Question

In this circuit shown, the potential drop across capacitor $C_1$ and $C_2$ respectively is (assuming the two diodes are ideal)

• A. $6\text{V},18\text{V}$
• B. $16\text{V},8\text{V}$
• C. $18\text{V},6\text{V}$
• D. $16\text{V},16\text{V}$

Answer 1

The correct option is B $16\text{V},8\text{V}$

The diode ${\text{D}}_1$ is reverse biased (open circuit), but the diode ${\text{D}}_2$ is forward (short circuit).

$\therefore$ The given circuit can be redrawn as shown below.


We know that,

$V=\frac{Q}{C}$

As both the capacitors are in series, so charge on both the capacitors will be the same.

$\therefore$ The potential of the battery divides across the two capacitors in the inverse ratio of their capacities.

$\frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{8}{4}=\frac{2}{1}$

$\Rightarrow V_1=\frac{2}{3}V=\frac{2}{3}\times 24=16\text{V}$

$\Rightarrow V_2=\frac{1}{3}V=\frac{1}{3}\times 24=8\text{V}$

Hence, option $\left(\text{B}\right)$ is correct.