Question

In the circuit given below, the charge in $\mu C$, on the capacitor having $5\mu F$ 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, $q=CV$
$\Rightarrow q=2.1\times 6\mu C$
$\Rightarrow q=12.6\mu C$
Potential across $3\mu F$ capacitance is
$V=\frac{12.6}{3}=4.2Volt$
Potential across $2$ and $5$ combination in parallel is
$6-4.2=1.8V$
So, $q^{\prime }=(1.8)(5)=9\mu C$