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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No current through $R$, so potential difference across
$A B$ is$V=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}+\frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}}=\frac{\frac{3}{1}+\frac{2}{1}+\frac{1}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=2 V$
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