Impedance
Resistance opposes current the same way at every frequency. Reactance opposes current in a frequency-dependent way, but introduces a 90-degree phase shift. In real circuits you almost always have both present at the same time — a resistor in series with a capacitor, or an inductor coil with both reactance and resistance. Impedance is the unified concept that combines resistance and reactance into a single quantity describing the total opposition to AC current at a specific frequency. It is measured in ohms, just like resistance, but it carries both a magnitude and a phase angle.
The impedance triangle. Resistance (R) and net reactance (X = XL − XC) are at right angles to each other. The impedance magnitude (Z) is the hypotenuse: Z = √(R² + X²). The phase angle θ is tan-1(X/R).
View LargerWhy You Cannot Simply Add R and X
When you have two resistors in series, you simply add them: Rtotal = R1 + R2. Why can you not do the same with resistance and reactance?
The reason is phase. Resistance causes voltage and current to be in phase (0° between them). Reactance causes a 90° phase shift between voltage and current. Two voltages that are 90° out of phase cannot be added arithmetically — their instantaneous values do not simply combine the way two in-phase voltages do.
Think of it geometrically: if two forces act at right angles to each other on an object, you cannot find the total force by adding their magnitudes. You use the Pythagorean theorem: total force = √(F₁² + F₂²). Impedance works the same way. Resistance and reactance are perpendicular quantities. Their combination (impedance) is found using the Pythagorean theorem.
The Impedance Triangle
Draw a right triangle with resistance (R) along the horizontal axis and reactance (X) along the vertical axis. The hypotenuse of this right triangle is the impedance magnitude (Z). By the Pythagorean theorem:
Z = √(R² + X²)
where X is the net reactance = XL − XC
Z is measured in ohms (Ω)
The angle between the impedance (hypotenuse) and the resistance (horizontal side) is the phase angle θ (theta). It tells you the phase difference between the total voltage across the circuit and the current through it.
θ = arctan(X / R) = tan-1(X / R)
Positive θ means the circuit is net inductive (voltage leads current)
Negative θ means the circuit is net capacitive (current leads voltage)
θ = 0° means the circuit is purely resistive (voltage and current in phase)
The impedance triangle gives two complete pieces of information about the circuit: its magnitude (how much total opposition to current) and its angle (the phase relationship between voltage and current). Together these constitute the full impedance of the circuit.
Complex Notation: R + jX
Electrical engineers write impedance using complex numbers: Z = R + jX, where j = √(−1) is the imaginary unit (engineers use j instead of the mathematician's i to avoid confusion with current i). This notation keeps track of both the resistive and reactive parts simultaneously.
- The real part (R) is the resistance — the part that dissipates power
- The imaginary part (jX) is the reactance — inductive reactance is positive (+jXL), capacitive reactance is negative (−jXC)
For a circuit with a resistor, inductor, and capacitor in series:
Z = R + j(XL − XC) = R + jX
where X = XL − XC (net reactance)
|Z| = √(R² + X²) (impedance magnitude, in ohms)
θ = tan-1(X/R) (phase angle, in degrees)
You do not need to work extensively with complex number arithmetic for most ham radio applications — the magnitude formula and phase angle formula handle the calculations you will encounter. But understanding that Z = R + jX is the complete representation of impedance helps make sense of impedance specifications and Smith chart readings.
Series RLC Impedance
In a series RLC circuit, a resistor, inductor, and capacitor are connected end-to-end so the same current flows through all three. The impedance of this circuit is found in three steps:
- Calculate XL = 2πfL
- Calculate XC = 1/(2πfC)
- Find net reactance X = XL − XC, then Z = √(R² + X²)
A series RLC circuit. The same current I flows through all three components. The voltages across individual components (VR, VL, VC) are phase-shifted relative to each other and cannot be added arithmetically — they are added as phasors.
View LargerA series circuit contains R = 50 Ω, L = 2.0 µH, and C = 25 pF. Find the impedance at 7 MHz.
Step 1: XL = 2π × 7 × 106 × 2.0 × 10-6 = 87.96 Ω ≈ 88.0 Ω
Step 2: XC = 1 / (2π × 7 × 106 × 25 × 10-12) = 1 / (1.0996 × 10-3) = 909.5 Ω
Step 3: X = XL − XC = 88.0 − 909.5 = −821.5 Ω (net capacitive)
Step 4: Z = √(50² + 821.5²) = √(2500 + 674,862) = √677,362 = 823.0 Ω
Phase angle: θ = arctan(−821.5 / 50) = arctan(−16.43) = −86.5°
This circuit is strongly capacitive at 7 MHz (C dominates), presenting about 823 ohms of impedance with the current leading the voltage by 86.5°.
For the same circuit (R = 50 Ω, L = 2.0 µH, C = 25 pF), find the frequency at which XL = XC (resonance).
Set 2πfL = 1/(2πfC): f = 1/(2π√LC)
f = 1/(2π√(2.0 × 10-6 × 25 × 10-12))
f = 1/(2π√(5 × 10-17)) = 1/(2π × 2.236 × 10-8) = 22.5 MHz
At 22.5 MHz, XL = XC, so X = 0 and Z = √(50² + 0²) = 50 Ω purely resistive. This is resonance, covered in depth in M07G.
Your 50-ohm transceiver is connected to an antenna that presents an impedance of 75 − j25 Ω at 14.2 MHz. What is the magnitude of this antenna impedance?
Z = √(75² + 25²) = √(5625 + 625) = √6250 = 79.06 Ω
The antenna is not 50 ohms — it is 79 ohms with a capacitive component (−j25). An antenna tuner would need to transform this to 50 + j0 Ω for a perfect match. The phase angle is: θ = arctan(−25/75) = −18.4° (slightly capacitive).
Series RLC Impedance Calculator
Series RLC Impedance Calculator
Enter R, L, C, and frequency. The calculator computes XL, XC, net reactance X, impedance magnitude Z, and phase angle θ.
Ohm's Law for AC Circuits
Once you have impedance Z, Ohm's Law for AC works exactly like DC Ohm's Law, but using impedance instead of resistance:
V = I × Z I = V / Z Z = V / I
Use RMS values for V and I, and the result gives the correct RMS relationship.
A 14.2 MHz signal with an amplitude of 10 V RMS is applied across a series circuit with R = 50 Ω, XL = 75 Ω, and XC = 25 Ω. Find the current.
X = XL − XC = 75 − 25 = 50 Ω
Z = √(50² + 50²) = √5000 = 70.71 Ω
I = V / Z = 10 / 70.71 = 141.4 mA RMS
Phase angle: θ = arctan(50/50) = arctan(1) = 45° (inductive, voltage leads current by 45°)
Impedance in Ham Radio
The 50-ohm standard. The characteristic impedance of coaxial cable and the standard load impedance for amateur radio transmitters is 50 ohms. This is a purely resistive impedance — 50 + j0 Ω. When your radio, feedline, and antenna are all matched to 50 ohms, maximum power transfers from transmitter to antenna. Any reactive component in the impedance causes energy to be reflected back toward the transmitter, visible as elevated SWR.
Antenna impedance. A half-wave dipole presents about 73 Ω at its fundamental resonant frequency — close to 50 Ω, which is why dipoles work well directly connected to 50-ohm coax without a tuner. Off resonance, the dipole's impedance becomes complex — it includes reactive components that grow as you move away from the resonant frequency. The antenna tuner's job is to add equal and opposite reactance to cancel the antenna's reactive component, and then transform the resistive part to 50 Ω.
Input impedance of a receiver. HF receivers present a 50-ohm resistive input impedance (or 50 Ω matched to the LNA's noise figure). When you connect an antenna with a non-50-ohm impedance to the receiver input, the mismatch reduces the power transferred from antenna to receiver by an amount determined by the SWR at the input connector.
Impedance of a PI network. The output tank circuit of a linear amplifier — typically a PI (pi) network — is designed to transform the relatively high output impedance of the final amplifier tube or transistor (which may be hundreds of ohms) down to 50 ohms. The PI network does this by presenting specific inductive and capacitive reactances that, combined with the source impedance, produce 50 ohms at the output port.
Frequently Asked Questions
What does it mean when an antenna has a purely resistive impedance?
A purely resistive impedance means Z = R + j0 — the reactive part is zero. This happens at the antenna's resonant frequency. The antenna is neither capacitive nor inductive at that exact frequency. All the power delivered to the antenna is either radiated as radio waves or dissipated as heat in the conductor resistance (which you want to minimize). A purely resistive antenna impedance also means zero SWR if the resistance equals the feedline's characteristic impedance — for example, 50 + j0 Ω connected to 50-ohm coax gives a perfect 1:1 SWR.
Is impedance the same as resistance?
No. Resistance is a special case of impedance where the reactive part is zero (Z = R + j0). Impedance is the more general concept that includes both resistive and reactive components. Resistance is always positive and frequency-independent. Impedance can have any magnitude and any phase angle from −90° to +90°, and changes with frequency whenever reactive components are present. In DC circuits, impedance equals resistance because there is no reactance at zero frequency.
Why does 50 ohms appear everywhere in amateur radio?
50 ohms was chosen as a compromise between two competing considerations for coaxial cable: minimum loss (which occurs at about 77 ohms for a standard air-dielectric coax) and maximum power handling capacity (which peaks at about 30 ohms). 50 ohms splits the difference. It was standardized for military and industrial RF equipment during World War II and has been the default ever since. Some television equipment uses 75-ohm coax (closer to the minimum-loss optimum), which is why RF transformers between TV and ham radio equipment are sometimes needed.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.