A $2 \mu \mathrm{F}$ capacitor $C_{1}$ is charged to a voltage $100 \mathrm{~V}$ and a $4 \mu \mathrm{F}$ capacitor $C_{2}$ is charged to a voltage $50 \mathrm{~V}$. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?

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Aicharya · Accepted Answer

Charge $Q_{1}$ and $C_{1}=C_{1} V_{1}=2 \times 10^{-6} \times 100=2 \times 10^{-4} \mathrm{C}$. Charge $Q_{2}$ on $C_{2}=C_{2} V_{2}=4 \times 10^{-6} \times 50=$ $2 \times 10^{-4} \mathrm{C}$. Total charge $Q=Q_{1}+Q_{2}=4 \times 10^{-4} \mathrm{C}$. Total energy before connection is $\begin{aligned} E_{1} &=\frac{1}{2} C_{1} V_{1}^{2}+\frac{1}{2} C_{2} V_{2}^{2} \ &=\frac{1}{2} \times 2 \times 10^{-6} \times(100)^{2}+\frac{1}{2} \times 4 \times 10^{-6} \times(50)^{2} \ &=1.5 \times 10^{-2} \mathrm{~J} \end{aligned}$ The common potential difference $V$ after connection is given by $C_{1} V+C_{2} V=Q$ or $V=\frac{Q}{C_{1}+C_{2}}$ Therefore, total energy after connection is $\begin{aligned} E_{2} &=\frac{1}{2}\left(C_{1}+C_{2}\right) V^{2}=\frac{1}{2} \times \frac{Q^{2}}{\left(C_{1}+C_{2}\right)} \ &=\frac{1}{2} \times \frac{\left(4 \times 10^{-4}\right)^{2}}{(2+4) \times 10^{-6}}=1.33 \times 10^{-2} \mathrm{~J} \end{aligned}$ $\therefore \quad$ Loss of energy $=E_{1}-E_{2}=0.17 \times 10^{-2} \mathrm{~J} .$ Hence the correct choice is (d). .

A \(2 \mu \mathrm{F}\) capacitor \(C_{1}\) is charged to a voltage \(100 \mathrm{~V}\) and a \(4 \mu…

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