Question

An ideal inductor with an inductance value of 1/3 H is connected to a 50 Hz sinusoidal AC voltage source. The energy stored in the inductor is 6 J. The value of the maximum power delivered to the inductor is \(\underline{\hspace{1cm}}\) W.

• A. 1200$\pi $
• B. 600$\pi $
• C. 1200
• D. 0

Hint:

Use the formula \( P_{\text{max}} = \frac{1}{2} L I^2 \omega \) to calculate maximum power delivered to an inductor in an AC circuit.

Answer 1

The Correct Option is A

The energy stored in the inductor is given by: \[ E = \frac{1}{2} L I^2 \] Where: - \( L = \frac{1}{3} \) H (inductance) - \( E = 6 \) J (energy stored) We can calculate the current using the energy formula: \[ 6 = \frac{1}{2} \times \frac{1}{3} \times I^2 \] \[ I^2 = \frac{6 \times 2 \times 3}{1} = 36 \] \[ I = 6 \text{ A} \] Now, the maximum power delivered to the inductor is given by the formula: \[ P_{\text{max}} = \frac{1}{2} L I^2 \omega \] Where: - \( \omega = 2\pi f = 2\pi \times 50 = 100\pi \) rad/s
- \( L = \frac{1}{3} \) H
- \( I = 6 \) A
Substituting the values: \[ P_{\text{max}} = \frac{1}{2} \times \frac{1}{3} \times 6^2 \times 100\pi = \frac{1}{2} \times \frac{1}{3} \times 36 \times 100\pi = 1200\pi \text{ W} \] Thus, the maximum power delivered to the inductor is 1200π W, which corresponds to (A).