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# Electric Cells and Batteries: Solved Example Problems

Physics : Current Electricity: Electric Cells and Batteries: Solved Example Problems

Determination of internal resistance: Solved Example Problems

EXAMPLE 2.17

A battery has an emf of 12 V and connected to a resistor of 3ŌĆå╬®. The current in the circuit is 3.93ŌĆåA. 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, ╬Š = 12 V, R = 3 Ōä”

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

V = IR = 3.93 ├Ś 3 = 11.79 V

The internal resistance of the battery,

r =  |╬Š ŌĆōV / V| R = | 12 ŌłÆ11 .79 /11 .79 |  ├Ś 3 = 0.05 ╬®

The power delivered by the battery P = I╬Š = 3.93 ├Ś 12 = 47.1 W

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

The remaining power = (47.1ŌĆåŌĆōŌĆå46.3) P = 0.772 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I2 r).

## Cells in series: Solved Example Problems

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 ╬Šeq = n╬Š = 4 9 = 36 V

ii) Equivalent internal resistance req = nr = 4 ├Ś 0.1 = 0.4 ╬®

iii) Total current I = n╬Š / R +nr

= [4 ├Ś9] / 10 + ( 4 ├Ś0.1)

= [4 ├Ś9] /  [10 +0 .4] = 36 /10.4

I = 3.46 A

iv) Potential difference across external resistance V = IR = 3.46 ├Ś 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

## Cells in parallel: Solved Example Problems

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 ╬Šeq = 5 V

ii) Equivalent internal resistance,

Req = r/n = 0 .5/4 = 0.125╬®

iii) total current, I Ōēł 0.5 A

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

v) Current from each cell, I ŌĆ▓ = I/n

I ŌĆ▓ = 0.5/4 = 0.125 A

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12th Physics : Current Electricity : Electric Cells and Batteries: Solved Example Problems |