A uniform conducting wire AB of length 5 , m | vedclass

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(B) The total resistance of the primary circuit is $R_{total} = R_{wire} + R_{external} + r = 5 \, \Omega + 4 \, \Omega + 1 \, \Omega = 10 \, \Omega$.

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$3$

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(B) The total resistance of the primary circuit is $R_{total} = R_{wire} + R_{external} + r = 5 \, \Omega + 4 \, \Omega + 1 \, \Omega = 10 \, \Omega$.The current flowing through the wire $AB$ is $I = \frac{V}{R_{total}} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$.The potential difference across the length of $3 \, m$ of the wire $AB$ is $V_{AB'} = I 	imes R_{AB'}$, where $R_{AB'}$ is the resistance of the $3 \, m$ segment.Since the wire is uniform, the resistance per unit length is $\lambda = \frac{5 \, \Omega}{5 \, m} = 1 \, \Omega/m$.Thus, $R_{AB'} = 1 \, \Omega/m 	imes 3 \, m = 3 \, \Omega$.The potential difference across the balancing length is $V_{AB'} = 1 \, A 	imes 3 \, \Omega = 3 \, V$.In the secondary circuit, two cells of $EMF$ $E$ are connected in parallel. The equivalent $EMF$ of the parallel combination is $E_{eq} = E$.At the balancing point, the potential difference across the balancing length must equal the $EMF$ of the secondary circuit.Therefore, $E = V_{AB'} = 3 \, V$.

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)

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