Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.

Solution

Physics sir · 3 steps

Circuit (b) can only match the pure resistor at resonance, so find ω = 1/√LC.

1. Power in (a). With only the resistor, the average power is Pₐ = V²/R, where V = 220 V is the rms voltage.
2. Power in (b). In the series LCR circuit, Iᵣₘₛ = V/Z, so P_b = Iᵣₘₛ² R = (V² R)/Z². Both circuits have the same V and the same R = 5 Ω, so Pₐ = P_b only when Z = R. That needs X_L = X_C, which is resonance.

   ω = 1/√LC = 1/√(0.1 × 40 × 10⁻⁶) = 1/√(4 × 10⁻⁶) = 1/(2 × 10⁻³)

3. So. ω = 500 rad s⁻¹, and the answer is 500.

The common slip: 80. Dividing by 2π to get the frequency in hertz gives about 80, but the question asks for the angular frequency in rad/s, which is 500.

The idea to keep. Average power in a series LCR circuit is V²R/Z², which is at most V²/R. It reaches that value, equal to a bare resistor's power, only at resonance, where Z = R.

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