An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.
(Take π²=10)

Solution

Physics sir · 4 steps

Maximum current means resonance, so set X_L = X_C; the 50 ohm resistor plays no part in finding L.

1. Condition for maximum. Current amplitude is I₀ = V₀/Z with Z = √(R² + (X_L - X_C)²). R is fixed, so Z is smallest, and I₀ largest, when X_L = X_C. That is resonance.

   ω L = 1/(ω C) ⇒ L = 1/(ω² C)

2. Find omega squared. ω = 2π f = 2π × 2000 = 4000π rad s⁻¹. So ω² = 16 × 10⁶ π² = 16 × 10⁷ = 1.6 × 10⁸ s⁻², using π² = 10.
3. Put in numbers. ω² C = 1.6 × 10⁸ × 62.5 × 10⁻⁹ = 10. So L = 1/10 = 0.1 H.
4. So. L = 0.1 H = 100 mH.

The common slip: 4000. Using f in place of ω = 2π f gives L = 4 H, that is 4000 mH, not 100.

The idea to keep. In a series LCR circuit, maximum current always means resonance: ω² = 1/LC. The resistance only decides how tall the peak is, not where it sits.

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