ElectricalAC CircuitsmediumFE ExamCourseworkPE / Advanced

Real Power and Power Factor in an R-L Load

The series R-L circuit from before has $R = 30\ \Omega$ , $X_L = 40\ \Omega$ , and carries $2.4\ \text{A}$ rms from a $120\ \text{V}$ rms source.

Find the real (average) power dissipated.

Given

R	30 ohm	Resistance
XL	40 ohm	Inductive reactance
I	2.4 A	Current (rms)
V	120 V	Source voltage (rms)

Hint 1

Which circuit element actually dissipates energy over a full cycle?

Hint 2

An ideal inductor returns all stored energy each cycle — its average power is zero.

Hint 3

P = I²R, using only the resistance. Cross-check with VI·cosθ.

Worked solution — try the problem first

Only the resistance dissipates real power. An ideal inductor stores energy in its magnetic field and returns it each cycle — its average power is zero.

$P = I^2 R = (2.4)^2(30) = 5.76 \times 30 = 172.8\ \text{W}$

$\boxed{P = 172.8\ \text{W}}$

Cross-check via power factor:

$|Z| = 50\ \Omega, \qquad \text{pf} = \frac{R}{|Z|} = \frac{30}{50} = 0.6\ \text{lagging}$

$P = VI\cos\theta = 120 \times 2.4 \times 0.6 = 172.8\ \text{W} \ \checkmark$

The full power triangle:

Apparent power $S = VI = 288$ VA
Real power $P = 172.8$ W
Reactive power $Q = I^2X_L = 5.76 \times 40 = 230.4$ VAR

Check: $\sqrt{P^2 + Q^2} = \sqrt{172.8^2 + 230.4^2} = \sqrt{29{,}860 + 53{,}084} = \sqrt{82{,}944} = 288$ ✓

Utilities bill for real power but must size conductors for apparent power — which is why a low power factor attracts a penalty.

Concepts:Real vs apparent powerPower factorPower triangle

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