Question

In the circuit given below, the charge in $\mu C$, on the capacitor having capacitance $5\mu F$ is

• A. 4.5
• B. 9
• C. 7
• D. 15

Answer 1

Answer: (B)

Potential difference across the branch de is $6V$.
Net capacitance of de branch is $2.1\mu F$
So, $q=CV=2.1\times 6\mu C=12.6\mu C$
Potential across $3\mu F$ capacitance is
$V=\frac{12.6}{3}=4.2\text{ volt }$
Potential across 2 and 5 combination in parallel is $6-4.2=1.8V$
$\text{ So, }{q}^{\prime }=(1.8)(5)=9\mu C$