Question

When a current of 0.2 A is drawn from a battery, then potential difference between its terminals is 20 V and when a current of 2 A is drawn, then the potential difference drops to 16 V. The emf of the battery is:​

Answer 1

Explanation:

When a current of 0.2 A is drawn from a battery, then potential difference between its terminals is 20 V and when a current of 2 A is drawn, then the potential difference drops to 16 V.

Answer 2

Given: A current of 0.2 A is drawn from a battery, then potential difference between its terminals is 20 V and when a current of 2 A is drawn, then the potential difference drops to 16 V.

To find: The emf of the battery.

Solution:

• By the law of conservation of energy, internal resistance of a cell and back emf are related as,

        $V = E - Ir$

• Here, V is the potential difference across its plates, E is the emf, I is the current in the circuit and r is the internal resistance.

• When a current of 0.2 A is drawn from a battery and the potential difference between its terminals is 20 V,

        $20V = E - (0.2A)r$

• When a current of 2 A is drawn and the potential difference drops to 16 V,

        $16V = E - (2A)r$

• On equationg both the equations formed, we get,

        $E = 20.44 V$

Therefore, the emf of the battery is 20.44 V.