Question

What is the charge stored on each capacitor $C_1$ and $C_2$ in the circuit shown in figure?

• 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)

When the capacitors are fully charged, no current draws from cell. Thus, current will flow only through resistors.
Current in the circuit is $I=\frac{12}{6+3+3}=1A$
Potential across d and c is $V_{dc}=6I+3I=6+3=9V$
Net capacitance between d and c is $C_{dc}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 1}{2+1}=\left(2/3\right)\mu F$
Equivalent charge $Q_{dc}=C_{dc}{V}_{dc}=\left(2/3\right)\times 9=6\mu C$
As the capacitors are in series so the charge on each capacitor is equal to $Q_{dc}=6\mu C$