Impedance and admittance

YOU WILL BE ABLE TO

State the impedance of R, L, and C.
Generalise Ohm's law to V = ZI and use it.
Combine impedances in series and parallel; use impedance dividers.
Interpret impedance angle as phase shift between voltage and current.

Apply d/dt ⟺ jω to the three element laws and something wonderful happens: all three collapse into the SAME law. For a resistor, V = RI. For an inductor, v = L di/dt becomes V = jωL·I. For a capacitor, i = C dv/dt becomes I = jωC·V. Each element relates its voltage phasor to its current phasor by one complex constant — its impedance Z, in ohms — and Ohm's law is reborn with every tool that ever rested on it.

Z_R = R

Z_L = j\omega L

Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}

the three impedances

\mathbf{V} = Z\,\mathbf{I}

generalised Ohm's law

The worked example: one series loop, three impedances, solved exactly like a CA-2 resistor chain.

Read the angles as physics. Z_L = ωL∠90°: an inductor's voltage leads its current by 90° (current is the slow state variable; voltage jumps first). Z_C has angle −90°: a capacitor's current leads its voltage. A resistor is 0°: perfectly in step. And EVERYTHING transfers: series impedances add, parallel impedances combine by reciprocals, dividers divide, Thevenin equivalents exist, nodal and mesh analysis run unchanged — the entire CA-2 through CA-4 toolkit, now over complex numbers. Admittance Y = 1/Z (siemens) is conductance's generalisation, and earns its keep in nodal work exactly as G did.

WORKED EXAMPLE

A series RL at one frequency

In the figure with the capacitor replaced by a wire for this example: R = 3 kΩ, and at the working ω, Z_L = j4 kΩ. Source: 10∠0° V.
Total impedance: Z = 3 + j4 kΩ → magnitude 5 kΩ, angle 53.1° (that triangle will not leave us alone).
I = V/Z = 10∠0° / 5∠53.1° = 2∠−53.1° mA.
Translation: 2 mA amplitude, lagging the source by 53.1° — inductive circuits lag, and the angle says by exactly how much.

i(t) = 2 cos(ωt − 53.1°) mA

WORKED EXAMPLE

A series RC, and a preview

R = 1 kΩ in series with C = 1 µF at ω = 1000 rad/s: Z_C = −j/(1000 × 10⁻⁶) = −j1 kΩ.
Z = 1 − j1 kΩ → magnitude √2 ≈ 1.414 kΩ, angle −45°. Current LEADS by 45°.
Now take the output across the capacitor, as an impedance divider: V_C/V_s = Z_C/(R + Z_C) = −j/(1 − j) → magnitude 1/√2 ≈ 0.707, angle −45°.
At this ω the circuit passes 70.7% of the input to its output. Change ω and that fraction changes — a frequency-dependent divider is a FILTER, and Lesson 4 is waiting.

|V_C/V_s| = 1/√2 at ω = 1/RC — remember this number

PRACTICE

Find the impedance of a 100 mH inductor and of a 10 µF capacitor at mains frequency, ω = 377 rad/s.

A 300 Ω resistor is in series with an impedance of j400 Ω. Find the total impedance in polar form, and state whether current leads or lags the applied voltage.

For the series RC of the worked example (R = 1 kΩ, C = 1 µF), at what ω does |Z_C| equal R, and what is the output-across-C divider ratio there?

Two impedances in parallel: Z₁ = j100 Ω (inductive) and Z₂ = −j100 Ω (capacitive), at one frequency. What is the parallel combination, and why is the answer alarming?