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:
$$\begin{gathered} E\mathit{-}i\mathit{(}R\mathit{+}r\mathit{)}=0 \\ \Rightarrow i\mathit{=}\frac{E}{R\mathit{+}r} \\ \mathit{=}\frac{6}{R\mathit{+}1} \end{gathered}$$
Also,
$$\begin{gathered} V\mathit{=}E\mathit{-}ir \\ \Rightarrow 5.8=6-\frac{6}{R+1}\times 1 \\ \frac{6}{R+1}=0.2 \\ R+1=\frac{6}{0.2}=30 \\ R=29\mathrm{\Omega } \end{gathered}$$