In the circuit shown, the potential difference between $A$ and $B$ is ............. $V$
JEE MAIN 2019, Medium
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Potential difference across $AB$ will be equal to battery equivalent across $CD$.
$ V_{A B} =V_{C D}=\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{1}{1}+\frac{2}{1}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}} $
$=\frac{6}{3}=2\, V$
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