In a series LCR circuit, the inductance, capacitance and resistance are L = 100 mH, C = 100 μF and R = 10 Ω respectively. They are connected to an AC source of voltage 220 V and frequency of 50 Hz. The approximate value of current in the circuit will be ___________ A.

Solution

Physics sir · 4 steps

Find both reactances first. They almost cancel, so the circuit is near resonance and R alone sets the current.

1. Angular frequency. ω = 2π f = 2π × 50 = 314.2 rad s⁻¹.
2. Both reactances. X_L = ω L = 314.2 × 0.1 = 31.4 Ω. X_C = 1/(ω C) = 1/(314.2 × 100 × 10⁻⁶) = 31.8 Ω. The difference X_C - X_L is only about 0.4 Ω.

   Z = √(R² + (X_C - X_L)²) = √(10² + 0.4²) ≈ 10.01 Ω

3. Current. Take 220 V as the rms value. Then I = V/Z = 220/10.01 = 21.98 A.
4. So. The current is about 22 A. It is almost V/R, because the circuit is very close to resonance.

The common slip: 3. Adding R, X_L and X_C directly gives Z = 73.2 Ω and a current of 3 A. Reactances subtract from each other and then combine with R as √(R² + (X_L - X_C)²).

The idea to keep. In series LCR, X_L and X_C act against each other, so only their difference adds to R in quadrature. When they nearly cancel, Z ≈ R and I ≈ V/R.

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