Question

Two capacitors, $C_1=2\mu F$ and $C_2=8\mu F$ are connected in series across a $300V$ source. Then

• A. the charge on each capacitor is $4.8\times {10}^{-4}C$
• B. the energy stored in the system is $5.2\times {10}^{-2}J$
• C. the potential difference across $C_1$ is $60V$
• D. the potential difference across $C_2$ is $240V$

Answer 1

Answer: (A)

The equivalent capacitance is $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 8}{2+8}=1.6\mu F$.

Equivalent charge is $Q_{eq}=C_{eq}V=1.6\times {10}^{-6}\times 300=4.8\times {10}^{-4}C$

As the capacitors are in series so charge on each capacitor is $Q_{eq}=4.8\times {10}^{-4}C$.

The potential difference across $C_1$ is $V_1=\frac{Q_{eq}}{C_1}=\frac{4.8\times {10}^{-4}}{2\times {10}^{-6}}=240V$

The potential difference across $C_2$ is $V_2=\frac{Q_{eq}}{C_2}=\frac{4.8\times {10}^{-4}}{8\times {10}^{-6}}=60V$

Energy stored in the system is $U=\frac{1}{2}{C}_{eq}{V}^2=\frac{1}{2}\times 1.6\times {10}^{-6}\times (300{)}^2=7.2\times {10}^{-2}J$