In a Wheatstone bridge, all four arms have equal resistance $R$ . If the resistance of the galvanometer arm is also $R$ , the equivalent resistance of the combination is

$R$

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$2 R$

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$\frac{R}{2}$

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$\frac{R}{4}$

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Solution

The correct option is A
$R$

Step 1: Given Data
Let the Wheatstone bridge consist of four resistances $R_{1}, R_{2}, R_{3}$ and $R_{4}$ as shown in the given figure.
It also consists of a galvanometer or resistance $R_{g}$ .
From the question we have,
$R_{1} = R_{2} = R_{3} = R_{4} = R_{g} = R$
Step 2: Calculate the equivalent resistance
Since $R_{1}$ and $R_{2}$ are in series,
$R_{C A D} = R_{1} + R_{2}$
Since $R_{3}$ and $R_{4}$ are in series,
$R_{C B D} = R_{3} + R_{4}$
Assume the battery resistance $r_{i} = 0$ we can write,
$R_{C A D}$ and $R_{C B D}$ are in parallel.
$∴ \frac{1}{R_{e q}} = \frac{1}{R_{1} + R_{2}} + \frac{1}{R_{3} + R_{4}}$
$R_{e q} = \frac{(R_{1} + R_{2}) (R_{3} + R_{4})}{(R_{1} + R_{2}) + (R_{3} + R_{4})}$
$= \frac{(R + R) (R + R)}{(R + R) + (R + R)}$
$= \frac{4 R^{2}}{4 R}$
$= R$
Hence, the correct answer is option (A).

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