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(C) For a balanced meter bridge,the ratio of resistances is equal to the ratio of the lengths of the wire segments:$\frac{X}{6} = \frac{l_1}{l_2} = \f

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$4 \ \Omega, 1.0 \ A$

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(C) For a balanced meter bridge,the ratio of resistances is equal to the ratio of the lengths of the wire segments:$\frac{X}{6} = \frac{l_1}{l_2} = \frac{40 \ cm}{60 \ cm} = \frac{2}{3}$$X = 6 	imes \frac{2}{3} = 4 \ \Omega$Now,calculate the equivalent resistance of the circuit. The circuit consists of two parallel branches connected to the $5 \ V$ battery. The left branch has resistance $(X + R_{wire, left}) = 4 + (40 	imes 0.1) = 4 + 4 = 8 \ \Omega$. The right branch has resistance $(6 + R_{wire, right}) = 6 + (60 	imes 0.1) = 6 + 6 = 12 \ \Omega$.However,in a balanced meter bridge,the galvanometer branch carries no current. Thus,the total resistance $R_{eq}$ is the parallel combination of the two branches: $(4+4) \ \Omega$ and $(6+6) \ \Omega$.$R_{eq} = \frac{(8 	imes 12)}{(8 + 12)} = \frac{96}{20} = 4.8 \ \Omega$.Wait,re-evaluating the circuit: The battery is connected across the ends $A$ and $B$. The total resistance of the wire is $100 \ cm 	imes 0.1 \ \Omega/cm = 10 \ \Omega$. The upper branch has $(X + 6) = 10 \ \Omega$. The lower branch (wire) has $10 \ \Omega$. These are in parallel.$R_{eq} = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$.The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{5 \ V}{5 \ \Omega} = 1.0 \ A$.

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

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