Series RLC impedance: reactance, the impedance triangle, phase and resonance
Put a resistor, a coil and a capacitor in series on AC and the same current flows through all three, but their voltages are not in step: the resistor's is in phase with the current, the coil's a quarter cycle ahead, the capacitor's a quarter cycle behind. So the three oppositions cannot simply be added. The coil's and capacitor's reactances subtract from each other, and what is left combines with the resistance at right angles: |Z| = √(R² + (XL − XC)²), with the phase angle φ = atan((XL − XC)/R). This calculator does that at any frequency, gives the current and the voltage across each part for a source voltage, the real, reactive and apparent power, and finds the resonance.
How to use the impedance calculator
- Type R, L and C with their units as you like (100, 4.7k; 100m, 1u; 10u, 100n). Leave any one blank to leave it out: an RL or RC circuit is just this with a part missing.
- Type the frequency, and a source voltage if you want the current, the voltage across each part and the powers.
- The drawing shows the loop with each part's reactance and voltage, the current, and the impedance triangle. Click a part for its own notes. The Resonance button jumps to the frequency where L and C cancel; the Across frequency tab shows how the impedance changes from a tenth of the frequency to ten times it.
Reactance
XL = 2πfL XC = 1 / (2πfC)
A coil opposes a changing current, more at higher frequency: its reactance is proportional to f. A capacitor opposes a changing voltage, less at higher frequency: its reactance is inversely proportional to f. Both are measured in ohms and both limit the current like a resistor would, but neither uses power: the energy stored in the coil's field or the capacitor's charge comes back to the circuit every half cycle. The sign of the phase shift is what separates them, and it is why in series they subtract: X = XL − XC.
The impedance triangle
|Z| = √(R² + X²) φ = atan(X / R) Z = R + jX
Draw R along the bottom and X straight up (or down when capacitive); the hypotenuse is |Z| and the angle at the corner is φ. A positive angle means the circuit is inductive and the current lags the voltage; a negative one that it is capacitive and the current leads. At 0° the circuit behaves as a plain resistor, at ±90° as pure reactance. In complex form the impedance is R + jX, and the triangle is just that number drawn on the complex plane.
Current and the voltage across each part
Ohm's law holds with |Z| in place of R: I = V / |Z|. Each part then drops I × R, I × XL or I × XC. Those three voltages do not add up to the source voltage arithmetically; they add as phasors, VR along and VL − VC up, giving back V. Near resonance the coil's and capacitor's voltages can each be many times the source voltage while they cancel each other, so the parts must be rated for it.
Power and power factor
P = I²R Q = I²X S = V I cos φ = P / S = R / |Z|
Only the resistor dissipates real power (watts). The reactive power (var) is energy going back and forth between the source and the fields, and the apparent power (VA) is what the generator and the wiring must carry. The power factor, cos φ, is the fraction of the apparent power that does work; a motor at 0.8 draws 25 % more current than its watts suggest, which is why utilities charge for it and why capacitors are added to correct it.
Resonance
f0 = 1 / (2π√(LC)) Q = 2πf0L / R
At one frequency XL equals XC, they cancel, and the series impedance falls to R alone: the current is at its maximum and in phase with the voltage. Below f0 the capacitor dominates (capacitive, leading); above it the coil does (inductive, lagging). The quality factor Q says how sharp the dip is and how much the voltage across L and C exceeds the source: with Q = 10, a 10 V source puts 100 V across each.
Your circuit, step by step
- Reactances at 50 Hz: XL = 31.4 Ω, XC = 318 Ω, net X = −287 Ω (capacitive).
- Impedance: √(100 Ω² + −287 Ω²) = 303.8 Ω at −70.8°, power factor 0.329.
- Current: 10 V / 304 Ω = 32.9 mA; VR 3.29 V, VL 1.03 V, VC 10.5 V; P 108 mW, S 329 mVA.
- Resonance: 1 / (2π√(LC)) = 159 Hz, Q = 1.
Worked example: 100 Ω, 100 mH and 10 µF at 50 Hz
At 50 Hz the coil's reactance is 2π × 50 × 0.1 = 31.4 Ω and the capacitor's 1 / (2π × 50 × 10 µF) = 318 Ω. The net reactance is 31.4 Ω − 318 Ω = −287 Ω: capacitive. The impedance is √(100² + 287 Ω²) = 303.8 Ω at −70.8°, so the current leads. From a 10 V source 32.9 mA flows: 3.29 V across R, 1.03 V across L and 10.5 V across C, which add as phasors back to 10 V. The real power is only 108 mW of 329 mVA apparent, a power factor of 0.329. The resonance is at 1 / (2π√(0.1 × 10 µF)) = 159 Hz: there the impedance falls to 100 Ω, the current rises to 100 mA and the voltage across the coil and the capacitor each reach 10 V.
Questions
Why do the voltages across the parts add up to more than the supply?
Because they are not in phase. The coil's and capacitor's voltages point in opposite directions and cancel; only the phasor sum equals the supply. Each part still really has that voltage across it.
What is the difference between impedance and reactance?
Reactance is the opposition from a coil or capacitor alone, with a 90° phase shift. Impedance is the whole thing, resistance and reactance combined, with its magnitude and angle.
Does this work for a parallel circuit?
No: in parallel the voltages are shared and the currents add as phasors, so the admittances (1/Z) add instead. This calculator is for series R, L and C only.
What does a negative reactance mean?
That the capacitor's reactance is larger than the coil's: the circuit is capacitive, the current leads and the phase angle is negative. It is a sign convention, not a negative resistance.