An inductor of 0.5 mH, a capacitor of 20 μ F and resistance of 20 Ω are connected in series with a 220 V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is √x A. The value of x is ___________

Solution

Physics sir · 3 steps

Current in phase with emf means resonance, so only R limits the current. Then turn 220 V rms into peak.

1. Recognise resonance. Current is in phase with the emf only when X_L = X_C. The two reactances cancel, so the impedance is just Z = R = 20 Ω. The values of L and C only fix the resonant frequency. They do not change the current.
2. Rms current. An ac voltage of 220 V is an rms value. So Iᵣₘₛ = 220/20 = 11 A.

   I₀ = √2 Iᵣₘₛ = 11√2 = √(121 × 2) = √242 A

3. So. The amplitude is √242 A, so x = 242.

The common slip: 121. Treating 220 V as the peak voltage gives I₀ = 11 = √121 A, so x = 121. The 220 V is rms, so the peak current is √2 times larger.

The idea to keep. In phase means resonance, and at resonance Z = R. A quoted ac voltage is rms unless the question says otherwise, so multiply by √2 when it asks for amplitude.

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