Question: Q. 10. A potentiometer wire of length $1 \mathrm{~m}$ has a resistance $10 \Omega$. It is connected to $6 \mathrm{~V}$ battery in series with a resistance of $5 \Omega$. Determine the emf of the primary cell which gives a balance point at $40 \mathrm{~cm}$.
A [Delhi I, II, III 2014]
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Solution:
Ans. Current flowing in potentiometer wire,
$$ I=\frac{V}{R+R^{\prime}} $$
where, $R$ is the resistance of potentiometer wire and $R^{\prime}=5 \Omega$
$$ I=\frac{6}{10+5} \mathrm{~A}=0.4 \mathrm{~A} \quad 1 / 2 $$
Potential drop across the potentiometer wire
$$ \begin{aligned} V & =I R \ & =0.4 \times 10 \mathrm{~V}=4.0 \mathrm{~V} \quad 1 / 2 \end{aligned} $$
Potential gradient $k=V / l=4.0 \mathrm{~V} / \mathrm{m} \quad 1 / 2$
$\therefore$ Unknown emf of the cell $(E)=k l^{\prime} \quad 1 / 2$
$$ \begin{align*} & =4.0 \times 0.4 \mathrm{~V} \ & =1.6 \mathrm{~V} \end{align*} $$
[CBSE Marking Scheme, 2014]
