Question

In the given circuit the potential difference across the $4.5\mu F$ capacitor is

• A. $\frac{8}{3}\text{V}$
• B. $4\text{V}$
• C. $6\text{V}$
• D. $8\text{V}$

Answer 1

The correct option is D $8\text{V}$

$C=\frac{9\times 4.5}{13.5}\mu F$ or $C=3\mu F$
$Q=CV=3\times 12\mu C=36\mu C$
Potential difference across $4.5\mu F$
$=\frac{36}{4.5}\text{V}=8\text{V}$

Time saving solution
$\frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{9}{4.5}=\frac{2}{1}$
$V_1+V_2=12$
$V_1=8\text{V}$