The resistance of each arm of the Wheatstone | vedclass

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(C) Wheatstone bridge is balanced when the ratio of resistances in opposite arms is equal. Here, $\frac{P}{Q} = \frac{R}{S} = \frac{10}{10} = 1$.In a

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$10 \, \Omega$

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(C) Wheatstone bridge is balanced when the ratio of resistances in opposite arms is equal. Here, $\frac{P}{Q} = \frac{R}{S} = \frac{10}{10} = 1$.In a balanced Wheatstone bridge, no current flows through the galvanometer arm.However, the problem states that a resistance of $10 \, \Omega$ is connected in series with the galvanometer. Since no current flows through the galvanometer branch in a balanced state, this additional resistance does not affect the equivalent resistance of the circuit.The circuit effectively consists of two parallel branches, each containing two $10 \, \Omega$ resistors in series.Resistance of each branch = $10 \, \Omega + 10 \, \Omega = 20 \, \Omega$.Since these two branches are in parallel, the equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}$.Therefore, $R_{eq} = 10 \, \Omega$.

The resistance of each arm of the Wheatstone bridge is $10 \, \Omega$. A resistance of $10 \, \Omega$ is connected in series with a galvanometer. Then, the equivalent resistance across the battery will be:

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