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Electricity
Class – 10
Aditya Mishra
Email: adityakumar2009@gmail.com
Calendula Learning
RESISTANCE IN SERIES
R1 R2 R3
• Current through (I) the resistances is same
• Potential difference (V) across each resistor is different
I I
Calendula Learning
1 2 3
Total potential difference:
V = V + V + V
V = I R
1 1
V = I R
2 2
V = I R
3 3
V = I R
1 2 3
I R = I R + I R + I R
 
1 2 3
I R = I R + R + R
 
1 2 3
R = R + R + R
eq 1 2 3
R = R + R + R
------- (1)
------- (2)
------- (3)
------- (4)
------- (5)
Substituting equation 2,3,4 and 5 in equation 1
------- (6)
RESISTANCE IN SERIES
Calendula Learning
eq
R = R + R + R
eq
R = 3R
eq
R = R + R + R + .......
eq
R = n R
---------------- (7)
---------------- (8)
• Equivalent resistance for three resistances each having a value of
R in series combination :
• Equivalent resistance for ‘n’ resistances each having a value of
R in series combination :
RESISTANCE IN SERIES
Calendula Learning
RESISTANCE IN PARALLEL
• Potential difference ( V) across each resistance is same
• Current flowing ( I ) through each resistance is different
A R2
R3
R1
B
V
I1
I2
I3
I
Calendula Learning
EQUIVALENT RESISTANCE FOR PARALLEL
COMBINATION
1 2 3
Total current (I ) = I + I + I
According to ohms law,
V= I R
 1
1
V
I =
R

2
2
V
I =
R

3
3
V
I =
R

1 2 3
I = I + I + I
1 2 3
V V V V
= + +
R R R R
1 2 3
V 1 1 1
= V + +
R R R R
 
 
 
eq 1 2 3
1 1 1 1
= + +
R R R R
(2)

(3)

(4)

Substituting eq. 2, 3, 4 in eq. 5
(6)

(1)

R2
R3
R1
A B
Calendula Learning
PROBLEMS
eq 1 2 3
1 1 1 1
= + +
R R R R
eq 1 2
1 1 1
= +
R R R
2 1
eq 1 2 3
R +R
1 1
= +
R R R R
 
3 2 1 1 2
eq 1 2 3
R R +R R R
1
=
R R R R

3 2 3 1 1 2
eq 1 2 3
R R + R R R R
1
=
R R R R

1 2 3
eq
1 2 2 3 3 1
R R R
R =
R R + R R R R

eq
1 1 1 1 2
= + 2
R R R R R
 
 
 
 
1 2
eq
1 2
R R
R =
R + R
eq
1 1 1 1
= + ........
R R R R
n
 
   
 
eq
R
R
n

• For two resistances connected in parallel,
• For two resistances having same resistance
connected in parallel,
• For n resistances connected in parallel
Calendula Learning
Q.1 Find the equivalent resistance for the parallel combination of
two resistor 5 ohm and 10 ohm?
1
R 5
 
eq
R ?

2
R 10
 
1 2
eq
1 2
R R
R
R +R

We know that for resistances
Connected in parallel ,
eq 1 2
1 1 1
R R R
 
 
 
eq
5 10
R
5 + 10


eq
50 10
R =
15 3
 
eq
R 3.33
 
Given:
To Find :
Solution :
Ans :
PROBLEMS
Calendula Learning
Q.2 Three resistances of 2 ohm, 3 ohm and 5 ohm are connected
together. How the resistance should be connected to obtain
maximum resistance?
(a)
(b) (c)
Ans :
PROBLEMS
Calendula Learning
1
R = 2 Ω
2
R = 3 Ω
3
R = 5 Ω
R1 is parallel to R2
eq 1 2
1 1 1
= +
R R R
1 2
eq
1 2
R R
R =
R + R
eq
2 3
R =
2 + 3

eq
6
R =
5

3
R = 5 Ω
eq
6
R = Ω
5
(b)
eq1 eq 3
R = R + R
eq1
6
R = + 5
5
eq1
6 25
R =
5


eq1
31
R =
5

PROBLEMS
Calendula Learning

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03_Electricity.pptx

  • 1. Electricity Class – 10 Aditya Mishra Email: adityakumar2009@gmail.com Calendula Learning
  • 2. RESISTANCE IN SERIES R1 R2 R3 • Current through (I) the resistances is same • Potential difference (V) across each resistor is different I I Calendula Learning
  • 3. 1 2 3 Total potential difference: V = V + V + V V = I R 1 1 V = I R 2 2 V = I R 3 3 V = I R 1 2 3 I R = I R + I R + I R   1 2 3 I R = I R + R + R   1 2 3 R = R + R + R eq 1 2 3 R = R + R + R ------- (1) ------- (2) ------- (3) ------- (4) ------- (5) Substituting equation 2,3,4 and 5 in equation 1 ------- (6) RESISTANCE IN SERIES Calendula Learning
  • 4. eq R = R + R + R eq R = 3R eq R = R + R + R + ....... eq R = n R ---------------- (7) ---------------- (8) • Equivalent resistance for three resistances each having a value of R in series combination : • Equivalent resistance for ‘n’ resistances each having a value of R in series combination : RESISTANCE IN SERIES Calendula Learning
  • 5. RESISTANCE IN PARALLEL • Potential difference ( V) across each resistance is same • Current flowing ( I ) through each resistance is different A R2 R3 R1 B V I1 I2 I3 I Calendula Learning
  • 6. EQUIVALENT RESISTANCE FOR PARALLEL COMBINATION 1 2 3 Total current (I ) = I + I + I According to ohms law, V= I R  1 1 V I = R  2 2 V I = R  3 3 V I = R  1 2 3 I = I + I + I 1 2 3 V V V V = + + R R R R 1 2 3 V 1 1 1 = V + + R R R R       eq 1 2 3 1 1 1 1 = + + R R R R (2)  (3)  (4)  Substituting eq. 2, 3, 4 in eq. 5 (6)  (1)  R2 R3 R1 A B Calendula Learning
  • 7. PROBLEMS eq 1 2 3 1 1 1 1 = + + R R R R eq 1 2 1 1 1 = + R R R 2 1 eq 1 2 3 R +R 1 1 = + R R R R   3 2 1 1 2 eq 1 2 3 R R +R R R 1 = R R R R  3 2 3 1 1 2 eq 1 2 3 R R + R R R R 1 = R R R R  1 2 3 eq 1 2 2 3 3 1 R R R R = R R + R R R R  eq 1 1 1 1 2 = + 2 R R R R R         1 2 eq 1 2 R R R = R + R eq 1 1 1 1 = + ........ R R R R n         eq R R n  • For two resistances connected in parallel, • For two resistances having same resistance connected in parallel, • For n resistances connected in parallel Calendula Learning
  • 8. Q.1 Find the equivalent resistance for the parallel combination of two resistor 5 ohm and 10 ohm? 1 R 5   eq R ?  2 R 10   1 2 eq 1 2 R R R R +R  We know that for resistances Connected in parallel , eq 1 2 1 1 1 R R R       eq 5 10 R 5 + 10   eq 50 10 R = 15 3   eq R 3.33   Given: To Find : Solution : Ans : PROBLEMS Calendula Learning
  • 9. Q.2 Three resistances of 2 ohm, 3 ohm and 5 ohm are connected together. How the resistance should be connected to obtain maximum resistance? (a) (b) (c) Ans : PROBLEMS Calendula Learning
  • 10. 1 R = 2 Ω 2 R = 3 Ω 3 R = 5 Ω R1 is parallel to R2 eq 1 2 1 1 1 = + R R R 1 2 eq 1 2 R R R = R + R eq 2 3 R = 2 + 3  eq 6 R = 5  3 R = 5 Ω eq 6 R = Ω 5 (b) eq1 eq 3 R = R + R eq1 6 R = + 5 5 eq1 6 25 R = 5   eq1 31 R = 5  PROBLEMS Calendula Learning