A capacitor of capacity C is connected with a battery of potential V in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential V again, the energy given by the battery will be

Question

A capacitor of capacity C is connected with a battery of potential V in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential V again, the energy given by the battery will be (A) C V2 / 4 (B) C V2 / 2 (C) 3 C V2 / 4 (D) C V2

Tardigrade Admin · Accepted Answer

Extra charge, Q=(2 C V-C V)=C V flows through potential V of the battery. Thus, W=Q V=C V2

Q. A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be

$C V^{2} /4$

$C V^{2} /2$

$3 C V^{2} /4$

$C V^{2}$

Extra charge, $Q = (2 C V - C V) = C V$ flows through potential $V$ of the battery. Thus, $W = Q V = C V^{2}$

Q. A capacitor of capacity C is connected with a battery of potential V in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential V again, the energy given by the battery will be

CV2/4

CV2/2

3CV2/4

CV2