An alternating current (AC) circuit consists of a resistor of resistance R = 30\ \Omega, a… — JAMB UTME

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ connected in series across an AC source of RMS voltage $150\ \text{V}$ . What is the average power dissipated in the circuit in watts?

Answer: 270 W

Answer

The average power dissipated in the circuit is

270\ \text{W}

The total impedance of a series RLC circuit is

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = 50\ \Omega

. The RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150}{50} = 3\ \text{A}

. Because pure inductors and capacitors consume zero average power over a complete cycle, power is dissipated only across the resistor, giving

P = I_{\text{rms}}^2 R = 3^2 \times 30 = 270\ \text{W}

Step-by-Step Solution

Calculate net reactance

X = 40\ \Omega

Inductive and capacitive reactances oppose each other in phase, so net reactance is

X_L - X_C

Calculate total circuit impedance

Z = 50\ \Omega

Resistance and net reactance add in quadrature:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Calculate RMS current

I_{\text{rms}} = 3\ \text{A}

Ohm's law for AC circuits gives

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}

Calculate average power dissipated

P = 270\ \text{W}

Power is dissipated exclusively by resistance in an AC circuit:

P = I_{\text{rms}}^2 R

Key Concept

Power Dissipation in AC Circuits

Estimated Time:1m 30s

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