A capacitor of reactance 4 √3 Ω and a resistor of resistance 4 Ω are connected in series with an ac source of peak value 8 √2 V. The power dissipation in the circuit is __________ W.

Solution

Physics sir · 4 steps

Only the resistor burns power, so find the rms current and use Iᵣₘₛ² R.

1. Impedance. In a series RC circuit, Z = √(R² + X_C²) = √(16 + 48) = √64 = 8 Ω.
2. Rms values. The rms voltage is Vᵣₘₛ = 8√2/√2 = 8 V. So Iᵣₘₛ = Vᵣₘₛ/Z = 8/8 = 1 A.

   P = Iᵣₘₛ² R = Vᵣₘₛ Iᵣₘₛ cos φ

3. Power. The capacitor stores and returns energy, so it dissipates none. P = (1)² × 4 = 4 W. The check with power factor cos φ = R/Z = 1/2 agrees: 8 × 1 × 1/2 = 4 W.
4. So. The power dissipated in the circuit is 4 W.

The common slip: 8. Using the peak current √2 A in I² R gives 8 W, double the true average power.

The idea to keep. Average power in AC always uses rms values, and only resistance dissipates it. Convert the peak voltage to rms first, then use Iᵣₘₛ² R.

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