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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 minQB

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A capacitor $C$ is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitors is :

Options:

Answer:

Solution:

Heat Loss is given by,

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

$V_{1}$ is initial voltage of capacitor $C$ $= V_{0}$

$V_{2}$ is the initial voltage of capacitor $\frac{C}{2}$ $= 0$

$= \frac{c \times \frac{c}{2}}{2 (c + \frac{c}{2})} {(V_{0} - 0)}^{2} = \frac{C}{6} V_{0}^{2}$

$H = \frac{1}{6} C_{0} V_{0}^{2}$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2mℹ️ Source: QB

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