Question

The circuit diagram shows that resistors $2\Omega$, $4\Omega$ and $R\Omega$ connected to a battery of e.m.f. 2 V and internal resistance $3\Omega$. A main current of 0.25 A flows through the circuit. The p.d. across the internal resistance of the cell is :

• A. 0.75 V
• B. 2 V
• C. 0.5 V
• D. 1 V

Answer 1

The correct option is A 0.75 V

In the question it is given that the current in the circuit is 0.25 A and the internal resistance is given as 3 ohms.
Hence, the potential difference across the resistance is given as $V=IR=0.25A\times 3\Omega =0.75V$.
Hence, p.d. across the internal resistance of the cell is 0.75 V.