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Voltage

The capacitie...

Question

The capacities of two conductors are $C_{1}$ and $C_{2}$ and their respective potentials are $V_{1}$ and $V_{2}$ . If they are connected by a thin wire, then the loss of energy will be given by

\frac{C_{1} C_{2} (V_{1} - V_{2})^{2}}{2 (C_{1} + C_{2})}

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\frac{C_{1} C_{2} (V_{1} - V_{2})}{2 (C_{1} + C_{2})}

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\frac{C_{1} C_{2} (V_{1} + V_{2})}{C_{1} C_{2}}

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\frac{C_{1} C_{2} (V_{1} + V_{2})}{2 (C_{1} + C_{2})}

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Solution

The correct option is A $\frac{C_{1} C_{2} (V_{1} - V_{2})^{2}}{2 (C_{1} + C_{2})}$
Initial energy $(E_{i}) = \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2}$
Let $V$ be the common potential after connection
By law of conservation of charge total initial charge $=$ total final charge
$q_{1} + q_{2} = q'_{1} + q'_{2}$

$C_{1} V_{1} + C_{2} V_{2} = C_{1} V + C_{2} V$

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

Final energy $(E_{f}) = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

$= \frac{1}{2} (C_{1} + C_{2}) {[\frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}]}^{2}$

$\Rightarrow$ Energy loss $= E_{i} - E_{f}$

$= \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2} - \frac{1}{2} (C_{1} + C_{2}) {[\frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}]}^{2}$

$= \frac{C_{1} C_{2} (V_{1} - V_{2})^{2}}{2 (C_{1} + C_{2})}$

Thus, energy loss of capacitors is $\frac{C_{1} C_{2} (V_{1} - V_{2})^{2}}{2 (C_{1} + C_{2})}$
Hence, option (c) is correct.

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Voltage

Standard VIII Physics

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The capacities of two conductors are C1 and C2 and their respective potentials are V1 and V2. If they are connected by a thin wire, then the loss of energy will be given by

The capacities of two conductors are $C_{1}$ and $C_{2}$ and their respective potentials are $V_{1}$ and $V_{2}$ . If they are connected by a thin wire, then the loss of energy will be given by