7.6 Impedance in Rectangular & Polar Coordinates

7.6 Impedance in Rectangular & Polar Coordinates

Quick Answer: X_L = 2πfL, X_C = 1/(2πfC). Rectangular: Z = R + jX_L or R − jX_C. Polar: |Z| = √(R² + X²), θ = arctan(X/R). Series 100 Ω R + 100 Ω X_L → 141 Ω ∠ +45°. Series 300 Ω R + 400 Ω X_C → 500 Ω ∠ −53.1°. A polar plot shows magnitude and phase angle.

Official sub-topics 3-B-016 (Impedance Networks-1) and 3-B-017 (Impedance Networks-2) are the same underlying task in two notations, which is why they cluster into one teaching section. Twelve pool questions ride on four formulas.

Step 1 — get the reactance

The sign convention is worth fixing permanently: inductive is positive j, capacitive is negative j. "ELI the ICE man" is the classic aid — in an invertEr, voLtage leads current I (inductive, +); in a Capacitor, current I leads voltage E (capacitive, −).

Step 2 — series networks in rectangular form

For components in series, just add: Z = R + jX.

Three worked pool items:

0.1 µH in series with 20 Ω at 30 MHz X_L = 2π × 30 × 10⁶ × 0.1 × 10⁻⁶ = 2π × 3 = 18.85 Ω → Z = 20 + j19

1.0 mH in series with 200 Ω at 30 kHz X_L = 2π × 30 × 10³ × 1 × 10⁻³ = 2π × 30 = 188.5 Ω → Z = 200 + j188

10 µH in series with 40 Ω at 500 MHz X_L = 2π × 500 × 10⁶ × 10 × 10⁻⁶ = 2π × 5000 = 31,416 Ω → Z = 40 + j31400 Note how completely the reactance swamps the resistance here: at 500 MHz a 10 µH inductor is very nearly an open circuit, which is exactly why RF chokes work and why you never see a 10 µH inductor used as a series element at UHF.

0.001 µF in series with 400 Ω at 500 kHz X_C = 1/(2π × 5 × 10⁵ × 1 × 10⁻⁹) = 1/(3.1416 × 10⁻³) = 318.3 Ω → Z = 400 − j318

The unit-prefix discipline

Nearly every wrong answer in this sub-topic comes from a prefix slip, not from the physics. Convert everything to base units before you touch the calculator:

A useful sanity check: µH with MHz and mH with kHz both leave you multiplying by 2π and a small integer, because 10⁻⁶ × 10⁶ = 1 and 10⁻³ × 10³ = 1. If your answer is a factor of 1000 out, that is where it went.

Step 3 — parallel networks

Parallel is where candidates lose marks, because you cannot simply add. Use the product-over-sum rule with complex arithmetic, which is easiest in polar form:

0.01 µF in parallel with 300 Ω at 50 kHz X_C = 1/(2π × 5 × 10⁴ × 1 × 10⁻⁸) = 318.3 Ω Z = (300 × 318.3∠−90°) / (300 − j318.3) Denominator magnitude = √(300² + 318.3²) = 437.4, angle = −46.7° Z = 95,490 / 437.4 ∠ (−90° + 46.7°) = 218.3 ∠ −43.3° Convert back: 218.3 cos(−43.3°) = 159, 218.3 sin(−43.3°) = −150 → Z = 159 − j150

Note what happened: the resistive part of the answer, 159 Ω, is lower than the 300 Ω resistor. That is normal for a parallel network and is a useful check — a parallel combination is always smaller in magnitude than either branch alone.

Step 4 — polar form

|Z| = √(R² + X²) and θ = arctan(X / R).

Two habits that make these fast:

1. In a series circuit, combine reactances first. 600 Ω inductive and 300 Ω capacitive in series net to 300 Ω inductive — a single number — before you touch the resistance. That third row looks intimidating and is actually the 3-4-5 triangle.
2. Recognise the Pythagorean triples. 3-4-5 (300, 400, 500) and the 1-1-√2 case (100, 100, 141) cover most of the pool. If your magnitude is not landing on a clean number, re-check the arithmetic.

The sign of the angle tells you the character of the circuit: positive = net inductive, negative = net capacitive, zero = resonant or purely resistive (section 10.1).

What a polar plot shows

Using the polar coordinate system, what visual representation would you get of a voltage in a sinewave circuit? The plot shows the magnitude and phase angle.

That is the whole point of polar notation. Rectangular form (R + jX) tells you what the circuit is made of; polar form (|Z| ∠ θ) tells you what it does — how big the opposition is and how far current lags or leads voltage. Both describe the same complex number, and converting between them is exactly the trigonometry above.