Question

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

• A. $12V,12V$
• B. $16V,8V$
• C. $0,24V$
• D. $8V,0$

Answer 1

Answer: (B)

$16V,8V$

Here current flowing through the circuit is:
$I=\frac{E}{\frac{1}{C_1}+\frac{1}{C_2}}$

$I=\frac{24}{\frac{1}{4}+\frac{1}{8}}$

$I=\frac{24\times 8}{3}=64A$

Now, potential drop across $C_1$ is,
$V_{C_1}=\frac{I}{C_1}=\frac{64}{4}=16V$

And potential drop across $C_2$ is,
$V_{C_2}=\frac{I}{C_2}=\frac{64}{8}=8V$