Question

The potential difference across the terminals of a battery is 8.5V when there is a current of 3A in the circuit form the negative to the positive terminal. When the current is 2A in the reverse direction, the potential difference becomes 11V.
(i) What is the internal resistance of the battery?
(ii) What is the e.m.f of the battery?

Answer 1

Answer on Question #73946, Physics / Electric Circuits

The potential difference across the terminals of a battery is 8.5V when there is a current of 3A in the circuit form the negative to the positive terminal. When the current is 2A in the reverse direction, the potential difference becomes 11V.

(i) What is the internal resistance of the battery?

(ii) What is the e.m.f of the battery?

Solution:

Let A, B represent the terminals of cell

$$V _ {A} - E + I r = V _ {B}$$ $$V _ {A} - V _ {B} = E - I r$$

In first case

$$V _ {A} - V _ {B} = 8. 5 \mathrm {V}$$ $$I = 3 A$$

So,

$$8. 5 = E - 3 r$$

In second case

$$V _ {A} - E - I r = V _ {B}$$ $$V _ {A} - V _ {B} = E + I r$$

So,

$$1 1 = E + 2 r$$

We have system of equations (1) and (2).

Difference between (2) and (1)

$$1 1 - 8. 5 = E + 2 r - (E - 3 r)$$ $$2. 5 = 5 r$$

Thus, the internal resistance is

$$r = \frac {2 . 5}{5} = 0. 5 \Omega$$

The e.m.f is

$$E = 8. 5 + 3 r = 8. 5 + 3 \times 0. 5 = 1 0 V$$

Answer: (i) $0.5\Omega$ ; (ii) $10V$

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