SATHEE: current-electricity Question 56

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Question: Q. 10. A potentiometer wire of length $1 m$ has a resistance $10 Ω$ . It is connected to $6 V$ battery in series with a resistance of $5 Ω$ . Determine the emf of the primary cell which gives a balance point at $40 cm$ .

A [Delhi I, II, III 2014]

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Solution:

Ans. Current flowing in potentiometer wire,

$I = \frac{V}{R + R^{'}}$

where, $R$ is the resistance of potentiometer wire and $R^{'} = 5 Ω$

$I = \frac{6}{10 + 5} A = 0.4 A 1 / 2$

Potential drop across the potentiometer wire

$\begin{aligned} V & = I R & = 0.4 \times 10 V = 4.0 V 1 / 2 \end{aligned}$

Potential gradient $k = V / l = 4.0 V / m 1 / 2$

$∴$ Unknown emf of the cell $(E) = k l^{'} 1 / 2$

$\begin{aligned} = 4.0 \times 0.4 V & = 1.6 V \end{aligned}$

[CBSE Marking Scheme, 2014]

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