Question

A battery supplies 0.9 A current through a 2 \( \Omega \) resistor and 0.3 A current through a 7 \( \Omega \) resistor when connected one by one. The internal resistance of the battery is:

• A. 20 \( \Omega \)
• B. 120 \( \Omega \)
• C. 10 \( \Omega \)
• D. 0.5 \( \Omega \) \bigskip

Hint:

When resistors are connected to a battery, the current and voltage relations help to find the internal resistance using Ohm's law and the equality of battery voltage in both cases.

Answer 1

The Correct Option is D

Let the internal resistance of the battery be \( r \). 1. When the 2 \( \Omega \) resistor is connected to the battery: \[ V = I_1 \cdot (R + r) = 0.9 \cdot (2 + r) \] 2. When the 7 \( \Omega \) resistor is connected to the battery: \[ V = I_2 \cdot (R + r) = 0.3 \cdot (7 + r) \] Since the battery voltage \( V \) is the same in both cases, we can equate both expressions: \[ 0.9 \cdot (2 + r) = 0.3 \cdot (7 + r) \] Expanding both sides: \[ 1.8 + 0.9r = 2.1 + 0.3r \] Simplifying: \[ 0.9r - 0.3r = 2.1 - 1.8 \] \[ 0.6r = 0.3 \] \[ r = \frac{0.3}{0.6} = 0.5 \, \Omega \] Thus, the internal resistance of the battery is \( r = 0.5 \, \Omega \). \bigskip