Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is :

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Physics

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$V_{A} - V_{B} = \frac{\frac{6}{1} + \frac{3}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{6 + 1 .5}{1 .5} = \frac{7 .5}{1 .5} = 5 V$

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