A capacitor of capacity 5 μ F is charged by a battery of e.m.f. 10 V. At an instant, the potential difference across the capacitor is 4 V and the time rate of change of potential difference across the

Question

A capacitor of capacity 5 μ F is charged by a battery of e.m.f. 10 V. At an instant, the potential difference across the capacitor is 4 V and the time rate of change of potential difference across the capacitor is 0.6 V/s. Then the time rate at which energy is stored in the capacitor at that instant is

TestHub · Accepted Answer

The energy stored in a capacitor is given by $U = \frac{1}{2} C V^{2}$

The time rate at which energy is stored is the derivative of energy with respect to time:

$\frac{d U}{d t} = \frac{d}{d t} (\frac{1}{2} C V^{2}) = C V \frac{d V}{d t}$

Given $C = 5 μ F = 5 \times 1 0^{- 6}$ F, $V = 4$ V, and $\frac{d V}{d t} = 0.6$ V/s.

Substituting the given values:

$\frac{d U}{d t} = (5 \times 1 0^{- 6}) \times 4 \times 0.6$

$\frac{d U}{d t} = 12 \times 1 0^{- 6}$ J/s

Answer: $12 \times 1 0^{- 6} J / s$

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion