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The emf of driver cell in a potentiometer circuit is 10 V. The length of potentiometer wire is 1 m and its resistance is $9\; \Omega$ . A cell of emf 5 V is balanced on 5/9 m length of the wire. The internal resistance of the driver cell is
Option: 1
$1\; \Omega$

Option: 2
$0.1\; \Omega$

Option: 3
$2\; \Omega$

Option: 4
$0.5\; \Omega$

Answers (1)

best_answer

$i=\frac{10}{9+r}$

C is the balance point, So p.d. across AC = 5V

$\Rightarrow \frac{10}{9+r} \times 5 \Omega \quad \Rightarrow \quad r=1 \Omega$

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Divya Prakash Singh

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