Question

The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω drops to 5.8 V when connected across an external resistor. Find the resistance of the external resistor.

Answer 1

Given:-

Emf of the battery, E = 6 V

Internal resistance, r = 1 Ω

Potential difference, V = 5.8 V

Let R be the resistance of the external resistor.

Applying KVL in the above circuit, we get:-

\[E - i(R + r) = 0\]

\[ \Rightarrow i = \frac{E}{R + r}\]

\[ = \frac{6}{R + 1}\]

Also,

\[V = E - ir\]

\[ \Rightarrow 5 . 8 = 6 - \frac{6}{R + 1} \times 1\]

\[\frac{6}{R + 1} = 0 . 2\]

\[R + 1 = \frac{6}{0 . 2} = 30\]

\[R = 29 \Omega\]