In the network shown,the potential difference | vedclass

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(B) The circuit consists of three parallel branches connected to point $B$,which are then connected in series with resistor $R$ to point $A$.Since the

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$2$

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(B) The circuit consists of three parallel branches connected to point $B$,which are then connected in series with resistor $R$ to point $A$.Since there is no external circuit connected to the terminals $A$ and $B$ (it is an open circuit),no current flows through the resistor $R$.Therefore,the potential difference between $A$ and $B$ is equal to the equivalent $EMF$ of the three parallel branches.The equivalent $EMF$ $(E_{eq})$ of cells in parallel is given by the formula:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$Substituting the given values $E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V$ and $r_1 = r_2 = r_3 = 1 \, \Omega$:$E_{eq} = \frac{\frac{3}{1} + \frac{2}{1} + \frac{1}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} = \frac{3 + 2 + 1}{1 + 1 + 1} = \frac{6}{3} = 2 \, V$Thus,the potential difference between $A$ and $B$ is $2 \, V$.

In the network shown,the potential difference between $A$ and $B$ is ................. $V$ $(R = r_1 = r_2 = r_3 = 1 \Omega, E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V)$.

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