An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a resistor of resistance 50 Ω are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is ___________ W.

Solution

Physics sir · 4 steps

Only the resistor eats power, so find the current through the full impedance and use I²R.

1. Net reactance. The inductor and the capacitor push the phase in opposite directions, so their reactances subtract: X_L - X_C = 100 - 50 = 50 Ω.

   Z = √(R² + (X_L - X_C)²) = √(50² + 50²) = 50√2 Ω

2. RMS current. The 10 V is an rms value, so Iᵣₘₛ = Vᵣₘₛ/Z = 10/50√2 = 1/5√2 A. Squaring gives Iᵣₘₛ² = 1/50 A².
3. Power in resistor. An ideal inductor and capacitor take no average power, so all of it goes into R: P = Iᵣₘₛ² R = 1/50 × 50 = 1 W.
4. So. The average power dissipated is 1 W. You get the same result from P = Vᵣₘₛ Iᵣₘₛ cos φ with cos φ = R/Z = 1/√2.

The common slip: 2. Using P = V²/R with the full source voltage across the resistor gives 100/50 = 2 W. Only part of the 10 V appears across R, so the current must come from Z.

The idea to keep. In a series LCR circuit, average power is Iᵣₘₛ² R or Vᵣₘₛ Iᵣₘₛ cos φ. The reactances only limit the current. They never dissipate energy themselves.

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