Electric Cells and Batteries: Solved Example Problems

EXAMPLE 2.17

A battery has an emf of 12 V and connected to a resistor of 3 Ω. The current in the circuit is 3.93 A. Calculate (a) terminal voltage and the internal resistance of the battery (b) power delivered by the battery and power delivered to the resistor

Solution

The given values I = 3.93 A, ξ = 12 V, R = 3 Ω

(a) The terminal voltage of the battery is equal to voltage drop across the resistor

V = IR = 3.93 × 3 = 11.79 V

The internal resistance of the battery,

r =  |ξ –V / V| R = | 12 −11 .79 /11 .79 |  Ã— 3 = 0.05 Ω

The power delivered by the battery P = Iξ = 3.93 × 12 = 47.1 W

The power delivered to the resistor = I2 R = 46.3 W

The remaining power = (47.1 – 46.3) P = 0.772 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I2 r).

EXAMPLE 2.18

From the given circuit,

Find

i) Equivalent emf of the combination

ii) Equivalent internal resistance

iii) Total current

iv) Potential difference across external resistance

v) Potential difference across each cell

Solution

i) Equivalent emf of the combination ξeq = nξ = 4 9 = 36 V

ii) Equivalent internal resistance req = nr = 4 × 0.1 = 0.4 Ω

iii) Total current I = nξ / R +nr

 = [4 ×9] / 10 + ( 4 ×0.1)

= [4 ×9] /  [10 +0 .4] = 36 /10.4

I = 3.46 A

iv) Potential difference across external resistance V = IR = 3.46 × 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.

v) Potential difference across each cell V/n = 34.6/4 = 8 .65V

EXAMPLE 2.19

From the given circuit

Find

i) Equivalent emf

ii) Equivalent internal resistance

iii) Total current (I)

iv) Potential difference across each cell

v) Current from each cell

Solution

i) Equivalent emf ξeq = 5 V

ii) Equivalent internal resistance,

Req = r/n = 0 .5/4 = 0.125Ω

iii) total current, 

I ≈ 0.5 A

iv) Potential difference across each cell V = IR = 0.5 × 10 = 5 V

v) Current from each cell, I ′ = I/n

 I ′ = 0.5/4 = 0.125 A