A series LCR circuit has L=0.01 H, R=10 Ω and C=1 μ F and it is connected to ac voltage of amplitude (Vₘ) 50 V. At frequency 60 % lower than resonant frequency, the amplitude of current will be approximately :

Solution

Physics sir · 3 steps

Find the resonant ω first, take 40% of it, then get the net reactance at that new frequency.

1. Resonant frequency. ω₀ = 1/√LC = 1/√(0.01 × 10⁻⁶) = 10⁴ rad s⁻¹. A frequency 60% lower than this is 40% of it, so ω = 0.4 ω₀ = 4000 rad s⁻¹.
2. Two reactances. X_L = ω L = 4000 × 0.01 = 40 Ω and X_C = 1/(ω C) = 1/(4000 × 10⁻⁶) = 250 Ω. Below resonance the capacitor dominates, and the net reactance is 250 - 40 = 210 Ω.

   Z = √(R² + (X_C - X_L)²) = √(10² + 210²) = √44200 ≈ 210.2 Ω

3. So. Iₘ = Vₘ/Z = 50/210.2 ≈ 0.238 A, which is about 238 mA, option (c).

The idea to keep. Away from resonance the net reactance X_C - X_L sets the current, not R. Read "x% lower" carefully: 60% lower means 0.4 times the original, not 0.6 times.

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