1.9 Reactance and Impedance in AC Circuits

1.9 Reactance and Impedance in AC Circuits

ACMA Exam Focus: Syllabus items 4.19 and 4.20 — understand and apply inductive reactance and capacitive reactance, including how each varies with frequency, and understand impedance as the total opposition that a circuit offers to alternating current.

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Why alternating current needs a second kind of opposition

A resistor opposes current in the same way whatever the frequency: 100 Ω is 100 Ω at DC, at 1 kHz and at 28 MHz. Capacitors and inductors do not behave that way at all. Their opposition to alternating current depends on frequency, and that dependence is the entire basis of filtering, tuning, bypassing and coupling inside a transceiver.

The opposition offered by a reactive component is called reactance, symbol $X$, and like resistance it is measured in ohms (Ω). The essential difference is what happens to the energy: a resistor turns electrical energy into heat, whereas an ideal reactance stores energy in a field and hands it back to the circuit later in the same cycle. A reactance opposes current without dissipating power.

Capacitive reactance ($X_C$) — item 4.19

A capacitor passes alternating current because its plates charge and discharge in step with the applied waveform. The faster the waveform alternates, and the larger the capacitor, the more charge moves each second — and charge per second is current. So the opposition falls as either frequency or capacitance rises:

$X_C = \frac{1}{2 \pi f C}$

with $X_C$ in ohms, $f$ in hertz and $C$ in farads.

At DC, $f = 0$ and $X_C$ becomes infinite: the capacitor is an open circuit. That is what "a capacitor blocks DC" looks like as a number.

Worked example. A 0.01 µF (10 nF) ceramic capacitor bypasses RF to chassis on the 80 m band at 3.5 MHz: $X_C = \frac{1}{2\pi \times 3.5 \times 10^{6} \times 1 \times 10^{-8}} = \frac{1}{0.2199} \approx 4.5\ \Omega$ At 3.5 MHz that part is very nearly a short circuit, so RF is shunted straight to earth. Feed the same capacitor a 1 kHz audio tone and $X_C = \frac{1}{2\pi \times 1000 \times 1 \times 10^{-8}} \approx 15\,900\ \Omega$ — close to 16 kΩ, so the audio passes by untouched. One component, two completely different jobs, decided purely by frequency.

Inductive reactance ($X_L$) — item 4.19

Current through an inductor builds a magnetic field. A changing current means a changing field, and by Lenz's law the coil develops a back EMF (electromotive force) that opposes the change. The faster the change, the larger the back EMF, so the opposition rises with frequency and with inductance:

$X_L = 2 \pi f L$

with $X_L$ in ohms, $f$ in hertz and $L$ in henries.

At DC, $f = 0$ and $X_L = 0$: the coil is simply a piece of wire. That is why a radio-frequency choke drops almost no supply voltage while still blocking RF effectively.

Worked example. A 1 mH RF choke is fitted in a 13.8 V supply lead.

• At 50 Hz, the Australian mains frequency: $X_L = 2\pi \times 50 \times 0.001 = 0.31\ \Omega$, which is negligible.
• At 14 MHz: $X_L = 2\pi \times 14 \times 10^{6} \times 0.001 \approx 88,000\ \Omega$, or 88 kΩ — a very effective block.

The frequency trends side by side

Two sentences carry most of the marks in this part of the paper: capacitive reactance falls as frequency rises, and inductive reactance rises as frequency rises.

Phase, and why reactance cannot simply be added to resistance

In a resistor, voltage and current rise and fall together — they are in phase. In an ideal reactance they sit 90 degrees apart:

• In a capacitor, current leads voltage by 90 degrees, because current has to flow before the plates can build up any voltage.
• In an inductor, voltage leads current by 90 degrees, because the back EMF appears the instant the current tries to change.

The traditional mnemonic is CIVIL: in a Capacitor, I leads V; V leads I in an L.

Impedance ($Z$) — item 4.20

Impedance is the total opposition a circuit presents to alternating current, combining resistance and reactance in one figure. It is measured in ohms and carries the symbol $Z$.

Because reactance is 90 degrees out of phase with resistance, ohms of reactance and ohms of resistance cannot be added arithmetically. Inductive and capacitive reactance are 180 degrees apart from each other, so those two subtract, and whatever is left is combined with the resistance using Pythagoras:

$|Z| = \sqrt{R^2 + (X_L - X_C)^2}$

Worked example. A series circuit carries $R = 100\ \Omega$, $X_L = 200\ \Omega$ and $X_C = 125\ \Omega$.

1. Net reactance: $200 - 125 = 75\ \Omega$, inductive.
2. $|Z| = \sqrt{100^2 + 75^2} = \sqrt{10,000 + 5,625} = \sqrt{15,625} = 125\ \Omega$.

The answer, 125 Ω, is far below the 425 Ω you would get by adding the three figures blindly. That blind addition is the most common arithmetic error in this part of the syllabus.

Ohm's law still works once you have the impedance, in the form $I = V / Z$. Apply 25 V RMS to the circuit above and the current is $25 / 125 = 0.2$ A.

Where impedance shows up on air

• 50 Ω is the standard characteristic impedance of amateur coaxial feedline and of transceiver antenna sockets in VK, which is why every match is quoted against it.
• A half-wave dipole in the clear presents roughly 73 Ω at its feed point; a folded dipole about four times that, near 300 Ω.
• An antenna tuning unit does not change the antenna. It transforms whatever impedance appears at the shack end of the feedline into something close to the 50 Ω resistive load the transmitter wants to see.

Whenever the reactive part of that load is not zero, the transmitter is working into a partly reactive load and the standing wave ratio rises. Arranging for the inductive and capacitive reactance to cancel is the business of tuned circuits, which the next section takes up.