Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf V volt. Now disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy? (1) $25 \%$ (2) $75 \%$ (3) $0 \%$ (4) $50 \%$ - Noon Academy

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf V volt. Now disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy? (1) $25 \%$ (2) $75 \%$ (3) $0 \%$ (4) $50 \%$

Physics | Physics

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals a and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using battery of emf V volt. Now disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy? (1) $25 \%$ (2) $75 \%$ (3) $0 \%$ (4) $50 \%$

Answer (4)
$\mathrm{Sol}$ ,
Switch the key at point $c$
$\begin{array}{l} \frac{q_{0}-q}{C}=\frac{q}{C} \\ 2 q=q_{0} \\ q=\left(\frac{q_{0}}{2}\right) \\ U_{f}=\frac{1}{2}\left(\frac{q_{0}}{2}\right)^{2} \times \frac{1}{C}+\frac{1}{2}\left(\frac{q_{0}}{2}\right)^{2} \times \frac{1}{C} \\ U_{f}=\frac{q_{0}^{2}}{4 C} \\ =\frac{\left(\frac{1}{2}-\frac{1}{4}\right) C V^{2}}{\frac{1}{2} C V^{2}} \times 100 \\ =50 \% \end{array}$

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