A 5\; \mu F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5\; \mu F capacitor. If the energy change during the charge redistribution is \frac{X}{100}J then value of X to the nearest integer is __________
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

Initially

Q_0=CV_0

\text { Now battery is disconnected and a capacitor of } \frac{C}{2} \text { is connected to } C \text { then }

\begin{aligned} &\text { Total charge }=\mathbf{Q}_{0}\\ &C_{e q .}=\frac{3 C}{2} \end{aligned}

\begin{array}{l} \therefore \Delta E=\frac{Q_{0}^{2}}{2 C}-\frac{2 Q_{0}^{2}}{3 C \times 2} \Rightarrow \Delta E=\frac{Q_{0}^{2}}{2 C}\left(\frac{1}{3}\right) \\ \begin{aligned} \Delta E &=\frac{1}{6} \times 220 \times 220 \times 5 \times 10^{-6} \approx 4 \times 10^{-2} \mathrm{~J} \\ \therefore \quad X &=4 \end{aligned} \end{array}

 

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