AC Circuit Analysis — Impedance and Phasors

Use impedance and phasor analysis when a circuit is in sinusoidal steady state at a fixed frequency. Under that condition you can replace resistors, inductors, and capacitors with complex impedances and solve for magnitude and phase using algebra instead of differential equations. Ohm's law and Kirchhoff's laws still hold; the quantities are now complex so they track phase shift as well as size. If the circuit is switching, starting up, or driven by a non-sinusoidal waveform, this method does not apply.

The Procedure, Step By Step

1. Fix the frequency. Choose the angular frequency $\omega = 2\pi f$ and confirm the circuit is in sinusoidal steady state.
2. Replace each element by its impedance, using $Z_R = R$, $Z_L = j\omega L$, and $Z_C = 1/(j\omega C)$.
3. Solve in phasor form. Apply Ohm's law and Kirchhoff's laws to the complex impedances instead of the time-varying sinusoids.
4. Read the answer physically. Convert the phasor result into magnitude and phase, then decide whether current leads or lags voltage.

Two ideas make those steps work. Impedance carries both magnitude and phase, so for a sinusoid at angular frequency $\omega$,

$$Z_R = R$$ $$Z_L = j\omega L$$ $$Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}$$

with $j^2 = -1$. The resistor has no phase shift in this ideal model, the inductor gives positive imaginary impedance, and the capacitor gives negative imaginary impedance. Because $Z_L$ and $Z_C$ depend on $\omega$, the analysis always has to name the frequency.

A phasor, meanwhile, represents a sinusoid by its complex amplitude. A source written as

$$v(t) = V_m \cos(\omega t + \phi)$$

becomes a phasor with magnitude and angle. Once everything is in phasor form, Ohm's law keeps its structure, $\tilde{V} = \tilde{I} Z$, and Kirchhoff's laws keep their logic, except the quantities are complex and add with both real and imaginary parts.

A Full Example: Series RC Circuit

Suppose a series RC circuit has

$$R = 100\\ \Omega$$ $$C = 100\\ \mu\mathrm{F}$$ $$f = 50\\ \mathrm{Hz}$$

and the source phasor is

$$\tilde{V} = 10\angle 0^\circ\\ \mathrm{V}$$

using RMS values. Find the current phasor.

Fix the frequency:

$$\omega = 2\pi f = 2\pi(50) \approx 314\\ \mathrm{rad/s}$$

Replace the capacitor by its impedance:

$$Z_C = -\frac{j}{\omega C} = -\frac{j}{314(100 \times 10^{-6})} \approx -j31.8\\ \Omega$$

Add the series impedances:

$$Z = R + Z_C = 100 - j31.8\\ \Omega$$

Convert to magnitude and angle:

$$|Z| = \sqrt\{100^2 + 31.8^2\} \approx 104.9\\ \Omega$$ $$\angle Z = \tan^{-1}\left(\frac{-31.8}{100}\right) \approx -17.7^\circ$$

Solve in phasor form with Ohm's law $\tilde{I} = \tilde{V}/Z$:

$$|\tilde\{I\}| = \frac\{10\}\{104.9\} \approx 0.095\\ \mathrm\{A\}$$ $$\angle \tilde{I} = 0^\circ - (-17.7^\circ) = 17.7^\circ$$

So

$$\tilde{I} \approx 0.095\angle 17.7^\circ\\ \mathrm{A}$$

Read it physically: the current leads the source voltage by about $17.7^\circ$, which matches the picture of a capacitive circuit, where current leads voltage.

Where Each Step Goes Wrong

• At fixing the frequency: applying phasor analysis outside sinusoidal steady state. Switching, startup, or non-sinusoidal drive is not the whole story.
• At replacing elements: adding reactances like ordinary positive resistances. Inductive and capacitive parts carry signs in the imaginary direction; drop the sign and you predict the wrong phase and current.
• At solving in phasor form: forgetting that impedance is complex. You usually cannot add element magnitudes alone; add the complex impedances first, then take a magnitude.
• A consistency trap throughout: mixing peak and RMS values. Either convention works, but one calculation must use one convention. The phase relationships stay the same as long as you stay consistent.

Where Impedance And Phasors Are Used

This approach shows up in power systems, filter design, resonance problems, signal processing, and electronics where sinusoidal response matters. Even when the full system is more complicated, impedance and phasors are often the first clean model that makes the behavior understandable.

Run It Yourself

Repeat the example but keep $R$ fixed and double the frequency. Since $Z_C = -j/(\omega C)$, the capacitor's impedance magnitude becomes smaller, so the current magnitude increases and the phase angle moves closer to $0^\circ$. To explore another case with different values, solve a similar phasor problem with GPAI Solver.