FOUNDATION · CHAPTER 6 OF 7

Sinusoids, complex numbers and phasors

This chapter shows why a linear circuit driven at one frequency settles to that frequency, and turns each sinusoid into a complex number, the phasor. From it come impedance, average power, rms value and reactive power: the quantities used to analyse and rate ac circuits and power systems.

Updated 12 Oct 2026

What you will be able to do

You will see why the voltages and currents of a linear circuit driven at one frequency settle to that frequency. You will derive the phasor and impedance instead of memorising them. You will get average power and rms value from integrals over one period, and read reactive power as energy swapped with the fields.

Before you start

Chapter 1, Charge, current, voltage and power: what the quantities mean
Chapter 2, Why R, L and C behave as they do
Chapter 5, Differential equations for circuits
Complex-number algebra (recalled below)
$cos (α - β) = cos α cos β + sin α sin β$ , $2 cos α cos β = cos (α - β) + cos (α + β)$ , $cos^{2} α, sin^{2} α = \frac{1}{2} (1 \pm cos 2 α)$
The chain rule; the integral of a cosine over whole periods

Why one frequency is enough, and what a phasor is

Chapter 5 split a forced response into a natural response and a particular solution that follows the source. When every root of the characteristic equation has a negative real part, the natural response decays and, after a few time constants, only the particular solution is left: the sinusoidal steady state. A lossless L-C loop is the exception: its roots are $\pm j ω_{0}$ , with $ω_{0} = 1/ L C$ rad/s, and its natural response never dies. Here you find the particular solution for a cosine source.

Recap of complex numbers: $j$ is the imaginary unit, $j^{2} = - 1$ . A number $z = a + j b = ∣ z ∣ e^{j θ} = ∣ z ∣∠ θ$ has real part $Re {z} = a$ , $∣ z ∣ = a^{2} + b^{2}$ and $θ$ the angle of the point $(a, b)$ , equal to $arctan (b / a)$ when $a > 0$ . Multiplying multiplies sizes and adds angles; dividing divides sizes and subtracts angles. The conjugate $z^{*} = a - j b$ has angle $- θ$ .

EULER'S FORMULA

e^{j θ} = cos θ + j sin θ, cos θ = Re {e^{j θ}}

Holds for any real angle $θ$ in radians. So a cosine is the real part of a rotating complex number: $cos (ω t + θ) = Re {e^{j θ} e^{j ω t}}$ , with $ω$ in rad/s and $t$ in s.

A ROTATING EXPONENTIAL TURNS THE EQUATION INTO ALGEBRA

\frac{d}{d t} e^{j ω t} = j ω e^{j ω t} a_{2} x^{''} + a_{1} x^{'} + a_{0} x = F e^{j ω t} \Rightarrow x = \frac{F e ^{j ω t}}{a _{2} ( j ω ) ^{2} + a _{1} j ω + a _{0}}

Constant real $a_{2}, a_{1}, a_{0}$ (first order: $a_{2} = 0$ ), real $ω$ (rad/s); $x$ is any circuit variable, $x^{'} = d x / d t$ , $x^{''} = d^{2} x / d t^{2}$ , $F$ a complex constant. Substituting $x = X e^{j ω t}$ turns each derivative into a factor $j ω$ ; cancel $e^{j ω t}$ . The denominator is Chapter 5's characteristic polynomial at $s = j ω$ , nonzero when every root has a negative real part.

Write a cosine source of peak $V_{m}$ as the real part of a complex one: $V_{m} cos (ω t + θ) = Re {V_{m} e^{j θ} e^{j ω t}}$ . The circuit's ODE has real coefficients, so the real part passes through its sums and derivatives: solve with the complex source and take the real part at the end. By the result above, every variable is then a constant times $e^{j ω t}$ . Any other particular solution differs from this one by a natural response, which decays. So this is the one steady state, and every voltage and current in it has frequency $ω$ .

THE PHASOR, THE COEFFICIENT SCALED TO RMS

v (t) = Re {2 V e^{j ω t}}, V = \frac{V _{m}}{2} e^{j θ} = ∣ V ∣∠ θ \Rightarrow v (t) = 2 ∣ V ∣ cos (ω t + θ)

At one $ω$ (rad/s); $V_{m}$ is the peak (V). The phasor $V$ (bold) is the coefficient of $e^{j ω t}$ divided by $2$ , so $∣ V ∣ = V_{m} / 2$ : the rms value, the square root of the mean of $v^{2}$ over a period, as the power section shows. $θ$ is the phase of a cosine: write a sine as $cos (ω t - 9 0^{\circ})$ first. A plain $V$ means $∣ V ∣$ .

WATCH OUT FOR

Adding phasors of different frequencies. Test: phasors add only when they share one $ω$ . If the frequencies differ, find each response separately and add the time functions.

Impedance: Ohm's law for phasors

Use the passive sign convention of Chapter 1: the current enters the + terminal of the element's voltage. Put $v = 2 Re {V e^{j ω t}}$ and $i = 2 Re {I e^{j ω t}}$ into each element law of Chapter 2. Each derivative becomes a multiplication by $j ω$ inside the real part. If $Re {A e^{j ω t}} = Re {B e^{j ω t}}$ for all $t$ , then $A = B$ : at $ω t = 0$ the real parts agree, at $ω t = π /2$ the imaginary parts do. What is left is algebra on complex numbers.

ELEMENT LAWS AS PHASORS

V_{R} = R I, V_{L} = j ω L I, I_{C} = j ω C V_{C}

From $v = R i$ , $v = L d i / d t$ and $i = C d v / d t$ by the step above. Valid in the steady state at one $ω$ (rad/s), for linear fixed elements; $R$ in ohm, $L$ in H, $C$ in F. KCL and KVL hold for phasors by the same step, since each is a sum of same-frequency terms equal to zero.

IMPEDANCE

Z = \frac{V}{I} = R + j X, X_{L} = ω L, X_{C} = - \frac{1}{ω C}

At one $ω$ , passive convention. $Z$ , the resistance $R$ and the reactance $X$ are in ohm. $Z$ is a ratio of phasors, not a phasor: it stands for no sinusoid. In series the same $I$ flows and KVL adds the voltages, so impedances add; for R, L and C in series, $Z = R + j (ω L - 1/ ω C)$ .

Element	Z	Angle of Z	At ω = 0	At ω → ∞
R	R	0°	R	R
L	jωL	+90°	short circuit	open circuit
C	1/(jωC)	−90°	open circuit	short circuit

The angle of $Z$ is $ϕ$ , the angle by which voltage leads current, because dividing phasors subtracts their angles. An inductor has $ϕ = + 9 0^{\circ}$ : its current lags. A capacitor has $ϕ = - 9 0^{\circ}$ : its current leads.

WORKED EXAMPLE

A source $V = 10∠ 0^{\circ}$ V (rms) at $ω = 1000$ rad/s drives R = 3 ohm in series with L = 4 mH. Find $i (t)$ in the steady state.

Solution

$X_{L} = ω L = 1000 \times 0.004 = 4$ ohm. We use $Z_{L} = j ω L$ because only the steady state is wanted.
The same current flows in both elements, so impedances add: $Z = 3 + j 4 = 5∠53.1 3^{\circ}$ ohm.
$I = V / Z = 10∠ 0^{\circ} /5∠53.1 3^{\circ} = 2∠ - 53.1 3^{\circ}$ A (rms). Dividing divides sizes and subtracts angles.
Back to time with $i (t) = 2 ∣ I ∣ cos (ω t + θ)$ : $i (t) = 22 cos (1000 t - 53.1 3^{\circ}) \approx 2.83 cos (1000 t - 53.1 3^{\circ})$ A.

Answer: i(t) = 2.83 cos(1000t − 53.13°) A. Current lags voltage by 53.13°, so φ = +53.13°.

Power over one cycle

Power is $p = v i$ . With the passive convention it is the power absorbed, and $p < 0$ means the element delivers power at that instant. It changes within a cycle, so you ask two things: its average, and the size of the part that flows in and back out. Take $v = 2 V cos ω t$ and $i = 2 I cos (ω t - ϕ)$ , where $V = ∣ V ∣$ and $I = ∣ I ∣$ are rms sizes and $ϕ$ is the angle by which voltage leads current. Let $T = 2 π / ω$ .

INSTANTANEOUS POWER

p = 2 V I cos ω t cos (ω t - ϕ) = V I [cos ϕ + cos (2 ω t - ϕ)]

Holds for any load whose $v$ and $i$ are sinusoids at one $ω$ , passive convention. Uses $2 cos α cos β = cos (α - β) + cos (α + β)$ with $α = ω t$ , $β = ω t - ϕ$ . $V$ and $I$ are rms values (V, A), $t$ in s. The power has a constant part and a part at twice the source frequency.

AVERAGE POWER AND RMS VALUE

P = \frac{1}{T} \int_{0}^{T} p d t = V I cos ϕ, V_{rms}^{2} = \frac{1}{T} \int_{0}^{T} v^{2} d t

Sinusoidal steady state. The $cos (2 ω t - ϕ)$ term integrates to zero over a period; $P$ is in W. For a resistor $p = v^{2} / R$ , so $P = V_{rms}^{2} / R$ : $V_{rms}$ is the dc voltage that heats $R$ equally. For $v = V_{m} cos ω t$ , $cos^{2} = \frac{1}{2} (1 + cos 2 ω t)$ gives a mean of $V_{m}^{2} /2$ , so $V_{rms} = V_{m} / 2$ : the $2$ in the phasor.

REACTIVE POWER

p = P (1 + cos 2 ω t) + Q sin 2 ω t, Q = V I sin ϕ

Expand $cos (2 ω t - ϕ) = cos ϕ cos 2 ω t + sin ϕ sin 2 ω t$ . For a load with $P \geq 0$ the $P$ term never goes negative; it averages $P$ . The $Q$ term averages zero: power that flows in and back out. $Q$ is in var; $V I$ , in VA, is the apparent power. Lagging ( $ϕ > 0$ ) gives $Q > 0$ .

REACTIVE POWER AS SWAPPED ENERGY

i = 2 I sin ω t : w_{L} = \frac{1}{2} L i^{2} = \frac{1}{2} L I^{2} (1 - cos 2 ω t) p = v i = \frac{d w _{L}}{d t} = ω L I^{2} sin 2 ω t \Rightarrow P = 0, Q_{L} = ω L I^{2} = ω W_{peak}

Inductor $L$ (H), rms current $I$ (A), passive convention. Then $v = L d i / d t = 2 ω L I cos ω t$ , so $ϕ = + 9 0^{\circ}$ and $Q = V I$ . $p$ is the rate of change of the stored energy $w_{L}$ (J), which repeats, so $P = 0$ . $w_{L}$ swings from 0 to $W_{peak} = L I^{2}$ : in each quarter period $L I^{2}$ flows into the field, in the next it flows back.

WATCH OUT FOR

Taking $P = V I$ for any load. Test: $P = V I cos ϕ$ equals $V I$ only when $ϕ = 0$ , which needs a purely resistive $Z$ : an R alone, or reactances that cancel at that $ω$ .

WORKED EXAMPLE

A load has $V = 100∠ 0^{\circ}$ V and $I = 5∠ - 3 0^{\circ}$ A (rms, passive convention). Find $P$ and $Q$ and say what the load is.

Solution

Voltage leads current by $0^{\circ} - (- 3 0^{\circ}) = 3 0^{\circ}$ , so $ϕ = + 3 0^{\circ}$ .
$P = V I cos ϕ = 100 \times 5 \times cos 3 0^{\circ} = 433$ W, the average of $p$ over a period. $cos ϕ = 0.866$ is the power factor.
$Q = V I sin ϕ = 500 \times 0.5 = 250$ var: the amplitude of the $Q sin 2 ω t$ term in $p$ , the power that flows in and back out.
$P > 0$ , so the load absorbs power on average. $Q > 0$ means current lags, so the load is resistive and inductive.

Answer: P = 433 W, Q = 250 var; a resistive-inductive load with lagging power factor cos 30° = 0.866.

Check yourself

CHECK YOURSELF 1

A linear circuit of fixed R, L and C is driven by $v_{s} = 10 cos (100 t)$ V. After the natural response has died away, which statement is true?

(A)
Every voltage and current is sinusoidal at 100 rad/s, with its own size and phase
(B)
The inductor current has frequency 200 rad/s
(C)
The capacitor voltage keeps decaying forever
(D)
Each current has the source's size but a different frequency

Answer and solution

Answer: (A)

With the natural response gone, only the particular solution remains.
Sums and derivatives keep the factor $e^{j ω t}$ , so that solution has the source frequency, 100 rad/s, everywhere.
Only size and phase change from one variable to another.

CHECK YOURSELF 2

A capacitor C = 100 micro F carries a sinusoidal current at $ω = 1000$ rad/s. Find $∣ Z_{C} ∣$ in ohm.

Answer and solution

Answer: $10$ (accepted range $\pm 0.05$ )

$Z_{C} = 1/ (j ω C)$ .
$∣ Z_{C} ∣ = 1/ (1000 \times 100 \times 1 0^{- 6}) = 1/0.1 = 10$ ohm.

CHECK YOURSELF 3One or more options are correct

An ideal capacitor C is in the sinusoidal steady state at $ω > 0$ , with the passive sign convention. Which statements are true?

(A)
Its current leads its voltage by 90°
(B)
Its average power P is zero
(C)
Its reactive power Q is positive
(D)
$∣ Z_{C} ∣$ halves when $ω$ doubles

Answer and solution

Answer: (A), (B), (D)

$Z_{C} = 1/ (j ω C)$ has angle $- 9 0^{\circ}$ , so $ϕ = - 9 0^{\circ}$ : current leads by $9 0^{\circ}$ (A true).
$P = V I cos (- 9 0^{\circ}) = 0$ (B true).
$Q = V I sin (- 9 0^{\circ}) = - V I$ , never positive (C false).
$∣ Z_{C} ∣ = 1/ (ω C)$ , so doubling $ω$ halves it (D true).

CHECK YOURSELF 4

Find the rms value, in V, of $v (t) = 6 cos ω t + 8 sin ω t$ V, with $ω > 0$ .

Answer and solution

Answer: $7.071$ (accepted range $\pm 0.02$ )

Write $8 sin ω t = 8 cos (ω t - 9 0^{\circ})$ , so both terms are cosines at one $ω$ .
rms phasor: $V = (6 - j 8) / 2$ , of size $36 + 64 / 2 = 10/ 2$ .
$V_{rms} = 10/ 2 = 7.07$ V; in time, $v = 10 cos (ω t - 53.1 3^{\circ})$ V.

Sinusoids, complex numbers and phasors

What you will be able to do

Before you start

Why one frequency is enough, and what a phasor is

Impedance: Ohm's law for phasors

Power over one cycle

Check yourself

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