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Question

The potential difference between points a and b of circuits shown in the figure is:

214440_61f27ac5857c4620a710e3fe67415c02.png

A
(E1+E2C1+C2)C2
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B
(E1E2C1+C2)C2
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C
(E1+E2C1+C2)C1
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D
(E1E2C1+C2)C1
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Solution

The correct option is B (E1+E2C1+C2)C1
As the potential across both capacitors is different, they are in series.
Thus, net capacitance of series combination Cnet=C1C2C1+C2
Also, total emf of the circuit is Et=E1+E2
Charge on each capacitor is q=CnetEt=C1C2C1+C2.(E1+E2)
Thus, potential difference between a and b, Vab= qC2
Vab=C1C2C1+C2.(E1+E2)1C2=(E1+E2)(C1+C2)C1

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