Question

The figure shows a network of seven capacitors. If the charge on the $5\mu F$ capacitor is $10\mu C$, the potential difference between the points $A$ and $C$ is given as $\frac{x}{3}V$.

Answer 1

Potential across $5\mu F$ is $V_5=\frac{q_5}{C_5}=\frac{10}{5}=2V$

As $(3,5,4)$ capacitors are in parallel so

charge on $3\mu F$ is $q_3=C_3{V}_5=3\times 2=6\mu C$ and

charge on $4\mu F$ is $q_4=C_3{V}_5=4\times 2=8\mu C$.

As $C_2$ and $(3,5,4)$ are in series so charge on $C_2$ and group (3,4,5) are same.

So charge on $C_2$ is $q_2=10+6+8=24\mu C$ and potential across it $V_2=\frac{24}{4}=6V$

Thus , $V_{AB}=V_5+V_2=2+6=8V$

Equivalent capacitance of lower branch of circuit is $C_{eq}=\frac{3(4+2)}{3+(4+2)}=2\mu F$

and $Q_{eq}=C_{eq}{V}_{AB}=2\left(8\right)=16\mu C$

$\therefore V_{AC}=\frac{Q_{eq}}{3}=\frac{16}{3}$ thus $x=16$