Question

A capacitor of capacitance 5μF is charged by a battery of emf 10V. At an instant of time, the potential difference across the capacitors is 4V and the time rate of change of potential difference across the capacitor is 0.6 Vs^-1. Then the time rate at which energy is stored the capacitor at the instant is

• A. 12μW
• B. 3μW
• C. Zero
• D. 30μW

Answer 1

The Correct Option is A

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

Differentiating with respect to time: 
\( \frac{dU}{dt} = C V \frac{dV}{dt} \) 

Substituting the given values: 
\( C = 5 \times 10^{-6} \, \text{F}, \quad V = 4 \, \text{V}, \quad \frac{dV}{dt} = 0.6 \, \text{V/s} \) 

\( \frac{dU}{dt} = (5 \times 10^{-6}) \cdot 4 \cdot 0.6 = 12 \times 10^{-6} \, \text{W} = 12\,\mu\text{W} \) 

Thus, the solution is \( {12\,\mu\text{W}} \).

Answer 2

The energy stored in a capacitor is given by: \[ E = \frac{1}{2} C V^2 \] The rate at which energy is stored is: \[ \frac{dE}{dt} = C \times V \times \frac{dV}{dt} \] Given: - \( C = 5 \times 10^{-6} \, \text{F} \) - \( V = 4 \, \text{V} \) - \( \frac{dV}{dt} = 0.6 \, \text{V/s} \) Thus: \[ \frac{dE}{dt} = (5 \times 10^{-6}) \times 4 \times 0.6 = 12 \times 10^{-6} \, \text{W} = 12 \, \mu\text{W} \] Thus, the rate at which energy is stored is \(12\ µW\).