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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.
Solution
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,
\[V = E - ir\]
\[ \Rightarrow 5 . 8 = 6 - \frac{6}{R + 1} \times 1\]
\[\frac{6}{R + 1} = 0 . 2\]
\[R + 1 = \frac{6}{0 . 2} = 30\]
\[R = 29 \Omega\]
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