Skip to content
Electrical Engineering

Sinusoidal steady state: phasors and impedance

Why sinusoids, the phasor and its operations, impedance and admittance of resistor, inductor and capacitor, a complete example and the phasor method step by step.

Updated 20 October 2026 3 min read 0 comments

The relations between voltage and current in two-terminal elements hold at every instant. With constant quantities they are written with algebra; when quantities vary in time, capacitors and inductors bring in derivatives, and the circuit equations become differential. There is, however, one case, the most important of all, in which you return to algebra: when every source is sinusoidal at the same frequency and the circuit has reached steady state. With complex numbers, phasors, derivatives become multiplications and all the methods for resistive circuits apply again. It is the language of the 50 Hz power grid, of audio and of radio.

Why sinusoids

In a linear circuit, if every source is sinusoidal with angular frequency \omega, at steady state every voltage and current is a sinusoid with the same angular frequency: only amplitude and phase change. The reason is that differentiating or integrating a sinusoid gives another sinusoid of the same frequency, shifted by 90°. Moreover, by Fourier series and transforms, any signal is a sum of sinusoids: knowing the response to a sinusoid of any frequency means being able to compute the response to any signal.

The phasor

A sinusoid v(t) = V_M\cos(\omega t + \varphi) is determined, once \omega is known, by two numbers: amplitude and phase. They are gathered into a single complex number, the phasor:

    \[ \bar V = V\,e^{j\varphi} = V\angle\varphi,\qquad v(t) = \sqrt2\,\operatorname{Re}\{\bar V e^{j\omega t}\} \]

In circuit theory the letter j is used for the imaginary unit (i is reserved for current) and the phasor’s magnitude is usually the RMS value V = V_M/\sqrt2, so that the power formulas have no \frac12 factors.

A rotating phasor and the sinusoid traced by its projection
The phasor is a vector in the complex plane rotating at angular velocity ω; its projection on an axis traces the sinusoid in time. The magnitude gives the amplitude, the initial angle the phase.

Operations. Adding sinusoids of the same frequency means adding their phasors as vectors, with the parallelogram rule. Differentiating means multiplying the phasor by j\omega (amplitude times \omega, phase advanced by 90°); integrating means dividing by j\omega.

Example. v_1 = 10\sqrt2\cos(\omega t) and v_2 = 10\sqrt2\cos(\omega t + 90°): the phasors are 10 and 10j, the sum 10 + 10j = 14.1\angle 45°, that is v_1 + v_2 = 14.1\sqrt2\cos(\omega t + 45°). Adding the amplitudes would have given 20: wrong, because the peaks do not occur at the same instant.

Impedance and admittance

For each element the ratio of the voltage phasor to the current phasor is the impedance:

    \[ \bar Z = \frac{\bar V}{\bar I} = R + jX \]

ElementImpedanceVoltage-current phase shift
resistorR0
inductorj\omega L+90° (voltage leads)
capacitor\frac{1}{j\omega C} = -\frac{j}{\omega C}−90° (voltage lags)

The real part R is the resistance, the imaginary part X the reactance: positive for inductive behaviour, negative for capacitive. The inverse of impedance is admittance \bar Y = 1/\bar Z = G + jB, handy for parallel connections.

With impedances all the results for resistive circuits hold: Kirchhoff’s laws for phasors, series and parallel, dividers, nodal analysis, superposition, Thevenin and Norton. You just work with complex numbers.

A complete example

A 230 V, 50 Hz source feeds a resistance R = 30 Ω in series with an inductor L = 127 mH. The reactance is X_L = \omega L = 2\pi\cdot 50\cdot 0.127 \approx 40 Ω, so

    \[ \bar Z = 30 + 40j = 50\angle 53.1°\ \Omega \]

The current has an RMS value of 230/50 = 4.6 A and lags the voltage by 53.1°:

    \[ \bar I = \frac{230\angle 0°}{50\angle 53.1°} = 4.6\angle{-53.1°}\ \text{A} \]

The voltages across the components are \bar V_R = R\bar I = 138\angle{-53.1°} V and \bar V_L = jX_L\bar I = 184\angle 36.9° V. Added as magnitudes they would give 322 V, more than the source’s 230 V: it looks like a paradox, but the two phasors are 90° apart, and indeed \sqrt{138^2 + 184^2} = 230.

The method in practice

  1. Choose a phase reference (usually a source at zero phase).
  2. Replace the sources with their phasors and the components with their impedances at angular frequency \omega.
  3. Solve the circuit with the methods for resistive circuits, in complex arithmetic.
  4. Go back to time: \bar I = I\angle\varphi becomes i(t) = \sqrt2\,I\cos(\omega t + \varphi).

Careful: the method holds only at steady state and with sources of the same frequency. With different frequencies you solve one circuit per frequency and add the results in time (superposition), never the phasors. Transients need the methods of the previous lesson or the Laplace transform.

The following lessons apply this method to power in sinusoidal steady state, to frequency response and to three-phase systems.

Related articles

Subscribe
Notify of
guest
0 Comments
Oldest
Newest Most Voted
0
Would love your thoughts, please comment.x
()
x