Question

A battery of e.m.f. 12 V and internal resistance 0.5 \( \Omega \) is connected to a 9.5 \( \Omega \) resistor through a key. The ratio of potential difference between the two terminals of the battery, when the key is open to that when the key is closed, is:

• A. 1.05
• B. 1
• C. 0.95
• D. 1.1

Hint:

When the key is closed, the total resistance increases due to the internal resistance of the battery, leading to a lower potential difference across the external resistor.

Answer 1

The Correct Option is B

The setup. When the key is open, the battery’s full e.m.f. \( \epsilon \) is available across the resistor. When the key is closed, the total resistance is the sum of the internal resistance of the battery and the external resistor. The potential difference across the terminals of the battery when the key is open is the e.m.f., and when the key is closed, it is reduced due to the internal resistance.
Let \( V_{\text{open}} \) be the potential difference when the key is open, and \( V_{\text{closed}} \) be the potential difference when the key is closed. Using the voltage divider rule, we find: \[ \frac{V_{\text{closed}}}{V_{\text{open}}} = \frac{R}{R + r} \] Where \( R \) is the external resistance (9.5 \( \Omega \)) and \( r \) is the internal resistance (0.5 \( \Omega \)). Final Step. Thus, the ratio of the potential difference is 1, corresponding to option .