Question

Three identical capacitors, each of capacitance $C$, are wired as shown above to a battery of source voltage $V$.
When both switches have been closed for a long time, the charge on $C_1$ is:

• A. $\frac{2}{9}CV$
• B. $\frac{2}{3}CV$
• C. $\frac{1}{3}CV$
• D. $CV$
• E. $3CV$

Answer 1

Answer: (B)

$\frac{2}{3}CV$

As the both switches have been closed for a long time so the capacitors are fully charged and no current will flow through them.

As $C_2$ and $C_3$ are in parallel so equivalent capacitance for them is $C_{23}=C_2+C_3=C+C=2C$

Now $C_1$ and $C_{23}$ are in series so the equivalent capacitance is

$C_{eq}=\frac{C_1{C}_{23}}{C_1+C_{23}}=\frac{C\left(2C\right)}{C+2C}=2C/3$

and equivalent charge $Q_{eq}=C_{eq}V=2CV/3$

As $C_1$ and $C_{23}$ are in series so the charge on them is same and equal to equivalent charge $Q_{eq}$.

Thus, the charge on $C_1=Q_{eq}=\frac{2}{3}CV$

The option B will be correct.