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Kirchhoff's Junction Law

The potential...

Question

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

(\frac{E_{1} + E_{2}}{C_{1} + C_{2}}) C_{2}

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(\frac{E_{1} - E_{2}}{C_{1} + C_{2}}) C_{2}

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(\frac{E_{1} + E_{2}}{C_{1} + C_{2}}) C_{1}

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(\frac{E_{1} - E_{2}}{C_{1} + C_{2}}) C_{1}

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Solution

The correct option is B $(\frac{E_{1} + E_{2}}{C_{1} + C_{2}}) C_{1}$
As the potential across both capacitors is different, they are in series.
Thus, net capacitance of series combination $C_{n e t} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$
Also, total emf of the circuit is $E_{t} = E_{1} + E_{2}$
Charge on each capacitor is $q = C_{n e t} E_{t} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . (E_{1} + E_{2})$
Thus, potential difference between $a$ and $b$ , $V_{a b} =$ $\frac{q}{C_{2}}$
$⟹ V_{a b} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} . (E_{1} + E_{2}) \frac{1}{C_{2}} = \frac{(E_{1} + E_{2})}{(C_{1} + C_{2})} C_{1}$

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