In order to measure the internal resistance $r_{1}$ of a cell of emf $E$ , a meter bridge of wire resistance $R_{0} = 50 Ω$ , a resistance $R_{0} / 2$ , another cell of emf $E / 2$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l = 72 cm$ , then the value of $r_{1} (in Ω)$ is .

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Solution

We know that total length of meter bridge wire is $100 cm$ .
Given:
$l = 72 cm; x = 100 - l = 28 cm$

Since, $(l + x)$ has resistance of $50 Ω$ ,

$∴ R_{l} = 36 Ω; R_{x} = 14 Ω$

Let current $i$ is flowing in the circuit for the given diagram.

$∴ i = \frac{E}{R_{0} + (R_{0} / 2) + r_{1}} = \frac{E}{75 + r_{1}}$

Now for potential difference across $PQ$ ,

$i (\frac{R_{0}}{2} + R_{x}) = \frac{E}{2}$

$i \times (25 + 14) = \frac{E}{2}$

Putting the value of $i$ , we get

$\frac{E}{75 + r_{1}} \times 39 = \frac{E}{2}$

$\Rightarrow r_{1} = 3 Ω$

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