First slide

Capacitance

Question

In the figure shown, the potential drop across each capacitor is (assume diodes to be ideal)

Moderate

A

12V, 12V
B

16V, 8V
C

Zero, 24V
D

8V, Zero

Solution

The diode connected parallel to the battery is reverse bias. So, current will not pass through it. So, total emf divided among $C_{1}$ and $C_{2}$ and in the inverse ratio of their capacitances.
$V = \frac{2}{C}$
$\frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}}$
$V_{1} = \frac{V C_{2}}{C_{1} + C_{2}}, V_{2} = \frac{V C_{1}}{C_{1} + C_{2}}$

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