Put a resistor, inductor and capacitor in series or in parallel, pick a frequency, and this complex impedance calculator returns |Z|, phase, R + jX, admittance and the equivalent series and parallel parts. Leave a part blank for an RC, RL or LC network, and add a voltage to get the current, power and the share each part takes.
Leave blank for no resistor.
Leave blank for no inductor.
Leave blank for no capacitor.
Optional: current, power and the voltage or current of each part.
|Z| at 1 kHz
1.827 kΩ
Ideal parts. A real capacitor is close to C, ESR and ESL in series, which you can enter here as series R, L and C. A real inductor has winding resistance in series and winding capacitance across both, a mixed network this calculator does not model.
Each part at 1 kHz
| Part | Impedance | Phase |
|---|---|---|
| R | 1 kΩ | 0° |
| L | 62.83 Ω | +90° |
| C | 1.592 kΩ | −90° |
In series, add them as complex numbers: Z = R + j(2πfL − 1/(2πfC)), then |Z| = √(R² + X²) and φ = arctan(X / R). In parallel, add their admittances and invert. For 4 Ω, 3 mH and 800 µF in series at 200 Hz, XL = 3.77 Ω and XC = 994.7 mΩ, so Z = 4 Ω + j2.775 Ω, |Z| = 4.868 Ω at 34.8° (inductive).
The impedance equation for each ideal part gives a purely real (R) or purely imaginary (L and C) value. Combine them like resistors, but with complex numbers.
In series the impedances add, which gives the series RLC impedance formula:
In parallel the admittances (Y = 1 / Z) add, which gives the parallel RLC impedance formula:
The same impedance can be drawn as a series or a parallel pair at one frequency:
With an RMS voltage V across the network: I = V ÷ |Z|, P = V² × G, reactive power = −V² × B, and the power factor is cos φ.
XL = 2π × 200 × 3 mH = 3.77 Ω and XC = 1 ÷ (2π × 200 × 800 µF) = 994.7 mΩ. So Z = 4 Ω + j2.775 Ω, |Z| = 4.868 Ω and φ = 34.8°, inductive because XL is larger. This is OpenStax Example 15.2, which gives 4.87 Ω and 0.607 rad (34.8°). The same impedance in parallel form is 5.925 Ω across 8.541 Ω of inductive reactance.
In parallel: Y = 1/250 + j(2π × 60 × 1.5 µF − 1/(2π × 60 × 650 mH)), so |Z| = 187.8 Ω at 41.3° and the source supplies 639 mA. The branches carry 480 mA, 489.7 mA and 67.86 mA; the inductor and capacitor currents partly cancel. Real power is 57.6 W at a power factor of 0.751.
In series, the same parts give |Z| = 1.544 kΩ at −80.7° and put 137.46 V across the capacitor, more than the 120 V supply. Lessons in Electric Circuits works both circuits and gets 0.639 A in parallel and 137.46 V across the capacitor in series.
Model the capacitor as 100 nF in series with 10 mΩ of ESR and 0.5 nH of ESL (illustrative values; take real ones from the part’s datasheet or impedance curve). It self-resonates at 1 ÷ (2π √(0.5 nH × 100 nF)) = 22.51 MHz. At 1 MHz it is still capacitive: |Z| = 1.588 Ω at −89.6°. At 100 MHz it is an inductor: |Z| = 298.4 mΩ at 88.1°. This is why decoupling at high frequencies depends on mounting inductance as much as on capacitance.
The two connections of the same parts against frequency. Their corner, where R equals XC, is 1 / (2πRC) = 1.592 kHz.
| Frequency | XC | Series |Z| | Series φ | Parallel |Z| | Parallel φ |
|---|---|---|---|---|---|
| 10 Hz | 159.2 kΩ | 159.2 kΩ | −89.6° | 1 kΩ | −0.4° |
| 100 Hz | 15.92 kΩ | 15.95 kΩ | −86.4° | 998 Ω | −3.6° |
| 1 kHz | 1.592 kΩ | 1.88 kΩ | −57.9° | 846.7 Ω | −32.1° |
| 1.592 kHz | 1 kΩ | 1.414 kΩ | −45.0° | 707.1 Ω | −45.0° |
| 10 kHz | 159.2 Ω | 1.013 kΩ | −9.0° | 157.2 Ω | −81.0° |
| 100 kHz | 15.92 Ω | 1 kΩ | −0.9° | 15.91 Ω | −89.1° |
| 1 MHz | 1.592 Ω | 1 kΩ | −0.1° | 1.592 Ω | −89.9° |
The same 100 nF with 10 mΩ ESR and 0.5 nH ESL in series (illustrative values). Well below self-resonance it follows the ideal curve; close to it the ESL cancels part of the capacitive reactance, and above it the part is an inductor.
| Frequency | Ideal |Z| | With ESR and ESL | Phase | Behaves as |
|---|---|---|---|---|
| 100 kHz | 15.92 Ω | 15.92 Ω | −90.0° | capacitor |
| 1 MHz | 1.592 Ω | 1.588 Ω | −89.6° | capacitor |
| 10 MHz | 159.2 mΩ | 128.1 mΩ | −85.5° | capacitor |
| 22.51 MHz (SRF) | 70.71 mΩ | 10 mΩ | 0.0° | resistor (ESR) |
| 50 MHz | 31.83 mΩ | 125.6 mΩ | 85.4° | inductor |
| 100 MHz | 15.92 mΩ | 298.4 mΩ | 88.1° | inductor |
| 1 GHz | 1.592 mΩ | 3.14 Ω | 89.8° | inductor |
The calculator treats R, L and C as ideal. Real parts are small networks of their own: a capacitor is C with ESR and ESL in series, an inductor is L with winding resistance in series and winding capacitance across it, and a resistor has a little inductance and capacitance too. A capacitor fits this calculator directly, as series R, L and C (example 3). An inductor’s winding capacitance sits across its series R and L, a mixed network that is neither all series nor all parallel, so simulate that one. Parasitics also depend on frequency, DC bias and temperature: ceramic X7R and X5R capacitors lose capacitance with DC bias, and ESR rises with skin effect, so take values from the datasheet at your operating point. For a whole circuit with sources and loads, simulate it.
The AC equivalent of resistance: the ratio of voltage to current, written Z = R + jX in ohms. R is the part that dissipates power; the reactance X comes from inductors (positive) and capacitors (negative) and shifts the current’s phase instead.
Z = V / I = R + jX, in ohms, where the reactance X is +2πfL for an inductor and −1/(2πfC) for a capacitor. The magnitude is |Z| = √(R² + X²) and the phase is arctan(X / R).
Add the parts as complex numbers: Z = R + j(2πfL − 1/(2πfC)). Then |Z| = √(R² + X²) and the phase is arctan(X / R). For 4 Ω, 3 mH and 800 µF at 200 Hz, X = 3.77 Ω − 994.7 mΩ = 2.775 Ω, so |Z| = 4.868 Ω at 34.8°.
Add admittances instead: Y = 1/R + j(2πfC − 1/(2πfL)), then Z = 1 / Y. For 250 Ω, 650 mH and 1.5 µF at 60 Hz this gives |Z| = 187.8 Ω at 41.3°, lower than any single branch’s impedance.
It is how far the voltage leads the current. A positive angle means the network is inductive (current lags); a negative angle means it is capacitive (current leads); 0° is purely resistive. The power factor is the cosine of this angle.
1.592 kΩ at −90°, from 1 / (2π × 1 kHz × 100 nF). In series with 1 kΩ the total is 1.88 kΩ at −57.9°; in parallel with it, 846.7 Ω at −32.1°.
Rp = (R² + X²) / R and Xp = (R² + X²) / X, the same impedance at that one frequency. The 4 Ω + j2.775 Ω network above is equivalent to 5.925 Ω in parallel with 8.541 Ω of inductive reactance. An LCR meter’s series and parallel modes (Cs/Cp, Ls/Lp) report these two equivalents.
Every real capacitor has series inductance (ESL). Above its self-resonant frequency 1 / (2π √(ESL × C)) the inductance dominates and the part behaves as an inductor. A 100 nF capacitor with 0.5 nH of ESL self-resonates at 22.51 MHz.
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