Learn › Topic 7

Alternating Current and Phasors: Sine Waves, RMS and Impedance

8 min read · bitGateLab Team

The power that reaches your home, the signal from a microphone and the output of a function generator have something in common: they change over time following, or approximating, a sine wave. To analyze circuits with this kind of signal you need three tools: knowing how to read the parameters of a sinusoid, understanding what the RMS value means, and working with phasors and impedances. With them, an alternating current (AC) circuit is solved with the same Ohm's and Kirchhoff's laws you already use for direct current, only with complex numbers.

The sinusoidal signal and its parameters

A sinusoidal voltage is written like this:

v(t) = V_p · sin(ωt + φ)

Amplitude or peak value (V_p): the maximum value the signal reaches, measured from zero. The peak-to-peak value is V_pp = 2·V_p.
Frequency (f): how many complete cycles occur in one second, in hertz (Hz). The period is the duration of one cycle: T = 1/f.
Angular frequency (ω): ω = 2πf, in radians per second. This is the form in which frequency appears inside the formulas.
Phase (φ): tells you how far the wave leads or lags a reference. It is expressed in degrees or radians. One full cycle is 360°.

Phase on its own does not say much; what matters is the phase difference between two signals of the same frequency, for example between the voltage and the current of a component.

Effective (RMS) value

The RMS (root mean square) value of an alternating voltage is the DC voltage that would deliver the same average power to a resistor. For a pure sinusoid:

V_rms = V_p / √2 ≈ 0.707 · V_p

The relationship V_p/√2 holds only for sinusoids. A symmetric square wave of amplitude V_p has V_rms = V_p, and a triangle wave has V_p/√3. AC multimeters that are not "true RMS" assume a sinusoid and give wrong readings for other waveforms.

Example: the wall outlet. In Mexico, household mains power is 127 V RMS at 60 Hz.

Peak value: V_p = 127 · √2 = 179.6 V.
Peak-to-peak value: V_pp = 2 · 179.6 = 359.2 V.
Period: T = 1/60 = 16.67 ms.
Angular frequency: ω = 2π · 60 = 377 rad/s.

So the outlet voltage is v(t) = 179.6 · sin(377t) V. When someone says "127 volts", they mean the RMS value, not the peak.

Phasors: the sinusoid as a complex number

In a linear circuit driven at a single frequency, all the voltages and currents are sinusoids of that same frequency. The only things that change from one signal to another are the amplitude and the phase. A phasor stores exactly those two pieces of information in a complex number:

v(t) = V_p · sin(ωt + φ)   →   V = V_p ∠ φ

You can write it in polar form (magnitude and angle) or rectangular form (real and imaginary parts):

V_p ∠ φ = V_p·cos φ + j·V_p·sin φ

Here j = √(−1); electronics uses j instead of i so that it is not confused with current. The advantage of phasors is that differentiating with respect to time becomes multiplying by jω. That turns the differential equations of inductors and capacitors into algebraic equations. You can use peak values or RMS values in phasors, as long as you do not mix them in the same calculation.

Impedance of R, L and C

The impedance Z is the AC version of resistance: Z = V/I, measured in ohms. It is a complex number: its real part is the resistance and its imaginary part is the reactance X.

Component	Impedance	Reactance	Phase relationship
Resistor	Z = R	0	V and I in phase
Inductor	Z = jωL	X_L = ωL	V leads I by 90°
Capacitor	Z = 1/(jωC) = −j/(ωC)	X_C = 1/(ωC)	V lags I by 90°

Notice the frequency dependence: the inductor's reactance grows with f, and the capacitor's decreases. At DC (f = 0) the inductor is a short circuit and the capacitor an open circuit; at very high frequencies the opposite happens.

Example: reactances at 1 kHz. A 10 mH inductor and a 1 µF capacitor at f = 1 kHz (ω = 6283 rad/s):

X_L = 6283 · 0.010 = 62.8 Ω, so Z_L = j62.8 Ω.
X_C = 1/(6283 · 1×10⁻⁶) = 159.2 Ω, so Z_C = −j159.2 Ω.

If you double the frequency to 2 kHz, X_L rises to 125.7 Ω and X_C falls to 79.6 Ω.

Complete example: a series RC circuit in AC

Impedances in series add just like resistances, and Ohm's law is applied with phasors.

Given: a sinusoidal source of 5 V peak at 100 Hz drives a 1 kΩ resistor in series with a 1 µF capacitor. Find the current and the voltages across R and across C.

Capacitor reactance: X_C = 1/(2π · 100 · 1×10⁻⁶) = 1591.5 Ω.
Total impedance: Z = 1000 − j1591.5 Ω.
Magnitude: |Z| = √(1000² + 1591.5²) = 1879.6 Ω. Angle: arctan(−1591.5/1000) = −57.9°. So Z = 1879.6 ∠ −57.9° Ω.
Current, taking the source as the reference (5 ∠ 0° V): I = 5 ∠ 0° / 1879.6 ∠ −57.9° = 2.66 ∠ +57.9° mA peak. The current leads the source voltage by 57.9°: the circuit is capacitive.
Voltage across R: V_R = R·I = 2.66 ∠ 57.9° V peak (in phase with the current).
Voltage across C: V_C = (−j1591.5)·I = 4.23 ∠ −32.1° V peak (lagging the current by 90°).
Check: √(2.66² + 4.23²) = 5.00 V. The phasor sum reproduces the source.

Notice that 2.66 + 4.23 = 6.89 V, more than the 5 V of the source. There is no contradiction: Kirchhoff's voltage law holds for phasors, not for the bare magnitudes, because the two voltages do not reach their peaks at the same time.

Phasor diagram taking V_R (and the current, which is in phase with it) as the horizontal reference. V_C is 90° behind, and the sum of the two is the source voltage, which lags the current by 57.9°.

Common mistakes

Mixing peak and RMS values in the same calculation. Choose one at the start and stick with it to the end.
Adding the magnitudes of voltages that have different phases. In AC you add phasors (real parts with real parts and imaginary parts with imaginary parts).
Using f instead of ω in X_L = ωL or X_C = 1/(ωC). Forgetting the 2π gives errors of more than a factor of six.
Setting the calculator to radians when you are working in degrees, or vice versa, when calculating arctangents.
Applying V_rms = V_p/√2 to waveforms that are not sinusoidal.

Try it in the simulator

Build the RC circuit from the example and compare what the simulator measures with the numbers you calculated by hand. On the virtual oscilloscope you will see the current leading and the two voltages out of phase.

Place a sine-wave voltage source and set it to 5 V amplitude and 100 Hz.
Connect a 1 kΩ resistor and a 1 µF capacitor in series, and close the circuit to ground.
Connect one oscilloscope channel to the source output and another to the node between R and C.
Observe the amplitude across the capacitor (close to 4.23 V) and its lag with respect to the source.
Run an AC analysis at 100 Hz and compare the magnitudes and phases with those of the example.
Change the frequency to 1 kHz and watch the capacitor voltage drop.

Exercises

A sinusoid measures 20 V peak to peak. What is its RMS value?

Show answer
V_p = 20/2 = 10 V. V_rms = 10/√2 = 7.07 V.
Calculate the reactance of a 10 nF capacitor at 10 kHz.

Show answer
X_C = 1/(2π · 10 000 · 10×10⁻⁹) = 1/(6.283×10⁻⁴) = 1591.5 Ω.
A series RL circuit has R = 100 Ω and L = 10 mH at 1 kHz. Find the impedance in polar form.

Show answer
X_L = 2π · 1000 · 0.01 = 62.83 Ω. |Z| = √(100² + 62.83²) = 118.1 Ω. Angle = arctan(62.83/100) = 32.1°. Z = 118.1 ∠ 32.1° Ω; the current lags the voltage by 32.1°.
At what frequency does the reactance of a 1 µF capacitor equal 1 kΩ?

Show answer
From X_C = 1/(2πfC), you get f = 1/(2π · X_C · C) = 1/(2π · 1000 · 1×10⁻⁶) = 159.2 Hz.