A series L-C-R circuit is designed to resonate at an angular frequency ω₀ = 10⁵ rad/s. The circuit draws 16W power from 120V source at resonance. The value of resistance 'R' in the circuit is _________ Ω.

Solution

Physics sir · 3 steps

At resonance the L and C cancel, so the circuit is just R and P equals V squared over R.

1. What resonance does. At resonance X_L = X_C, so the two reactances cancel and the impedance becomes Z = R. The power factor is 1, so the source sees a pure resistor.
2. Power in R. The current is V/R, so the power drawn is P = I²R = V²/R. Here V = 120 V is the rms value, as source voltages are by default.

   R = V²/P = 120²/16 = 14400/16

3. So. R = 900 Ω. The value ω₀ = 10⁵ rad s⁻¹ only tells you the circuit is at resonance; it is not needed in the calculation. The answer is 900.

The common slip: 450. Treating 120 V as the peak value and using Vᵣₘₛ = 120/√2 gives R = 7200/16 = 450 Ω, not 900.

The idea to keep. At resonance a series LCR circuit behaves like a pure resistor: Z = R and P = Vᵣₘₛ²/R. Take a source voltage as rms unless the question says peak.

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