﻟﻄﺎﻗ‬ ‫ﺖ‬ ‫وﻛﺎﻧ‬ ‫ﺔ‬ ‫ﺑﺒﻄﺎرﯾ‬ ‫ﻮﻟﺘﺎن‬ ‫ﻣﻮﺻ‬
3750×10-6
J‫اﻟﺜﺎﻧﯿﺔ‬ ‫وﻓﻲ‬7500×10-6
J‫ﻣﻘﺪار‬ ‫ﻣﺎ‬:
1-‫اﻟﺒﻄﺎرﯾ‬ ‫ﺟﮭﺪ‬ ‫ﻓﺮق‬‫ﺔ‬2-‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬ ‫اﻟﺸﺤﻨﺔ‬
‫اﻟﺤﻞ‬/
2
11 )V.(C
2
1
PE ∆= ⇒ 2
1
66
)V(103
2
1
103750 ∆××=× −−
2500
3
7500
3
37502
)V( 2
1 ==
×
=∆ ⇒ ∆V1 =50V
2
222 )V.(C
2
1
PE ∆= ⇒ 2
2
66
)V(106
2
1
107500 ∆××=× −−
2500
3
7500
)V( 2
2 ==∆ ⇒ ∆V2=50V
Q ∆V1 = ∆V2
‫ﺗﻮازي‬ ‫ﻓﺎﻟﺮﺑﻂ‬ ‫ﻟﺬﻟﻚ‬
∴ ∆VT =50V
2- Ceq =C1 + C2 =3 + 6=9µF
QT =Ceq . ∆VT =9 × 50 =450µC
‫ﻣﺜﺎل‬154/‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬3µF‫ﺟﮭﺪھﺎ‬ ‫وﻓﺮق‬150V‫ﺷﺤﻨﺘﮭﺎ‬ ‫ﺛﺎﻧﯿﺔ‬ ‫وﻣﺘﺴﻌﺔ‬900µC‫ﺚ‬‫ﺑﺤﯿ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫وﺻﻠﺘﺎ‬
‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﺖ‬‫ادﺧﻠ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﺴﻌﺘﯿﻦ‬‫ﻟﻠﻤﺘ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫اﻟﻤﺨﺘﻠﻔﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫رﺑﻄﺖ‬
‫ﻋﺰﻟﮭﺎ‬2‫ﺷﺤﻨﺘﮭﺎ‬ ‫ﻓﺄﺻﺒﺤﺖ‬360µF‫اﻟﻌﺎزل؟‬ ‫وﺿﻊ‬ ‫ﻗﺒﻞ‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬
‫اﻟﺤﻞ‬/
‫اﻟﻌﺎزل‬ ‫وﻗﺒﻞ‬ ‫اﻟﺘﻮﺻﯿﻞ‬ ‫ﺑﻌﺪ‬:
Q1=C1.∆V1=3×150=450µC
QT =Q2 – Q1 =900 – 450 = 450µC

69. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
68
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
QT = Q1 + Q2 ⇒ 450 = Q1 + 360 ⇒ Q1 =90µC
V30
3
90
C
Q
V
1
1
1
===∆ =∆V2
F12
30
360
V
Q
C
2
2
K2
µ==
∆
=∴ , C2k = k C2 ⇒ 12 =2C2 ⇒ C2 =6µF
‫ﻣﺜـﺎل‬155/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=6µF,C2=12µF(‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻲ‬‫ﻗﻄﺒ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫رﺑﻄﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬90V‫ﺪ‬‫اﻋﯿ‬ ‫ﻢ‬‫ﺛ‬ ‫ﺔ‬‫ﺑﺎﻟﻄﺎﻗ‬ ‫ﯿﺎع‬‫ﺿ‬ ‫ﺣﺪوث‬ ‫دون‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫وﻋﻦ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬
‫ﻣﻊ‬ ‫رﺑﻄﮭﻤﺎ‬‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫ﻟﻠﻤﺘﺴﻌﺘﯿﻦ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﻟﻤﺘﻤﺎﺛﻠﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫رﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﺑﻌﺾ‬:
1-‫ﺻﻔﯿﺤﺘﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻟﻜﻞ‬ ‫اﺣﺴﺐ‬.
2-‫ﺎ‬‫ﻋﺰﻟﮭ‬ ‫ﺖ‬‫ﺛﺎﺑ‬ ‫ﺎ‬‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫ﻋﺎزﻟ‬ ‫ﺎدة‬‫ﻣ‬ ‫ﻣﻦ‬ ‫ﻟﻮح‬ ‫ادﺧﻞ‬)k(‫ﺔ‬‫اﻟﻤﺠﻤﻮﻋ‬ ‫ﺪ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺒﺢ‬‫اﺻ‬ ‫ﺔ‬‫اﻟﺜﺎﻧﯿ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﯿﻦ‬‫ﺑ‬
20V‫اﻟﻌﺰ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺎ‬‫ل‬)k.(
‫اﻟﺤﻞ‬/
1-
4
1
12
3
12
12
12
1
6
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq=4µF
QT =Ceq × ∆VT =4 × 90 =360µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =360µC
‫ﺑﻌﺾ‬ ‫ﻣﻊ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫ﺑﻌﺪ‬:
QT =Q1 + Q2 =360 + 360 =720µC
Ceq =C1 + C2 = 6 + 12 =18µF
V40
18
720
C
Q
V
eq
T
T ===∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 =∆V2 =40V
Q1=C1 . ∆V1 =6 × 40 =240µC , Q2 =C2 . ∆V2 =12 × 40 =480µC
J10481024040
2
1
Q.V
2
1
PE 46
111
−−
×=×××=∆=
J10961048040
2
1
Q.V
2
1
PE 46
222
−−
×=×××=∆=
2-
‫اﻟﻌﺎزل‬ ‫ﺑﻌﺪ‬:
Q‫ﻋ‬ ‫ﻣﻔﺼﻮﻟﺔ‬ ‫اﻟﻤﺠﻤﻮﻋﺔ‬‫ﻟﺬﻟﻚ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫ﻦ‬QTk = QT =720µC
F36
20
720
V
Q
C
Tk
Tk
eqk
µ==
∆
=
Ceqk =C1 + C2k ⇒ 36 = 6 – C2k ⇒ C2k =36 – 6 =30µC
C2k = k C2 ⇒ 30 =k × 12 ⇒ 5.2
2
5
12
30
k ===

70. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
69
‫ﻣﺜﺎل‬156/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬ ‫ﺖ‬‫رﺑﻄ‬)C1=24µF,C2=8µF(‫ﺪھﺎ‬‫ﺟﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬‫إﻟ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬40V
‫اﺣﺴﺐ‬:
1-‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺟﮭﺪ‬ ‫وﻓﺮق‬ ‫ﺷﺤﻨﺔ‬.
2-‫ﺎن‬‫اﻟﻤﻮﺟﺒﺘ‬ ‫ﻔﯿﺤﺘﯿﮭﻤﺎ‬‫ﺻ‬ ‫وﺻﻠﺖ‬ ‫ﺑﺤﯿﺚ‬ ‫ﺑﻌﺾ‬ ‫ﻣﻊ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫أﻋﯿﺪ‬ ‫ﺛﻢ‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫وﻋﻦ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﻓﺼﻠﺖ‬ ‫اذا‬
‫واﻟﺴﺎﻟﺒﺘﺎن‬ ‫ﻣﻌﺎ‬‫ﻣﻨﮭﻤﺎ‬ ‫ﻟﻜﻞ‬ ‫اﻟﺠﮭﺪ‬ ‫وﻓﺮق‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻓﺎﺣﺴﺐ‬ ‫ﻣﻌﺎ‬.
‫اﻟﺤﻞ‬/
1-
6
1
24
4
24
31
8
1
24
1
C
1
C
1
C
1
21eq
==
+
=+=+= ⇒ Ceq =6µF
QT =Ceq . ∆VT =6 × 40 =240µC
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮاﻟﻲ‬ ‫اﻟﺮﺑﻂ‬QT = Q1 = Q2 =240µC
V10
24
240
C
Q
V
1
1
1
===∆ , V30
8
240
C
Q
V
2
2
2
===∆
2- QT =Q1 + Q2 =240 + 240 =480µC
Ceq =C1 + C2 =24 + 8 =32µF
V15
32
480
VT ==∆
Q‫ﻟﺬﻟﻚ‬ ‫ﺗﻮازي‬ ‫اﻟﺮﺑﻂ‬∆VT = ∆V1 = ∆V2 =15V
Q1 =C1 . ∆V1 =24 × 15 =360µC , Q2 =C2 . ∆V2 =120µC
‫ﻣﺜــﺎل‬157/‫ﺴﻌﺘﺎن‬ ‫ﻣﺘ‬)C1=12µF,C2=6µF(‫ﻰ‬ ‫اﻟ‬ ‫ﺎ‬ ‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬ ‫رﺑﻄ‬ ‫ﻢ‬ ‫ﺛ‬ ‫ﻀﮭﻤﺎ‬ ‫ﺑﻌ‬ ‫ﻊ‬ ‫ﻣ‬ ‫ﻮاﻟﻲ‬ ‫اﻟﺘ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﺎ‬ ‫رﺑﻄﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬36V‫ﺔ‬‫ﺑﺎﻟﻄﺎﻗ‬ ‫ﺿﯿﺎع‬ ‫ﺣﺪوث‬ ‫دون‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫وﻋﻦ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬ ،
‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫ﻟﻠﻤﺘﺴﻌﺘﯿﻦ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﻟﻤﺘﻤﺎﺛﻠﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫رﺑﻄﺖ‬ ‫ﺑﺤﯿﺚ‬ ‫ﺑﻌﺾ‬ ‫ﻣﻊ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫اﻋﯿﺪ‬ ‫ﺛﻢ‬‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬
‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬.
‫اﻟﺤﻞ‬/
C96166V.CQ,C1921612V.CQ
V16
18
288
C
Q
V
F18612CCC
C288144144QQQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﺘﻮﺻﯿﻞ‬
QQC144364V.CQ
F4
3
12
)12(6
612
612
612
CC
C.C
C
2211
eq
T
21eq
21T
21TeqT
21
21
eq
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
µ=+=+=
==µ=×=∆=
µ==
+
×
=
+
×
=
+
=

71. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
70
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬158/‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬)C1=20µF,C2=30µF(‫ﻰ‬‫اﻟ‬ ‫ﺎ‬‫ﻣﺠﻤﻮﻋﺘﮭﻤ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻀﮭﻤﺎ‬‫ﺑﻌ‬ ‫ﻊ‬‫ﻣ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫رﺑﻄﺘ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﺑﻄﺎرﯾﺔ‬25V‫ﺣﺪوث‬ ‫دون‬ ‫اﻟﺒﻄﺎرﯾﺔ‬ ‫وﻋﻦ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﻓﺼﻠﺖ‬ ‫ﻓﺎذا‬ ،‫ﺔ‬‫ﺑﺎﻟﻄﺎﻗ‬ ‫ﺿﯿﺎع‬
‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﻣﻘﺪار‬ ‫ﻓﻤﺎ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫ﻟﻠﻤﺘﺴﻌﺘﯿﻦ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﻟﻤﺘﻤﺎﺛﻠﺔ‬ ‫اﻟﺼﻔﺎﺋﺢ‬ ‫رﺑﻄﺖ‬ ‫ﺑﺤﯿﺚ‬ ‫ﺑﻌﺾ‬ ‫ﻣﻊ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫اﻋﯿﺪ‬ ‫ﺛﻢ‬
‫اﻟﺮﺑﻂ‬ ‫ﺑﻌﺪ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﻛﻞ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬.
‫اﻟﺤﻞ‬/
C3601230V.CQ,C2401220V.CQ
V12
50
600
C
Q
V
F503020CCC
C600300300QQQ
‫ﺪ‬‫ﻌ‬‫ﺑ‬‫اﻟﺘﻮﺻﯿﻞ‬
QQC3002512V.CQ
F12
510
600
)32(10
3020
3020
3020
CC
C.C
C
2211
eq
T
21eq
21T
21TeqT
21
21
eq
µ=×=∆=µ=×=∆=
===∆
µ=+=+=
µ=+=+=
==µ=×=∆=
µ=
×
=
+
×
=
+
×
=
+
=
‫ﻣﺜﺎل‬159/‫ﺴ‬‫اﻟﻤﺘ‬ ‫ﺔ‬‫ﻗﯿﻤ‬ ‫ﺚ‬‫ﺛﻠ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺖ‬‫ﻛﺎﻧ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫رﺑﻄﺘ‬ ‫اذا‬ ‫ﺴﻌﺘﺎن‬‫ﻣﺘ‬‫ﻰ‬‫ﻋﻠ‬ ‫ﺎ‬‫رﺑﻄﺘ‬ ‫واذا‬ ‫ﻰ‬‫اﻷوﻟ‬ ‫ﻌﺔ‬
‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫ﻛﺎﻧﺖ‬ ‫اﻟﺘﻮازي‬3μF‫اﻟﺴﻌﺔ‬ ‫ﻣﻘﺪار‬ ‫ﺟﺪ‬‫ﻟﻜﻞ‬‫اﻟﻤﺘﺴﻌﺘﯿﻦ‬ ‫ﻣﻦ‬.
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮاﻟﻲ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
21
21
eq
CC
C.C
C
+
= ⇒
21
21
1
CC
C.C
C
3
1
+
= ⇒ C1 + C2=3C
C1=2C2
‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬‫اﻟﺘﻮازي‬
Ceq=C1 + C2 ⇒ 3=2C2 + C2 ⇒ 3=3C2 ⇒ C2=1μF
Q C1=2C2 ⇒ C1=2×1=2μF
‫ـﺎل‬‫ـ‬‫ﻣﺜ‬160/‫ﺔ‬‫اﻟﻤﺘﻤﺎﺛﻠ‬ ‫ﺴﻌﺎت‬‫اﻟﻤﺘ‬ ‫ﻦ‬‫ﻣ‬ ‫ﺔ‬‫ﻣﺠﻤﻮﻋ‬‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺖ‬‫ﻛﺎﻧ‬ ‫ﺎذا‬‫ﻓ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺖ‬‫رﺑﻄ‬ ‫ﻢ‬‫ﺛ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺖ‬‫رﺑﻄ‬
‫ﻟﻠﺘﻮازي‬100‫اﻟﻤﺘﺴﻌﺎت؟‬ ‫ﻋﺪد‬ ‫ﻓﻤﺎ‬ ‫ﻟﻠﺘﻮاﻟﻲ‬ ‫اﻟﺴﻌﺔ‬ ‫ﺑﻘﺪر‬
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮاﻟﻲ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
n
C
C series)eq( = --------- (1)
‫اﻟﺘﻮازي‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
C(eq)parallel=nC --------- (2)
‫ﺑﻘﺴﻤﺔ‬)2(‫ﻋﻠﻰ‬)1(‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬
2
series)eq(
parallel)eq(
n
n
C
nC
C
C
==
2
series)eq(
series)eq(
n
C
C100
= ⇒ n2
=100 ⇒ n=10

72. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
71
‫ﻣﺜﺎل‬161/‫أي‬ ‫ﻌﺔ‬‫ﺳ‬ ‫ﻦ‬‫ﻣ‬ ‫ﻐﺮ‬‫اﺻ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺳﻌﺘﮭﻤﺎ‬ ‫ﻓﺎن‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺴﻌﺔ‬ ‫ﻣﺘﺴﺎوﯾﺘﻲ‬ ‫ﻣﺘﺴﻌﺘﯿﻦ‬ ‫وﺻﻞ‬ ‫ﻋﻨﺪ‬ ‫اﻧﮫ‬ ‫اﺛﺒﺖ‬
‫؟‬ ‫ﻣﻨﮭﻤﺎ‬ ‫واﺣﺪة‬
‫اﻟﺤﻞ‬/
2
C
n
C
Ceq == ⇒ C
2
1
Ceq =
‫ﻣﺜﺎل‬162/‫ﺔ‬‫أرﺑﻌ‬ ‫ﺴﺎوي‬‫ﺗ‬ ‫ﺎ‬‫ﻟﮭﻤ‬ ‫ﺔ‬‫اﻟﻤﻜﺎﻓﺌ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺎن‬‫ﻓ‬ ‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﺴﻌﺔ‬‫اﻟ‬ ‫ﺴﺎوﯾﺘﺎ‬‫ﻣﺘ‬ ‫ﻣﺘﺴﻌﺘﺎن‬ ‫وﺻﻠﺖ‬ ‫اذا‬ ‫اﻧﮫ‬ ‫اﺛﺒﺖ‬
‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫ﻟﻮ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫ﺳﻌﺘﮭﻤﺎ‬ ‫أﻣﺜﺎل‬.
‫اﻟﺤﻞ‬/
Cparallel =nC ⇒ Cparallel = 2C …….. (1)
n
C
Cseries = ⇒
2
C
Cseries = ……... (2)
‫ﻣﻌﺎدﻟﺔ‬ ‫ﺑﻘﺴﻤﺔ‬)1(‫ﻋﻠﻰ‬)2(‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬:
4
2
C
C2
C
C
series
parallel
==
∴ Cparallel = 4Cseries
‫ﻣﺜﺎل‬163/‫و‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﻣﺘﺴﻌﺘﺎن‬‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫ﺳﻌﺘﮭﻤﺎ‬ ‫ﻛﺎﻧﺖ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫ﺻﻠﺘﺎ‬9μF‫ﺖ‬‫ﻛﺎﻧ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫وﻋﻨﺪﻣﺎ‬
‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫ﺳﻌﺘﮭﻤﺎ‬2μF‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫اﺣﺴﺐ‬.
‫اﻟﺤﻞ‬/
‫ﻓﺎن‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬:
Ceq =C1 + C2 ⇒ 9 =C1 + C2 …….. (1)
‫ﻓﺎن‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫رﺑﻄﮭﻤﺎ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬:
21
21
eq
CC
C.C
C
+
= ⇒
21
21
CC
C.C
2
+
= ……..(2)
‫ﻣﻌﺎدﻟﺔ‬ ‫ﺑﺘﻌﻮﯾﺾ‬)1(‫ﻓﻲ‬)2(‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬:
9
C.C
2 21
= ⇒ C1.C2 =18 ⇒
2
1
C
18
C = ……… (3)
‫ﻣﻌﺎدﻟﺔ‬ ‫ﺑﺘﻌﻮﯾﺾ‬)3(‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬)1(‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬:
2
2
C
C
18
9 += ⇒ 2
22
C18C9 += ⇒ 018C9C 2
2
2
=+−
(C2 – 6)(C2 – 3)=0 ⇒ C2 =6µF or C2=3µF
‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﯾﺾ‬)1(‫ﻋﻠﻰ‬ ‫ﻧﺤﺼﻞ‬:
C1 =3µF or C=6µF

73. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
72
‫ﻣﺜﺎل‬164/‫ﻣﺘﺴﻌﺘﺎن‬C1 , C2‫ﺼﻠﺔ‬‫اﻟﻤﺤ‬ ‫ﺔ‬‫ﻗﯿﻤ‬ ‫ﺖ‬‫ﻛﺎﻧ‬ ‫ﻮاﻟﻲ‬‫اﻟﺘ‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫ﻣﺘﻰ‬1C
3
1
‫ﻮازي‬‫اﻟﺘ‬ ‫ﻰ‬‫ﻋﻠ‬ ‫ﻠﺘﺎ‬‫وﺻ‬ ‫واذا‬
‫اﻟﻤﺤﺼﻠﺔ‬ ‫ﻛﺎﻧﺖ‬3µF‫ﻗﯿﻤﺔ‬ ‫ﻣﺎ‬C1 , C2‫؟‬
‫ﻓﺎن‬ ‫ﻣﺘﻮاﻟﯿﺘﺎن‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﺗﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬:
21
21
eq
CC
C.C
C
+
= ⇒
21
21
1
CC
C.C
C
3
1
+
= ⇒ C1 + C2 =3C2
∴ C1=2C2
‫ﻓﺎن‬ ‫ﻣﺘﻮازﯾﺘﺎن‬ ‫اﻟﻤﺘﺴﻌﺘﺎن‬ ‫ﺗﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬:
Ceq = C1 + C2 3=2C2 + C2 ⇒ C2 =1μF
C1=2μF
‫ﻣﺜﺎل‬165/‫ﺟﮭﺪه‬ ‫ﻓﺮق‬ ‫ﻣﺼﺪر‬ ‫اﻟﻰ‬ ‫ووﺻﻠﺘﺎ‬ ‫ﺑﻌﻀﮭﻤﺎ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫ﻣﺘﻰ‬ ‫ﻣﺘﺴﻌﺘﺎن‬100V‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺖ‬‫ﻛﺎﻧ‬
‫ﻟﻠﻤﺠﻤﻮﻋﺔ‬ ‫اﻟﻜﻠﯿﺔ‬300µC‫ﺔ‬‫ﻟﻠﻤﺠﻤﻮﻋ‬ ‫ﺔ‬‫اﻟﻜﻠﯿ‬ ‫ﺸﺤﻨﺔ‬‫اﻟ‬ ‫ﺖ‬‫ﻛﺎﻧ‬ ‫ﺼﺪر‬‫اﻟﻤ‬ ‫ﺑﻨﻔﺲ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬ ‫وﺻﻠﺘﺎ‬ ‫وﻣﺘﻰ‬1600µC‫ﺪ‬‫ﺟ‬
‫؟‬ ‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻣﻘﺪار‬
‫اﻟﺤﻞ‬/
‫اﻟﺘﻮاﻟﻲ‬ ‫رﺑﻂ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬:
F3
100
300
V
Q
C
t
t
eq
µ==
∆
=
21
21
eq
CC
C.C
C
+
= ⇒
21
21
CC
C.C
3
+
=
‫اﻟﺘﻮازي‬ ‫رﺑﻂ‬ ‫ﺣﺎﻟﺔ‬ ‫ﻓﻲ‬
F16
100
1600
V
Q
C
t
t
eq
µ==
∆
=
Ceq=C1 + C2 ⇒ 16 =C1 + C2 ⇒ C1 =16 – C2
22
22
CC16
C)C16(
3
+−
−
= ⇒ 048C16C 2
2
2
=+− ⇒ (C2 – 12)(C2 – 4)=0
C2=12μF or C2 =4μF
C1=16 – 12=4μF or C1=16 – 4=12μF
‫ﻣﺜﺎل‬166/‫ﻣﻨﮭﻤﺎ‬ ‫ﻛﻞ‬ ‫ﺳﻌﺔ‬ ‫ﻣﺘﻤﺎﺛﻠﺘﺎن‬ ‫ﻣﺘﺴﻌﺘﺎن‬ ‫ﻟﺪﯾﻚ‬C‫ﻣﺮة‬ ‫رﺑﻄﺘﺎ‬‫ﺔ‬‫اﻟﻌﻼﻗ‬ ‫ﺎ‬‫ﻓﻤ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫وأﺧﺮى‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬
‫اﻟﺤﺎﻟﺘﯿﻦ؟‬ ‫ﻓﻲ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫ﺑﯿﻦ‬
Ceq1=C + C =2C (‫)ﺗﻮازي‬
2
C
C2
C
C
2
2eq == (‫)ﺗﻮاﻟﻲ‬
‫اﻟﻌﻼﻗﺔ‬ ‫ﻹﯾﺠﺎد‬ ‫ﺑﺎﻟﻘﺴﻤﺔ‬
2
C
C2
C
C
2eq
1eq
= ⇒ 4
C
C
2eq
eq1
= ⇒ Ceq1=4Ceq2
‫ﻟﻠﺘﻮاﻟﻲ‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫أﻣﺜﺎل‬ ‫أرﺑﻌﺔ‬ ‫ﻟﻠﺘﻮازي‬ ‫اﻟﻤﻜﺎﻓﺌﺔ‬ ‫اﻟﺴﻌﺔ‬ ‫ان‬ ‫أي‬
2
sires)eq(
parallel)eq(
n
C
C
= ⇒ C(eq)parallel =n2
C(eq)sires
∴ Cparallel = 4Cseries

74. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
73
‫واﻟﺘﻔﺮﯾﻎ‬ ‫اﻟﺸﺤﻦ‬:
‫ﻣﺜﺎل‬167/‫ﺪارھﺎ‬‫ﻣﻘ‬ ‫ﺔ‬‫ﻣﻘﺎوﻣ‬ ‫ﻦ‬‫ﻣ‬ ‫ﺎﻟﻒ‬‫ﺗﺘ‬ ‫ﺮﺑﻂ‬‫اﻟ‬ ‫ﺔ‬‫ﻣﺘﻮاﻟﯿ‬ ‫ﺮة‬‫داﺋ‬)200Ω(‫ﻌﺘﮭﺎ‬‫ﺳ‬ ‫ﺴﻌﺔ‬‫وﻣﺘ‬)50µF(‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫وﺑﻄﺎرﯾ‬
‫ﻗﻄﺒﯿﮭﺎ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬)20V(‫اﺣﺴﺐ‬ ‫اﻟﺪاﺋﺮة‬ ‫وﻏﻠﻖ‬ ‫ﻟﻔﺘﺢ‬ ‫وﻣﻔﺘﺎح‬:
1-‫اﻟﻤﻔﺘﺎح‬ ‫ﻏﻠﻖ‬ ‫ﻟﺤﻈﺔ‬ ‫اﻟﺸﺤﻦ‬ ‫ﻟﺘﯿﺎر‬ ‫اﻻﻋﻈﻢ‬ ‫اﻟﻤﻘﺪار‬.
2-‫اﻟﻤﻔﺘﺎح‬ ‫اﻏﻼق‬ ‫ﻣﻦ‬ ‫ﻣﺪة‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬)‫اﻟﺸﺤﻦ‬ ‫ﻋﻤﻠﯿﺔ‬ ‫اﻛﺘﻤﺎل‬ ‫ﺑﻌﺪ‬.(
3-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
4-‫اﻟﺼﻔﯿﺤﺘﯿﻦ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺒﻌﺪ‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬0.2cm.
5-‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬‫اﻟﻤﺘﺴﻌﺔ‬.
‫اﻟﺤﻞ‬/
1- A1.0
200
20
R
V
I
battery
==
∆
=
2- ∆Vc = ∆Vbattery =20V
3- Q = C . ∆V =50 × 20 =1000µC
4- m/V5000
104.0
20
d
V
E 2
=
×
=
∆
= −
5- J01.010100020
2
1
Q.V
2
1
PE 6
=×××=∆= −
‫ﻣﺜﺎل‬168/‫ﻓﻲ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﯿﺔ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻓﻲ‬ ‫اﻟﻤﻮﺿﺤﺔ‬ ‫اﻟﻤﻌﻠﻮﻣﺎت‬ ‫ﻣﻦ‬
‫اﺣﺴﺐ‬ ‫اﻟﺸﻜﻞ‬:
1-‫ﻟﺘﯿﺎ‬ ‫اﻻﻋﻈﻢ‬ ‫اﻟﻤﻘﺪار‬‫اﻟﻤﻔﺘﺎح‬ ‫اﻏﻼق‬ ‫ﻟﺤﻈﺔ‬ ‫اﻟﺸﺤﻦ‬ ‫ر‬.
2-‫اﻟﻤﻔﺘﺎح‬ ‫اﻏﻼق‬ ‫ﻣﻦ‬ ‫ﻣﺪة‬ ‫ﺑﻌﺪ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺠﮭﺪ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬
)‫اﻟﺸﺤﻦ‬ ‫ﻋﻤﻠﯿﺔ‬ ‫اﻛﺘﻤﺎل‬ ‫ﺑﻌﺪ‬.(
3-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬ ‫أي‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬.
4-‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬.
‫اﻟﺤﻞ‬/
1- A8.0
25
20
R
V
I battery
==
∆
=
2- ∆Vc = ∆Vbattery =20V
3- Q = C . ∆V =80 × 20 =1600µC
4- J101610160020
2
1
Q.V
2
1
PE 36 −−
×=×××=∆=
‫ﻣﺜــﺎل‬169/‫ﮫ‬ ‫ﻣﻘﺎوﻣﺘ‬ ‫ﺼﺒﺎح‬ ‫ﻣ‬ ‫ﻰ‬ ‫ﻋﻠ‬ ‫ﻮي‬ ‫ﺗﺤﺘ‬ ‫ﺮﺑﻂ‬ ‫اﻟ‬ ‫ﺔ‬ ‫ﻣﺘﻮاﻟﯿ‬ ‫ﺔ‬ ‫ﻛﮭﺮﺑﺎﺋﯿ‬ ‫ﺮة‬ ‫داﺋ‬)r =10Ω(‫ﺪارھﺎ‬ ‫ﻣﻘ‬ ‫ﺔ‬ ‫وﻣﻘﺎوﻣ‬
)R=30Ω(‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﻓﺮق‬ ‫ﻣﻘﺪار‬ ‫وﺑﻄﺎرﯾﺔ‬)∆Vbattery=12V(‫ﺴﻌ‬‫ﻣﺘ‬ ‫ﺪاﺋﺮة‬‫اﻟ‬ ‫ﻲ‬‫ﻓ‬ ‫ﺖ‬‫رﺑﻄ‬‫ﺼﻔﯿﺤﺘﯿﻦ‬‫اﻟ‬ ‫ذات‬ ‫ﺔ‬
‫ﺳﻌﺘﮭﺎ‬ ‫اﻟﻤﺘﻮازﯾﺘﯿﻦ‬)20µF. (‫ﺔ‬‫اﻟﻤﺨﺘﺰﻧ‬ ‫ﺔ‬‫اﻟﻜﮭﺮﺑﺎﺋﯿ‬ ‫ﺔ‬‫واﻟﻄﺎﻗ‬ ‫ﺴﻌﺔ‬‫اﻟﻤﺘ‬ ‫ﻔﯿﺤﺘﻲ‬‫ﺻ‬ ‫ﻦ‬‫ﻣ‬ ‫أي‬ ‫ﻲ‬‫ﻓ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫ﻣﻘﺪار‬ ‫ﻣﺎ‬
‫اﻟﻤﺘﺴﻌﺔ‬ ‫رﺑﻂ‬ ‫ﻟﻮ‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻣﺠﺎﻟﮭﺎ‬ ‫ﻓﻲ‬:
1-‫اﻟﻤﺼﺒﺎح‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬.
2-‫ﻧﻔﺴﮭﺎ‬ ‫اﻟﺪاﺋﺮة‬ ‫ﻓﻲ‬ ‫واﻟﺒﻄﺎرﯾﺔ‬ ‫واﻟﻤﻘﺎوﻣﺔ‬ ‫اﻟﻤﺼﺒﺎح‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬)‫ﻓ‬ ‫ﺑﻌﺪ‬‫ﺪاﺋﺮة‬‫اﻟ‬ ‫ﻋﻦ‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺼﻞ‬‫ﻰ‬‫اﻷوﻟ‬‫ﺎ‬‫واﻓﺮاﻏﮭ‬
‫ﺷﺤﻨﺘﮭﺎ‬ ‫ﺟﻤﯿﻊ‬ ‫ﻣﻦ‬.(

75. ‫اﻟﻤﺎدة‬ ‫ﻣﺪرس‬‫ﺗﻮﻣﺎن‬ ‫ﻣﺤﻲ‬ ‫ﺳﻌﯿﺪ‬
74
‫اﻟﺤﻞ‬/
A3.0
40
12
1030
12
rR
V
I
battery
==
+
=
+
∆
=
∆Vr=I . r =0.3 × 10 =3V
Q‫رﺑﻄ‬ ‫اﻟﻤﺘﺴﻌﺔ‬‫ﺖ‬‫ﻟﺬﻟﻚ‬ ‫اﻟﻤﺼﺒﺎح‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬∆Vc = ∆Vr = 3V
Q = C . ∆Vc =20 × 3 =60µF
J109109010603
2
1
Q.V
2
1
PE 566
c
−−−
×=×=×××=∆=
‫اﺗﻤﺎ‬ ‫ﺑﻌﺪ‬‫ﻓﺎن‬ ‫اﻟﻤﺘﺴﻌﺔ‬ ‫ﺷﺤﻦ‬ ‫م‬:2-
∆Vc = ∆Vbattery =12V
Q = C . ∆Vc =20 × 12 =240µC
J101441014401024012
2
1
Q.V
2
1
PE 566
c
−−−
×=×=×××=∆=
‫ﻣﺜﺎل‬170/‫اﻟﻤﻘﺎوﻣﺘﺎن‬ ‫رﺑﻄﺖ‬)r=5Ω,R=10Ω(‫اﻟ‬ ‫ﺎ‬‫رﺑﻄﺘ‬ ‫ﻢ‬‫ﺛ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬‫ﺎ‬‫ﻗﻄﺒﯿﮭ‬ ‫ﯿﻦ‬‫ﺑ‬ ‫ﺪ‬‫اﻟﺠﮭ‬ ‫ﺮق‬‫ﻓ‬ ‫ﺔ‬‫ﺑﻄﺎرﯾ‬ ‫ﻰ‬
30V‫ﻓﻲ‬ ‫اﻟﻤﺨﺘﺰﻧﺔ‬ ‫اﻟﺸﺤﻨﺔ‬ ‫اﺣﺴﺐ‬‫اي‬‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﺘﺴﻌﺔ‬ ‫ﺻﻔﯿﺤﺘﻲ‬ ‫ﻣﻦ‬20µF‫رﺑﻄﺖ‬ ‫ﻟﻮ‬
1-‫اﻟﻤﻘﺎوﻣﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮازي‬ ‫ﻋﻠﻰ‬)5Ω(2-‫اﻟﻤﺠﻤﻮﻋﺔ‬ ‫ﻣﻊ‬ ‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬.
‫اﻟﺤﻞ‬/
C6003020V.CQ
V30VV)2
C2001020V.CQ
VV1052rIV
A2
15
30
510
30
rR
V
I)1
c
Tc
c
cr
T
µ=×=∆=∴
=∆=∆
µ=×=∆=
∆==×==∆
==
+
=
+
∆
=