In the circuit shown, a meter bridge is in it | vedclass

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(C) The resistance of the wire segments $AC$ and $CB$ are calculated as follows:$R_{AC} = 0.1 \, \Omega/cm 	imes 40 \, cm = 4 \, \Omega$$R_{CB} = 0.1

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$4 \, \Omega, 1.0 \, A$

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(C) The resistance of the wire segments $AC$ and $CB$ are calculated as follows:$R_{AC} = 0.1 \, \Omega/cm 	imes 40 \, cm = 4 \, \Omega$$R_{CB} = 0.1 \, \Omega/cm 	imes 60 \, cm = 6 \, \Omega$In the balanced condition of the Wheatstone bridge, the ratio of resistances is equal:$\frac{X}{6 \, \Omega} = \frac{R_{AC}}{R_{CB}} = \frac{4 \, \Omega}{6 \, \Omega}$$\frac{X}{6} = \frac{4}{6} \implies X = 4 \, \Omega$The total resistance of the circuit is the sum of the two parallel branches:Branch $1$: $X + 6 \, \Omega = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega$Branch $2$: $R_{AC} + R_{CB} = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega$Since these two branches are in parallel, the equivalent resistance $R_{eq}$ is:$R_{eq} = \frac{10 \, \Omega 	imes 10 \, \Omega}{10 \, \Omega + 10 \, \Omega} = 5 \, \Omega$The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{5 \, V}{5 \, \Omega} = 1.0 \, A$.Thus, $X = 4 \, \Omega$ and $I = 1.0 \, A$.

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance of $0.1 \, \Omega/cm$. The value of the unknown resistance $X$ and the current drawn from the battery of negligible internal resistance are:

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