Choose the correct option: Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.

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KollegeApply · Accepted Answer

1 : 1 Explanation: The potential energy stored in the capacitor equals the work done in charging the capacitor. The capacitor is charged by working against the existing electric field between the two plates. Denote the charge stored by the capacitor of capacitance C when it is fully charged as Q. Denote the battery voltage as V. The potential energy U, stored in the capacitor, can be represented as: U = $\frac{1}{2}$ "CV"^2 The work done by battery W, in supplying charge Q can be represented as: W = QV $\Rightarrow$ W = (CV)V $\Rightarrow$ W = CV2 Therefore, the energy dissipated during charging H equals the difference between W and U. H = W – U H = "CV"^2 - $\frac{1}{2}$ "CV"^2 H = $\frac{1}{2}$ "CV"^2 The ratio of and is calculated as follows: `"U"/"H" = ($\frac{1}{2}$ "CV"^2)/($\frac{1}{2}$ "CV"^2) "U"/"H" = $\frac{1}{1}$` i.e. 1 : 1

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