A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8Ω, L = 24 mH and C = 60 μF. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.

Solution

Physics sir · 3 steps

At resonance X_L and X_C cancel, so the circuit is just R. Use the rms voltage, not the peak.

1. Resonance condition. At resonance X_L = X_C, so the impedance becomes Z = R = 8 Ω and the power factor is 1. The values of L and C only fix the resonant frequency; they do not change the power.
2. Convert to rms. Power depends on the rms voltage: Vᵣₘₛ = V₀/√2, so Vᵣₘₛ² = 250²/2 = 31250 V².

   P = Vᵣₘₛ²/R = V₀²/2R = 250²/(2 × 8) = 62500/16 = 3906.25 W

3. So. P ≈ 3.9 kW, which rounds to 4.

The common slip: 8. Using the peak voltage instead of the rms voltage gives 250²/8 = 7812.5 W, which rounds to 8, not 4.

The idea to keep. At resonance an LCR circuit behaves like a pure resistor, so P = Vᵣₘₛ²/R. Whenever a peak value is given, divide its square by 2 before finding power.

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