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POVEZIVANJE
ELEKTRIČNIH OTPORA,
KONDENZATORA
U KOLU ISTOSMJERNE STRUJE
I
VEZE OPRUGA
ISTOSMJERNA ELEKTRIČNA
STRUJA
ISTOSMJERNA ELEKTRIČNA STRUJA JE USMJERENO
KRETANJE NAELEKTRISANIH ČESTICA.
U provodnicima I.reda-metalima čestice su elektroni.
U provodnicima II. reda –tečnostima i gasovima čestice su joni.
ISTOSMJERNA ELEKTRIČNA
STRUJA U METALIMA
𝑅 Ω - el.otpor (1 Ω-om)
𝐼 𝐴 - el. struja (1A –amper)
I 𝑈 𝑉 - el.napon ( 1V- volt)
U G 𝑺 - el.vodljivost (1S-simens)
+ -
Tehnički smjer struje je od + ka -,iako se elektroni kreću od – ka +.
Napon je razlika potencijala: 𝑈 = (𝑉2 − 𝑉1) 𝑉
OHMOV zakon povezuje tri veličine: U=I∙R
ELEKTRIČNA VODLJIVOST je recipročna vrijednost el.otpora:
𝐺 =
1
𝑅
1
Ω
= 1𝑆
R
II.KIRHOFOVI ZAKONI
Elektromotorna sila izvora jednaka je zbiru svih padova napona u
zatvorenom električnom kolu.
𝑅1 𝑅 𝟐
𝟐1
Ovaj zakon je posljedica zakona o
𝑼1 𝑼𝟐 održanju energije.
Posljedica ovog zakona je serijska
U veza električnih otpora.
𝑈 = 𝑈1 + 𝑈2
SERIJSKA-REDNA VEZA
EL.OTPORA
𝑅1 𝑅𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝐼 = 𝑐𝑜𝑛𝑠𝑡.
𝐼 𝑈1 𝑈1 𝑈 = 𝑈1 + 𝑈2
𝐼 ∙ 𝑅 = 𝐼 ∙ 𝑅1 + 𝐼 ∙ 𝑅2 : 𝐼
U R = R1 + R2
1
𝐺
=
1
𝐺1
+
1
𝐺2
U KOLU ISTOSMJERNE STRUJE SERIJSKI SU POVEZANI OTPORI OD 1Ω I
2Ω NA NAPON OD 12V. KOLIKAJE STRUJAU KOLU,VODLJIVOST I PAD
NAPONANASVAKOM OTPORU?
𝑅1 𝑅𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝐼 = 𝑐𝑜𝑛𝑠𝑡.
𝐼 𝑈1 𝑈𝟐 𝑈 = 𝑈1 + 𝑈2
U R = R1 + R2= 1Ω+ 2Ω= 3Ω
𝑅1= 1Ω 𝐼 =
𝑈
𝑅
=
12𝑉
3Ω
= 4𝐴
𝑅𝟐= 2Ω
1
𝐺
=
1
𝐺1
+
1
𝐺2
U=12V 𝐺1 =
1
𝑅1
=
1
1Ω
= 1𝑆
I,G, 𝑈𝟏, 𝑈𝟐=?
𝐺𝟐 =
1
𝑅𝟐
=
1
𝟐Ω
=
1
𝟐
𝑆
𝐺 =
1
𝑅
=
1
3Ω
=
1
3
𝑆
𝐼 = 𝑐𝑜𝑛𝑠𝑡.
𝑈1 = 𝐼 ∙ 𝑅1 = 4𝐴 ∙ 1Ω = 4V
𝑈2 = 𝐼 ∙ 𝑅2 = 4𝐴 ∙ 2Ω = 8V
𝑈 = 𝑈1 + 𝑈2=12V
I.KIRHOFOV ZAKON
Algebarski zbir struja koje ulaze u čvor jednak je algebarskom zbiru
struja koje izlaze iz čvora.
za Ovaj zakon je posljedica zakona o
𝑼1 𝑼𝟐 održanju električnog naboja.
Posljedica ovog zakona je paralelna
U veza električnih otpora.
𝐼 = 𝐼1 + 𝐼2
PARALELNA VEZA
EL.OTPORA
𝑅1 𝐼 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝑈 = 𝑐𝑜𝑛𝑠𝑡.
𝐼𝟏 𝑹𝟐
𝐼 = 𝐼1 + 𝐼2
𝐼𝟐
𝑈
𝑅
=
𝑈
𝑅1
+
𝑈
𝑅2
: 𝑈
𝐼 + -
𝑈
1
𝑅
=
1
𝑅1
+
1
𝑅2
𝐺 = 𝐺1 + 𝐺2
U KOLU ISTOSMJERNE STRUJE PARALELNO SU POVEZANI OTPORI OD 1Ω I
2Ω NANAPONOD 12V. KOLIKAJE STRUJAU KOLU,VODLJIVOST I JAČINASTRUJE
KROZ SVAKI OTPOR?
𝑅1 𝐼 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝑈 = 𝑐𝑜𝑛𝑠𝑡.
𝐼𝟏 𝑹𝟐
𝐼 = 𝐼1 + 𝐼2
𝐼𝟐
1
𝑅
=
1
𝑅1
+
1
𝑅2
𝐼 + -
1
𝑅
=
1
1Ω
+
1
2Ω
𝑅 =
2Ω
3
𝐺 =
1
𝑅
=
3
2Ω
=
3
2
𝑆
𝑈 = 𝑐𝑜𝑛𝑠𝑡.
𝐼 = 𝐼1 + 𝐼2
𝐼 =
𝑈
𝑅
=
12𝑉
2
3
Ω
= 18𝐴
𝐼1 =
𝑈
𝑅1
=
12𝑉
1Ω
= 12𝐴
𝐼2 =
𝑈
𝑅2
=
12𝑉
2Ω
= 6𝐴
ELEKTRIČNI KAPACITET
Električni kapacitet je svojstvo provodnika da primi određenu količinu
elektriciteta.
𝐶 =
𝑄
𝑈
𝐶
𝑉
= 𝐹 𝐶 𝐹 -el. kapacitet (1F-farad)
Q 𝐶 -el. naboj (1C-kulon)
𝑈 𝑉 -el.napom (1V-volt)
Kondenzator stalnog kapaciteta C.
Pločasti kondenzator: 𝐶 =
𝜀𝑆
𝑙
𝐹
Permitivnost sredine: 𝜀 = 𝜀0∙𝜀𝑟 ; 𝜀0 = 8,85 ∙ 10−12 𝐶2
𝑁𝑚2 ; 𝜀𝑟 1
Sferni kondenzator: 𝐶 =
𝑟
𝑘
𝐹
S-površina ploča, l- udaljenost ploča ; r-poluprečnik sfere
SERIJSKA-REDNA VEZA
EL.KONDENZATORA
𝐶1 𝐶𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝑈1 𝑈𝟐 𝑄 = 𝑐𝑜𝑛𝑠𝑡.
𝑈 = 𝑈1 + 𝑈2
𝑄
𝐶
=
𝑄
𝐶1
+
𝑄
𝐶2
: 𝑄
1
𝐶
=
1
𝐶1
+
1
𝐶2
PARALELNA VEZA ELEKTRIČNIH
KONDENZATORA
𝐶1, 𝑄1 𝑈 = 𝑐𝑜𝑛𝑠𝑡.
𝑄 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝐶𝟐, 𝑄𝟐 𝑄 = 𝑄1 + 𝑄2
𝐶𝑈 = 𝐶1𝑈 + 𝐶2𝑈 ∶ 𝑼
+ -
U 𝐶 = 𝐶1 + 𝐶2
ELASTIČNA SILA-OPRUGE
Harmonijske oscilacije se vrše pod
uticajem sile koja je proporcionalna sa
elongacijom x.
Silu koja djeluje na oprugu –elastična
sila F=- kx
k-konstanta proporcionalnosti
x-elongacija,udaljenost od ravnotežnog
položaja
„-“ – sila je usmjerena KA ravnotežnom
položaju, tj. suprotnog je smjera od elongacije.
SERIJSKA-REDNA VEZA
OPRUGA
𝐹 = 𝑐𝑜𝑛𝑠𝑡.
𝑥 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝑘1 𝑥 = 𝑥1 + 𝑥2
𝐹
𝑘
=
𝐹
𝑘1
+
𝐹
𝑘2
:𝐹
𝑘2
1
𝑘
=
1
𝑘1
+
1
𝑘2
F
PARALELNA VEZA OPRUGA
𝐹 ≠ 𝑐𝑜𝑛𝑠𝑡.
𝑥 = 𝑐𝑜𝑛𝑠𝑡.
𝑘1 𝑘𝟐 𝐹 = 𝐹1 + 𝐹2
𝑘𝑥 = 𝑘1𝑥 + 𝑘2𝑥 ∶ 𝑥
F 𝑘 = 𝑘1 + 𝑘2
INTERESANTNO
AKO PAŽLJIVO POGLEDATE PREDHODNE SLAJDOVE
MOŽETE UOČITI SLJEDEĆE:
1. Konačne formule za serijsku vezu kondenzatora i
opruga imaju ISTI OBLIK ,tj. zbir recipročnih vrijednosti
datih veličina,a kod el.otpora to vrijedi za paralelnu
vezu.
2. Konačne formule za paralelnu vezu kondenzatora i
opruga imaju ISTI OBLIK ,tj. zbir algebarskih
vrijednosti datih veličina,a kod el.otpora to vrijedi
za serijsku vezu.
KOMBINOVANA VEZA
OTPORA
𝐼𝑎 𝑅1 𝑅𝟐 𝑅𝟒
I
I 𝐼𝟑 𝑅𝟑 I
U
Kod kombinovane veze PRVO se izračuna vrijednost
paralelne veze.
Ako u sklopu te veze ima serijska ,tada se prvo
izračuna vrijednost serijske,pa se onda izračuna
vrijednost paralelne.
Cilj je dobiti „čistu“ serijsku vezu.
SVI otpori su jednaki i imaju vrijednost 𝑅 = 1Ω,U= 10V.
1. Ra = R1 + R2= 1Ω+ 1Ω= 2Ω
Ra
R4
R3
𝑅𝒃
U
2.
1
𝑅𝑏
=
1
𝑅𝑎
+
1
𝑅3
1
𝑅𝑏
=
1
2Ω
+
1
1Ω
1
𝑅𝑏
=
2+1
2Ω
𝑅𝑏 =
2Ω
3
3. 𝑅𝑏 𝑅𝟒
I
U
𝑅 = 𝑅𝑏 + 𝑅𝟒
𝑅 =
2Ω
3
+ 1Ω
𝑅 =
5Ω
3
𝐼 =
𝑈
𝑅
=
10𝑉
5Ω
3
𝐼 = 6𝐴
PAD NAPONA I JAČINA STRUJE NA SVAKOM
OTPORNIKU:
𝑈 = 𝑈𝑏 + 𝑈4
𝑈𝑏 = 𝑅𝑏𝐼 =
2Ω
3
∙ 6𝐴 = 4𝑉
𝑈4 = 𝑈 − 𝑈𝑏 = 6𝑉
𝑅𝑏 je ukupni otpor paralelne veze 𝑅𝒂 i 𝑅𝟑, pa je napon te veze 4V,
dok je napon na 𝑅𝟒, 6V.
Kroz otpor 𝑅𝟒 protiče ukupna struja od 6A.
Kroz otpor 𝑅𝒃 protiče struja 𝐼𝑏 =
𝑈𝑏
𝑅𝑏
=
4𝑉
2
3
Ω
= 6𝐴 i grana se i protiče
kroz otpor 𝑅𝑎 ,koji čini serijska veza otpora 𝑅𝟏 i 𝑅𝟐 i vrijednost
struje kroz te otpore je: 𝐼𝑎 =
4𝑉
2Ω
= 2𝐴
Kroz otpor 𝑅3 protiče struja: 𝐼3 =
𝑈𝑏
𝑅3
=
4𝑉
1Ω
= 4𝐴
𝐼 = 𝐼𝑎 + 𝐼3 = 2𝐴 + 4𝐴 = 6𝐴
Pad napona na 𝑅1 i 𝑅2 je :𝑈1 = 𝑈2 = 𝐼𝑎∙ 𝑅1 = 𝐼𝑎∙ 𝑅2 = 2𝐴 ∙ 1Ω = 2V
Prof.fizike Mirjana Drmić-Malić
Zadatke za kondenzatore i opruge uradite na isti način
kao i za otpore i obavezno uvrštavajte mjerne jedinice tih
veličina.
Ljepotu fizike čine mjerne jedinice kada se uredno
uvrste,skrate... i dobije fizikalni rezultat.
PPT PREZENTACIJE:
Mjerenja u fizici---
https://www.slideshare.net/esterahadasa1/mjerenja-u-fizici
Abeceda fizikeppt---
https://www.slideshare.net/esterahadasa1/abeceda-fizikeppt-
234475459
Skalarne i vektorske veličine u fizici---
https://www.slideshare.net/esterahadasa1/skalarne-i-
vektorske-veliine-u-fizici-234475582
SRETNO!

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Povezivanje R,C,k

  • 1. POVEZIVANJE ELEKTRIČNIH OTPORA, KONDENZATORA U KOLU ISTOSMJERNE STRUJE I VEZE OPRUGA
  • 2. ISTOSMJERNA ELEKTRIČNA STRUJA ISTOSMJERNA ELEKTRIČNA STRUJA JE USMJERENO KRETANJE NAELEKTRISANIH ČESTICA. U provodnicima I.reda-metalima čestice su elektroni. U provodnicima II. reda –tečnostima i gasovima čestice su joni.
  • 3. ISTOSMJERNA ELEKTRIČNA STRUJA U METALIMA 𝑅 Ω - el.otpor (1 Ω-om) 𝐼 𝐴 - el. struja (1A –amper) I 𝑈 𝑉 - el.napon ( 1V- volt) U G 𝑺 - el.vodljivost (1S-simens) + - Tehnički smjer struje je od + ka -,iako se elektroni kreću od – ka +. Napon je razlika potencijala: 𝑈 = (𝑉2 − 𝑉1) 𝑉 OHMOV zakon povezuje tri veličine: U=I∙R ELEKTRIČNA VODLJIVOST je recipročna vrijednost el.otpora: 𝐺 = 1 𝑅 1 Ω = 1𝑆 R
  • 4. II.KIRHOFOVI ZAKONI Elektromotorna sila izvora jednaka je zbiru svih padova napona u zatvorenom električnom kolu. 𝑅1 𝑅 𝟐 𝟐1 Ovaj zakon je posljedica zakona o 𝑼1 𝑼𝟐 održanju energije. Posljedica ovog zakona je serijska U veza električnih otpora. 𝑈 = 𝑈1 + 𝑈2
  • 5. SERIJSKA-REDNA VEZA EL.OTPORA 𝑅1 𝑅𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝐼 = 𝑐𝑜𝑛𝑠𝑡. 𝐼 𝑈1 𝑈1 𝑈 = 𝑈1 + 𝑈2 𝐼 ∙ 𝑅 = 𝐼 ∙ 𝑅1 + 𝐼 ∙ 𝑅2 : 𝐼 U R = R1 + R2 1 𝐺 = 1 𝐺1 + 1 𝐺2
  • 6. U KOLU ISTOSMJERNE STRUJE SERIJSKI SU POVEZANI OTPORI OD 1Ω I 2Ω NA NAPON OD 12V. KOLIKAJE STRUJAU KOLU,VODLJIVOST I PAD NAPONANASVAKOM OTPORU? 𝑅1 𝑅𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝐼 = 𝑐𝑜𝑛𝑠𝑡. 𝐼 𝑈1 𝑈𝟐 𝑈 = 𝑈1 + 𝑈2 U R = R1 + R2= 1Ω+ 2Ω= 3Ω 𝑅1= 1Ω 𝐼 = 𝑈 𝑅 = 12𝑉 3Ω = 4𝐴 𝑅𝟐= 2Ω 1 𝐺 = 1 𝐺1 + 1 𝐺2 U=12V 𝐺1 = 1 𝑅1 = 1 1Ω = 1𝑆 I,G, 𝑈𝟏, 𝑈𝟐=?
  • 7. 𝐺𝟐 = 1 𝑅𝟐 = 1 𝟐Ω = 1 𝟐 𝑆 𝐺 = 1 𝑅 = 1 3Ω = 1 3 𝑆 𝐼 = 𝑐𝑜𝑛𝑠𝑡. 𝑈1 = 𝐼 ∙ 𝑅1 = 4𝐴 ∙ 1Ω = 4V 𝑈2 = 𝐼 ∙ 𝑅2 = 4𝐴 ∙ 2Ω = 8V 𝑈 = 𝑈1 + 𝑈2=12V
  • 8. I.KIRHOFOV ZAKON Algebarski zbir struja koje ulaze u čvor jednak je algebarskom zbiru struja koje izlaze iz čvora. za Ovaj zakon je posljedica zakona o 𝑼1 𝑼𝟐 održanju električnog naboja. Posljedica ovog zakona je paralelna U veza električnih otpora. 𝐼 = 𝐼1 + 𝐼2
  • 9. PARALELNA VEZA EL.OTPORA 𝑅1 𝐼 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝑈 = 𝑐𝑜𝑛𝑠𝑡. 𝐼𝟏 𝑹𝟐 𝐼 = 𝐼1 + 𝐼2 𝐼𝟐 𝑈 𝑅 = 𝑈 𝑅1 + 𝑈 𝑅2 : 𝑈 𝐼 + - 𝑈 1 𝑅 = 1 𝑅1 + 1 𝑅2 𝐺 = 𝐺1 + 𝐺2
  • 10. U KOLU ISTOSMJERNE STRUJE PARALELNO SU POVEZANI OTPORI OD 1Ω I 2Ω NANAPONOD 12V. KOLIKAJE STRUJAU KOLU,VODLJIVOST I JAČINASTRUJE KROZ SVAKI OTPOR? 𝑅1 𝐼 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝑈 = 𝑐𝑜𝑛𝑠𝑡. 𝐼𝟏 𝑹𝟐 𝐼 = 𝐼1 + 𝐼2 𝐼𝟐 1 𝑅 = 1 𝑅1 + 1 𝑅2 𝐼 + - 1 𝑅 = 1 1Ω + 1 2Ω 𝑅 = 2Ω 3 𝐺 = 1 𝑅 = 3 2Ω = 3 2 𝑆
  • 11. 𝑈 = 𝑐𝑜𝑛𝑠𝑡. 𝐼 = 𝐼1 + 𝐼2 𝐼 = 𝑈 𝑅 = 12𝑉 2 3 Ω = 18𝐴 𝐼1 = 𝑈 𝑅1 = 12𝑉 1Ω = 12𝐴 𝐼2 = 𝑈 𝑅2 = 12𝑉 2Ω = 6𝐴
  • 12. ELEKTRIČNI KAPACITET Električni kapacitet je svojstvo provodnika da primi određenu količinu elektriciteta. 𝐶 = 𝑄 𝑈 𝐶 𝑉 = 𝐹 𝐶 𝐹 -el. kapacitet (1F-farad) Q 𝐶 -el. naboj (1C-kulon) 𝑈 𝑉 -el.napom (1V-volt) Kondenzator stalnog kapaciteta C. Pločasti kondenzator: 𝐶 = 𝜀𝑆 𝑙 𝐹 Permitivnost sredine: 𝜀 = 𝜀0∙𝜀𝑟 ; 𝜀0 = 8,85 ∙ 10−12 𝐶2 𝑁𝑚2 ; 𝜀𝑟 1 Sferni kondenzator: 𝐶 = 𝑟 𝑘 𝐹 S-površina ploča, l- udaljenost ploča ; r-poluprečnik sfere
  • 13. SERIJSKA-REDNA VEZA EL.KONDENZATORA 𝐶1 𝐶𝟐 𝑈 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝑈1 𝑈𝟐 𝑄 = 𝑐𝑜𝑛𝑠𝑡. 𝑈 = 𝑈1 + 𝑈2 𝑄 𝐶 = 𝑄 𝐶1 + 𝑄 𝐶2 : 𝑄 1 𝐶 = 1 𝐶1 + 1 𝐶2
  • 14. PARALELNA VEZA ELEKTRIČNIH KONDENZATORA 𝐶1, 𝑄1 𝑈 = 𝑐𝑜𝑛𝑠𝑡. 𝑄 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝐶𝟐, 𝑄𝟐 𝑄 = 𝑄1 + 𝑄2 𝐶𝑈 = 𝐶1𝑈 + 𝐶2𝑈 ∶ 𝑼 + - U 𝐶 = 𝐶1 + 𝐶2
  • 15. ELASTIČNA SILA-OPRUGE Harmonijske oscilacije se vrše pod uticajem sile koja je proporcionalna sa elongacijom x. Silu koja djeluje na oprugu –elastična sila F=- kx k-konstanta proporcionalnosti x-elongacija,udaljenost od ravnotežnog položaja „-“ – sila je usmjerena KA ravnotežnom položaju, tj. suprotnog je smjera od elongacije.
  • 16. SERIJSKA-REDNA VEZA OPRUGA 𝐹 = 𝑐𝑜𝑛𝑠𝑡. 𝑥 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝑘1 𝑥 = 𝑥1 + 𝑥2 𝐹 𝑘 = 𝐹 𝑘1 + 𝐹 𝑘2 :𝐹 𝑘2 1 𝑘 = 1 𝑘1 + 1 𝑘2 F
  • 17. PARALELNA VEZA OPRUGA 𝐹 ≠ 𝑐𝑜𝑛𝑠𝑡. 𝑥 = 𝑐𝑜𝑛𝑠𝑡. 𝑘1 𝑘𝟐 𝐹 = 𝐹1 + 𝐹2 𝑘𝑥 = 𝑘1𝑥 + 𝑘2𝑥 ∶ 𝑥 F 𝑘 = 𝑘1 + 𝑘2
  • 18. INTERESANTNO AKO PAŽLJIVO POGLEDATE PREDHODNE SLAJDOVE MOŽETE UOČITI SLJEDEĆE: 1. Konačne formule za serijsku vezu kondenzatora i opruga imaju ISTI OBLIK ,tj. zbir recipročnih vrijednosti datih veličina,a kod el.otpora to vrijedi za paralelnu vezu. 2. Konačne formule za paralelnu vezu kondenzatora i opruga imaju ISTI OBLIK ,tj. zbir algebarskih vrijednosti datih veličina,a kod el.otpora to vrijedi za serijsku vezu.
  • 19. KOMBINOVANA VEZA OTPORA 𝐼𝑎 𝑅1 𝑅𝟐 𝑅𝟒 I I 𝐼𝟑 𝑅𝟑 I U Kod kombinovane veze PRVO se izračuna vrijednost paralelne veze. Ako u sklopu te veze ima serijska ,tada se prvo izračuna vrijednost serijske,pa se onda izračuna vrijednost paralelne. Cilj je dobiti „čistu“ serijsku vezu. SVI otpori su jednaki i imaju vrijednost 𝑅 = 1Ω,U= 10V.
  • 20. 1. Ra = R1 + R2= 1Ω+ 1Ω= 2Ω Ra R4 R3 𝑅𝒃 U 2. 1 𝑅𝑏 = 1 𝑅𝑎 + 1 𝑅3 1 𝑅𝑏 = 1 2Ω + 1 1Ω 1 𝑅𝑏 = 2+1 2Ω 𝑅𝑏 = 2Ω 3
  • 21. 3. 𝑅𝑏 𝑅𝟒 I U 𝑅 = 𝑅𝑏 + 𝑅𝟒 𝑅 = 2Ω 3 + 1Ω 𝑅 = 5Ω 3 𝐼 = 𝑈 𝑅 = 10𝑉 5Ω 3 𝐼 = 6𝐴
  • 22. PAD NAPONA I JAČINA STRUJE NA SVAKOM OTPORNIKU: 𝑈 = 𝑈𝑏 + 𝑈4 𝑈𝑏 = 𝑅𝑏𝐼 = 2Ω 3 ∙ 6𝐴 = 4𝑉 𝑈4 = 𝑈 − 𝑈𝑏 = 6𝑉 𝑅𝑏 je ukupni otpor paralelne veze 𝑅𝒂 i 𝑅𝟑, pa je napon te veze 4V, dok je napon na 𝑅𝟒, 6V. Kroz otpor 𝑅𝟒 protiče ukupna struja od 6A. Kroz otpor 𝑅𝒃 protiče struja 𝐼𝑏 = 𝑈𝑏 𝑅𝑏 = 4𝑉 2 3 Ω = 6𝐴 i grana se i protiče kroz otpor 𝑅𝑎 ,koji čini serijska veza otpora 𝑅𝟏 i 𝑅𝟐 i vrijednost struje kroz te otpore je: 𝐼𝑎 = 4𝑉 2Ω = 2𝐴
  • 23. Kroz otpor 𝑅3 protiče struja: 𝐼3 = 𝑈𝑏 𝑅3 = 4𝑉 1Ω = 4𝐴 𝐼 = 𝐼𝑎 + 𝐼3 = 2𝐴 + 4𝐴 = 6𝐴 Pad napona na 𝑅1 i 𝑅2 je :𝑈1 = 𝑈2 = 𝐼𝑎∙ 𝑅1 = 𝐼𝑎∙ 𝑅2 = 2𝐴 ∙ 1Ω = 2V Prof.fizike Mirjana Drmić-Malić
  • 24. Zadatke za kondenzatore i opruge uradite na isti način kao i za otpore i obavezno uvrštavajte mjerne jedinice tih veličina. Ljepotu fizike čine mjerne jedinice kada se uredno uvrste,skrate... i dobije fizikalni rezultat. PPT PREZENTACIJE: Mjerenja u fizici--- https://www.slideshare.net/esterahadasa1/mjerenja-u-fizici Abeceda fizikeppt--- https://www.slideshare.net/esterahadasa1/abeceda-fizikeppt- 234475459 Skalarne i vektorske veličine u fizici--- https://www.slideshare.net/esterahadasa1/skalarne-i- vektorske-veliine-u-fizici-234475582 SRETNO!