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 \(\dfrac{X_L}{R}\).
Show Hint
In AC circuits, remember:
average energy in an inductor depends on \(I_{\text{rms}}\),
while power loss depends only on resistance.
Concept:
For an AC circuit containing an inductor and a resistor:
Average energy stored in an inductor:
\[
U = \frac{1}{2} L I_{\text{rms}}^2
\]
Power dissipated in a resistor:
\[
P = I_{\text{rms}}^2 R
\]
Inductive reactance:
\[
X_L = \omega L
\]
Step 1: Use Energy Stored in Inductor
Given:
\[
U = 16\,\text{J}
\]
\[
16 = \frac{1}{2} L I_{\text{rms}}^2
\Rightarrow L I_{\text{rms}}^2 = 32
\quad \cdots (1)
\]
Step 2: Use Power Dissipated in Resistance
Given:
\[
P = 32\,\text{W}
\]
\[
32 = I_{\text{rms}}^2 R
\Rightarrow I_{\text{rms}}^2 = \frac{32}{R}
\quad \cdots (2)
\]
Step 3: Find Inductance \(L\)
Substitute (2) into (1):
\[
L \cdot \frac{32}{R} = 32
\Rightarrow L = R
\]
Step 4: Calculate \(\dfrac{X_L}{R}\)
\[
\frac{X_L}{R} = \frac{\omega L}{R} = \omega
\]
Given frequency:
\[
f = 50\,\text{Hz}
\Rightarrow \omega = 2\pi f = 100\pi \approx 314
\]
\[
\boxed{\dfrac{X_L}{R} = 314}
\]
