Question

In the given circuit, energy stored in the inductor is \(16\,\text{J}\) and power dissipated in resistance is \(32\,\text{W}\). Find the value of \(\dfrac{X_L}{R}\).

• A. \(314\)
• B. \(328\)
• C. \(335\)
• D. \(340\)

Hint:

In AC circuits, remember: average energy in an inductor depends on \(I_{\text{rms}}\), while power loss depends only on resistance.

Answer 1

The Correct Option is A

To solve this problem, we need to find the ratio \(\dfrac{X_L}{R}\) for the given circuit, where the energy stored in the inductor is \(16\,\text{J}\) and the power dissipated in the resistor is \(32\,\text{W}\).

1. Energy stored in the inductor:

\(U = \dfrac{1}{2} L I^2 = 16\,\text{J}\)

Solving for current:

\(I = \sqrt{\dfrac{2U}{L}}\)

1. Power dissipated in the resistor:

\(P = I^2 R = 32\,\text{W}\)

Substituting the value of \(I^2\):

\(32 = \dfrac{2U}{L} \, R\)

1. Inductive reactance:

\(X_L = \omega L\)

where \(\omega = 2\pi f\) and \(f = 50\,\text{Hz}\).

\(X_L = 100\pi L\)

1. Determination of the ratio \(\dfrac{X_L}{R}\):

R = \dfrac{32L}{2U} = \dfrac{32L}{32} = L

\(\dfrac{X_L}{R} = \dfrac{100\pi L}{L} = 100\pi\)

\(\dfrac{X_L}{R} = 314\)

Therefore, the correct answer is: \(314\).