Question: Q. 4. Draw a circuit for determining internal resistance of a cell using a potentiometer. Explain the principle on which this method is based.
U] [SQP II 2017]
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Solution:
Ans.
(ii) If key $K_{2}$ is open and key $K_{1}$ is closed and null point is obtained at a distance $l_{1}$ from $Q$,
$$ \begin{equation*} E=K l_{1} \tag{i} \end{equation*} $$
If key $K_{2}$ is closed and key $K_{1}$ is open and null point is obtained at a distance $l_{2}$ from $Q$,
$$ \begin{equation*} V=K l_{2} \tag{ii} \end{equation*} $$
On dividing equation (i) by (ii)
$$ \begin{align*} \frac{E}{V} & =\frac{l_{1}}{l_{2}} \ \frac{R+r}{R} & =\frac{l_{1}}{l_{2}} \ r & =\left(\frac{l_{1}-l_{2}}{l_{2}}\right) R \end{align*} $$
where,
$R=$ shunt resistance in parallel with the cell, $l_{1}$ and $l_{2}=$ balancing length without and with shunt $r=$ internal resistance of the cell
[CBSE Marking Scheme, 2017]
