An ac voltage V=220(2×10³t) V is applied to a series LCR circuit. The current amplitude in the circuit is: [L=10 mH, C=25 µF, R=100 Ω]
- A.2.2 A✓
- B.5.5 A
- C.11.0 A
- D.22.0 A
Correct Answer
(A) 2.2 A
Solution & Explanation
Read the source: ω = 2×10³ rad s⁻¹, V₀ = 220 V, with L = 10 mH, C = 25 µF, R = 100 Ω. Inductive reactance X_L = ωL = (2×10³)(10×10⁻³) = 20 Ω. Capacitive reactance X_C = 1/(ωC) = 1/[(2×10³)(25×10⁻⁶)] = 1/(0.05) = 20 Ω. Since X_L = X_C, the reactances cancel — the circuit is at resonance. Impedance Z = √(R² + (X_L − X_C)²) = √(R² + 0) = R = 100 Ω. Current amplitude i₀ = V₀/Z = 220/100 = 2.2 A. Key point: spotting X_L = X_C (resonance) is the whole trick — the reactive parts drop out and only R survives, giving answer (A), 2.2 A.
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