Why Are Phasors Used in Circuit Analysis?

Phasors turn differentiation into multiplication when we seek a sinusoidal response at a fixed frequency. Their use depends on the circuit being linear and time invariant, and on separating that response from any transient behavior.

Encoding amplitude and phase

Fix an angular frequency $\omega>0$ and use peak-amplitude phasors. A sinusoid can be written as

v(t)=V_m\cos(\omega t+\phi) =\operatorname{Re}\{\widetilde V e^{j\omega t}\}, \qquad \widetilde V=V_me^{j\phi},\quad j^2=-1.

The phasor $\widetilde V$ is constant; all time dependence sits in $e^{j\omega t}$ . Differentiating the whole expression gives

\frac{dv}{dt}=\operatorname{Re}\{j\omega\widetilde V e^{j\omega t}\}.

Thus $d/dt$ acts as multiplication by $j\omega$ on this family of signals. The derivative of the constant phasor alone would be zero and is not what the method uses.

If $\widetilde V=a+jb$ , the real signal is $a\cos\omega t-b\sin\omega t$ . Its two real coefficients encode amplitude and phase, rather than two spatial components of the voltage. At a fixed nonzero frequency this encoding is unique when signals are compared for all times; taking a real part at one instant is not an injective operation.

From component laws to impedance

With passive sign convention, the ideal resistor, inductor, and capacitor laws yield

Component	Time-domain law	Phasor relation	Impedance
Resistor	$v=Ri$	$\widetilde V=R\widetilde I$	$R$
Inductor	$v=L\,di/dt$	$\widetilde V=j\omega L\widetilde I$	$j\omega L$
Capacitor	$i=C\,dv/dt$	$\widetilde I=j\omega C\widetilde V$	$1/(j\omega C)$

Kirchhoff's laws still impose the same linear sums on currents and voltages. Together with the component laws, they give algebraic equations for the phasors. MIT's sinusoidal steady-state notes develop this approach.

For a series RLC circuit driven by $v_s(t)=V_m\cos\omega t$ ,

\widetilde I=\frac{V_m}{R+j(\omega L-1/(\omega C))}.

As a numerical example, take $R=10\,\Omega$ , $L=0.1\,\mathrm H$ , $C=100\,\mu\mathrm F$ , $\omega=100\,\mathrm{rad/s}$ , and $V_m=10\,\mathrm V$ . Then

Z=10-j90\,\Omega,\qquad |\widetilde I|=\frac{10}{\sqrt{8200}}\,\mathrm A\approx0.1104\,\mathrm A.

The current leads the source voltage by $\arctan9\approx83.66^\circ$ . The sign agrees with the circuit's net capacitive reactance.

What the calculation leaves out

A linear circuit's complete response includes a forced response and a natural response determined by initial conditions. In a stable circuit, decaying transients leave the sinusoidal steady state. Phasor analysis at one frequency computes that sinusoidal part; it does not generally determine the initial-condition terms.

A bounded sinusoidal solution must also exist. An ideal undamped circuit driven exactly at a resonance can instead have a growing response, so a division by zero in the phasor equations must not be ignored.

Several input frequencies can be handled separately and their time-domain responses added by linearity. Nonlinear circuits can generate new frequencies, so a single-frequency phasor calculation does not describe their full response. A small-signal linearization may still be useful near a specified operating point.

The general linear-system calculation

For a state-space model $\dot x=Ax+Bu$ with constant real matrices, try

u(t)=\operatorname{Re}\{Ue^{j\omega t}\},\qquad x_p(t)=\operatorname{Re}\{Xe^{j\omega t}\}.

Substitution gives $(j\omega I-A)X=BU$ . If the matrix is invertible,

X=(j\omega I-A)^{-1}BU.

This is the same idea as impedance: complex exponentials convert a constant-coefficient differential equation into algebra. The homogeneous solution must still be included when initial conditions matter.