When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is __________ mH. (Take π=3)

Solution

Physics sir · 4 steps

DC sees only the resistance, AC sees the impedance. Take one from the other and you have the reactance.

1. DC gives R. With steady DC the current does not change, so the coil's inductance has no effect. Only its resistance limits the current: R = 20/5 = 4 Ω.
2. AC gives Z. On AC both the resistance and the inductive reactance oppose the current: Z = 20/4 = 5 Ω.

   Z² = R² + X_L² ⇒ X_L = √(5² - 4²) = 3 Ω

3. Find L. X_L = ω L = 2π f L, so L = 3/(2 × 3 × 50) = 3/300 = 0.01 H.
4. So. L = 0.01 H = 10 mH.

The common slip: 60. Writing X_L = fL instead of 2π f L gives L = 3/50 H, which is 60 mH, not 10.

The idea to keep. A coil on DC tells you its resistance. The same coil on AC tells you its impedance. Then X_L = √(Z² - R²), and always use ω = 2π f, not f.

Answer from the NTA final answer key. Solution by PrepWiser.