Single-phase AC circuits
Find AC current, phase angle and power
Solve a series RLC example with RMS values, use a phasor triangle, and check real and reactive power.
What you’ll learn
- Distinguish impedance from resistance and RMS values from peak values.
- Calculate current, power factor and real power, then check the result independently.
Before you start: Know Ohm’s law, square roots and basic trigonometry. Use a calculator in degree mode for the angles.
In this guide 6 sections
1. Decide what the numbers describe
Consider a resistor, an ideal inductor and an ideal capacitor connected in series. The source is sinusoidal and the circuit has reached steady state. At the source frequency, the resistance is 12 Ω, the inductive reactance is 25 Ω and the capacitive reactance is 9 Ω. The source voltage is 120 V RMS.
We are given reactances, not inductance and capacitance. That means we can begin the calculation without knowing the frequency. If a question gives L and C instead, calculate XL = 2πfL and XC = 1/(2πfC) first. A change of frequency changes those reactances, even when the components stay the same.
Write “RMS” beside the voltage before starting. Dividing an RMS voltage by impedance produces an RMS current. Dividing a peak voltage produces a peak current. Mixing the two conventions creates a √2 error that can survive several otherwise correct steps.
2. Build the impedance triangle
Inductive and capacitive effects oppose each other in the imaginary part of a series impedance. The net reactance is X = 25 − 9 = 16 Ω. Therefore Z = 12 + j16 Ω. Here j represents a quarter-turn in the complex plane; it is not another unit.
|Z| = √(12² + 16²) = √400 = 20 Ω.
The positive net reactance tells us that the circuit is inductive. Its impedance angle is φ = tan⁻¹(16 / 12) ≈ 53.13°.
3. Find current and explain its phase
The RMS current magnitude is I = 120 / 20 = 6 A. If the source voltage is our 0° reference, the current phasor is 6∠−53.13° A. The minus sign means that current lags the voltage. This is consistent with the positive inductive reactance found earlier.
Every series component carries this same current. Their voltage magnitudes are VR = 6 × 12 = 72 V, VL = 6 × 25 = 150 V and VC = 6 × 9 = 54 V. An individual reactive voltage can exceed the source voltage: these are phasor quantities, not three in-phase drops.
Check the source using the component voltages: √(72² + (150 − 54)²) = √14400 = 120 V. Simply adding 72 + 150 + 54 would fail this check.
4. Use power as an independent check
The power factor is cos φ = R / |Z| = 12 / 20 = 0.6 lagging. Apparent power is S = VI = 720 VA. Real power is P = VI cos φ = 432 W. For the ideal components in this problem, only the resistor consumes average power, and I²R = 6² × 12 = 432 W confirms the answer.
Net reactive power is Q = I²X = 6² × 16 = 576 var, positive for this inductive load. The power triangle also closes: √(432² + 576²) = 720 VA. Keep W, var and VA distinct so that the physical meaning of each result remains clear.
Try two changes before opening the answers
A. Make the circuit capacitive
Keep R = 12 Ω and XL = 25 Ω, but change XC to 41 Ω at the same frequency. With the same 120 V RMS source, find current and power factor.
Show the working
X = 25 − 41 = −16 Ω. The impedance magnitude remains 20 Ω, so current remains 6 A RMS. The impedance angle is now −53.13° and current leads the source. Power factor is 0.6 leading, real power is still 432 W, and net reactive power is −576 var.
B. Cancel the reactance
Set XC = 25 Ω while keeping R and the source unchanged. Predict current and real power.
Show the working
The net reactance is zero. Z = 12 Ω, current is 10 A RMS, and power factor is 1. Real power is 1200 W. The inductor and capacitor can each have a large voltage, but those two voltages cancel as phasors in the ideal series circuit.
Where this method applies
This calculation assumes linear components, one sinusoidal frequency and steady state. It does not describe a switching transient, a distorted current waveform or the losses of a real inductor. For a parallel network, combine branch admittances or currents instead of reusing this series formula.
Formula reference: OpenStax, University Physics Volume 2, section 15.3, and section 15.4 on AC power. The numerical example, diagram and practice variations on this page were prepared for Padhnus.