Question

The charge stored in each capacitor $C_1$ and $C_2$ in the circuit shown below are

• A. 6$\mu$C, 6$\mu$C
• B. 6$\mu$C, 3$\mu$C
• C. 3 $\mu$C, 6 $\mu$C
• D. 3 $\mu$C, 3$\mu$C

Answer 1

Answer: (A)

6$\mu$C, 6$\mu$C

The current in the circuit is $I=\frac{V}{6+6+2}=\frac{12}{12}=1$A.

So the potential across the capacitors will be $12-3=9V$.

The equivalent capacitance of these two capacitors in series is given by,

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\left(2\right)\left(1\right)}{2+1}=\frac{2}{3}\mu$F.

So the charge stored in each capacitor is $Q=C_{eq}V=\frac{2}{3}\times 9=6\mu$F.