How this instrument works
A series RLC circuit is a resistor, an inductor and a capacitor wired one after another around a single loop, carrying one shared current at every point. Each part reacts to that current differently: the resistor's voltage tracks current in step, the inductor's voltage runs a quarter cycle ahead, and the capacitor's voltage runs a quarter cycle behind. Rotate those three voltages onto a phasor diagram and the inductor's and capacitor's contributions point in exactly opposite directions, ninety degrees off from the resistor's. That geometry is why the impedance formula uses a square root of a sum of squares rather than plain addition: Z equals the Pythagorean combination of resistance and net reactance, with the two reactances already cancelled against each other before resistance ever joins in.
Both reactance terms fall out of the same differential equation Kirchhoff's voltage law writes for the loop: L di/dt + Ri + q/C = v(t). Assume a steady sine-wave drive and that equation collapses into ordinary algebra on complex numbers, with the inductor and capacitor each contributing an imaginary term of opposite sign; strip away that bookkeeping and what remains is the magnitude and phase this instrument reports. Audio engineers lean on exactly this arithmetic when sizing a loudspeaker crossover network, amateur radio operators use it to build antenna tuners that present a clean 50 ohm load to a transmitter, and power-supply designers use it to check that an input filter's impedance dip does not land on a frequency the circuit still needs to pass.
The formula assumes a single frequency, steady-state sine wave and linear, frequency-independent parts — assumptions real hardware only approximates. Where inductive reactance exactly equals capacitive reactance, the two cancel completely, net reactance drops to zero, and impedance falls to its minimum, bare resistance: that is the resonant frequency. Below resonance the capacitor dominates and the circuit looks capacitive, current leading voltage; above resonance the inductor takes over and the circuit looks inductive, current lagging instead. Push a square wave or any non-sinusoidal signal through the same three parts and no single impedance value describes the response any more — that calls for a full transient solution of the differential equation, not this steady-state formula.
- Enter the loop's Resistance in ohms — the value that would show on a meter with the source disconnected.
- Enter Inductance, H for the coil and Capacitance for the capacitor, in farads or a scaled unit such as microfarads.
- Set Drive frequency to the sine-wave frequency actually driving the circuit; every other figure on the page depends on this one.
- Read Inductive reactance and Capacitive reactance separately, then Impedance magnitude and Impedance phase angle for the combined result.
- A negative Impedance phase angle means the loop is capacitively dominated, with current leading voltage; positive means inductively dominated, current lagging instead.
Worked example — 50 Ω, 1 mH and 1 µF at 1,000 Hz
Take a 50 Ω resistor, a 1 mH inductor and a 1 µF capacitor wired in series and driven by a 1,000 Hz sine wave — a bench test typical of checking a crossover section or an EMI filter stage. First the reactances: X_L = 2π × 1000 × 0.001 = 6.283185 Ω, and X_C = 1 ⁄ (2π × 1000 × 0.000001) = 159.154943 Ω. The capacitor's reactance is more than twenty-five times the inductor's at this frequency, so its influence dominates everything that follows.
Impedance magnitude comes from Z = √(50² + (6.283185 − 159.154943)²) = √(2500 + 23369.774) = 160.840835 Ω, and the phase angle from φ = atan2(−152.871758, 50) = −1.254692 rad, about −71.9°. Both figures match the instrument's own readout to the last printed digit. The negative angle confirms current is leading voltage — this loop behaves capacitively, which makes sense since 1,000 Hz sits well under its own resonant frequency near 5,033 Hz, where X_L and X_C would finally trade places.
Questions
Why isn't the total impedance simply R plus X_L plus X_C?
Because resistance and reactance point in different directions on the phasor diagram — resistance in phase with current, reactance ninety degrees off it — so they combine as the two legs of a right triangle, not as terms in a plain sum. Adding them arithmetically overstates impedance every time; only the square root of R² plus net reactance squared gives the true figure.
What does the sign of the phase angle mean?
A negative angle means capacitive reactance exceeds inductive reactance: the circuit is net capacitive and current leads voltage, as in the worked example's −71.9°. A positive angle means inductive reactance dominates instead, current lagging voltage. Exactly zero happens only at resonance, when the two reactances cancel and the loop behaves as a pure resistor.
Why do X_L and X_C subtract instead of adding together?
Because they sit 180 degrees apart on the phasor diagram, not alongside each other. Inductive reactance pushes voltage ninety degrees ahead of current; capacitive reactance holds it ninety degrees behind. One is effectively the negative of the other's direction, so the net reactive term is their difference, and that difference is what gets squared against resistance.
What happens to impedance exactly at resonance?
It falls to its minimum possible value, bare resistance, because inductive and capacitive reactance are equal and cancel completely, leaving zero net reactance and a phase angle of exactly zero. For this circuit's 50 Ω, 1 mH and 1 µF, that minimum sits near 5,033 Hz — well above the 1,000 Hz worked example, where the capacitor still dominates.
Does the physical order of R, L and C around the loop matter?
No. Series impedances add regardless of which part comes first, because identical current flows through all three at every instant, and Kirchhoff's voltage law does not care about component order. Physical layout can still matter for stray capacitance or coupling between parts, but none of that is captured by this ideal formula.
Can this formula be used for a parallel RLC circuit instead?
No — a parallel arrangement needs admittances added rather than impedances, since each branch shares voltage instead of current. A parallel tank circuit peaks in impedance at resonance rather than dipping to a minimum, the opposite behaviour of the series case this instrument calculates.