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 \(\dfrac{X_L}{R}\).
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.
Updated On: Jan 29, 2026
- \(314\)
- \(328\)
- \(335\)
- \(340\)
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The Correct Option is A
Solution and Explanation
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: \[ f = 50\,\text{Hz} \Rightarrow \omega = 2\pi f = 100\pi \approx 314 \] \[ \boxed{\dfrac{X_L}{R} = 314} \]
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: \[ f = 50\,\text{Hz} \Rightarrow \omega = 2\pi f = 100\pi \approx 314 \] \[ \boxed{\dfrac{X_L}{R} = 314} \]
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