A series LCR circuit of R = 5Ω, L = 20 mH and C = 0.5 μF is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ × 10² W.

Solution

Physics sir · 3 steps

At resonance the inductor and capacitor cancel each other, so the circuit behaves like R alone.

1. Reactances cancel. At resonance X_L = X_C, so the impedance Z = √(R² + (X_L - X_C)²) drops to just R = 5 Ω. The values of L and C only fix the resonant frequency, ω₀ = 1/√LC = 10⁴ rad s⁻¹.
2. Current and phase. The current is I = V/R = 250/5 = 50 A. The current is in phase with the voltage, so the power factor is cos φ = 1.

   P = V²/R = (250)²/5 = 62500/5 = 12500 W

3. So. P = 12500 W = 125 × 10² W, so the blank is 125.

The common slip: 12500. Writing the power in watts gives 12500. The blank is already multiplied by 10², so the answer is 125.

The idea to keep. At resonance, treat the series LCR circuit as a pure resistor: Z = R, power factor 1, and P = V²/R. The values of L and C only set the frequency.

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