The self inductance of a choke coil is 10 mH. When it is connected with a 10V dc source, then the loss of power is 20 watt. When it is connected with 10 volt ac source loss of power is 10 watt. The frequency of ac source will be

Choke Coil Self-Inductance Problem

Given Data:

• Self-inductance, L = 10 mH = 10 × 10^-3 H
• Voltage, V = 10 V
• Power loss in DC, P_dc = 20 W
• Power loss in AC, P_ac = 10 W

Step 1: Find Resistance

In DC circuit (steady state), inductive reactance is zero:

P_dc = V²/R ⟹ R = V²/P_dc = 10²/20 = 5 Ω

Step 2: Find Inductive Reactance

Power in AC circuit:

P_ac = (V² × R)/(R² + X_L²)

Rearranging:

X_L² = (V² × R - P_ac × R²)/P_ac

X_L² = (10² × 5 - 10 × 5²)/10 = (500 - 250)/10 = 25

X_L = 5 Ω

Step 3: Calculate Frequency

Using X_L = 2πfL:

f = X_L/(2πL) = 5/(2π × 10 × 10^-3) = 5/0.0628 ≈ 79.58 Hz

Final Answer:

The frequency of the AC source is approximately 80 Hz.