Question

Capacitor $A$ is charged to a potential of $100V$ and capacitor $B$ is charged to a potential of $75V$. What are the charges on $A$ and $B$ after key $K$ is closed, as shown in figure?

• A. $\frac{250}{3}\mu C;\frac{500}{3}\mu C$
• B. $\frac{250}{4}\mu C;\frac{250}{3}\mu C$
• C. $\frac{500}{4}\mu C;\frac{500}{3}\mu C$
• D. $\frac{1250}{4}\mu C;\frac{1250}{3}\mu C$

Answer 1

Answer: (A)

$Q_A=5\times 100=500\mu C$,
$Q_B=10\times 75=750\mu C$
As negative plate of $A$ is connected to positive plate of $B$,
Therefore, net charge $=750-500=250\mu C$
$\therefore$ Common potential
$V=\frac{\text{ net charge }}{\text{ total capacity }}$
$=\frac{250}{5+10}=\frac{250}{15}$ volt.
$Q_A^{\prime }=C_{1}V=5\times \frac{250}{15}=\frac{250}{3}\mu C$
$Q_B^{\prime }=C_{2}V=10\times \frac{250}{15}=\frac{500}{3}\mu C$