Question

A capacitor of capacitance $C$, charged to potential difference $V$ is connected to another capacitor of capacitance $3C$, charged to potential difference $3V$ as shown in the figure. The final charges on capacitor of capacitance $C$ and $3C$ are respectively

• A. $\frac{5CV}{2},-\frac{15CV}{2}$
• B. $\frac{15CV}{2},\frac{5CV}{2}$
• C. $-\frac{5CV}{2},\frac{10CV}{2}$
• D. $\frac{5CV}{2},\frac{15CV}{2}$

Answer 1

The correct option is D $\frac{5CV}{2},\frac{15CV}{2}$

Initially charges on the capacitors are calculated using $Q=CV$ as shown in figure below.


From the law of conservation of charge,

$\text{Initial charge}\left(q_i\right)=\text{Final charge}\left(q_f\right)$

Let the final potential across both the capacitors be $E$. So,

$CV+9CV=CE+3CE$

Finally:


$\Rightarrow E=\frac{10V}{4}=\frac{5V}{2}$

So, the final charge on $C$

$q_C=\frac{5V}{2}\times C=\frac{5CV}{2}$

Charge on $3C$,

$q_{3C}=\frac{5V}{2}\times 3C=\frac{15CV}{2}$

Hence, option (a) is correct answer.