WBJEE · Physics · Capacitance
A capacitor of capacitance \(C\) is connected in series with a resistance \(R\) and \(D C\) source of emf \(E\) through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged. what amount of energy is dissipated in the resistance \(R ?\) 
- A \(\frac{1}{2} C E^{2}\)
- B 0
- C \(\mathrm{CE}^{2}\)
- D \(\frac{E^{2}}{R}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} C E^{2}\)
Step-by-step Solution
Detailed explanation
We know that, energy stored in the capacitor \[ =\frac{1}{2} C E^{2} \] and energy supplied by the source of emf \[ =C E^{2} \] \(\therefore\) Energy dissipated in resistance \(R\) = Energy supplied by the source of emf \(E\) - Energy stored in the capacitor…
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