Physics · Alternating current

Impedance of a series RLC circuit

Impedance combines resistance and frequency-dependent reactance in a series RLC circuit.

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Formula

Z = R^{2} + (ω L - \frac{1}{ω C})^{2}

Z and R in Ω; ω in rad/s; L in H; C in F.

Variables & units

Variables & units
Symbol	Meaning	Unit
$Z$	Magnitude of total impedance	Ω
$R$	Ohmic resistance	Ω
$ω$	Angular frequency	rad/s
$L$	Inductance	H
$C$	Capacitance	F

LaTeX

Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}

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Two impedance triangles with resistance 24 ohms: at 500 radians per second reactance is plus 24 ohms and magnitude 33.94 ohms; at 250 radians per second reactance is minus 12 ohms and magnitude 26.83 ohms. — Original series circuit: R = 24 Ω, L = 80 mH, C = 125 μF. Both axes use 5 px/Ω. Blue: ω = 500 rad/s, voltage leads current by 45°. Dashed: ω = 250 rad/s, voltage lags current by 26.565°. Signed net reactance determines direction, rather than adding positive reactances. Based on OpenStax §15.3.

Series RLC impedance calculator

Find the reactances, impedance, phase shift, rms current or real power of a series circuit of R, L and C on a sinusoidal AC supply.

Calculation model

Find

Resistance R

Resistance in the series circuit, e.g. 40 Ω; it must be greater than 0.

Inductance L

Inductance of the coil, e.g. 300 mH.

Capacitance C

Capacitance of the capacitor, e.g. 100 µF.

Frequency f

Frequency of the AC supply, e.g. 50 Hz on the mains.

Filled with example values. Type your own and the result updates straight away.

Result: Impedance |Z|

|Z| = 74.13 Ω

With R = 40 Ω, L = 300 mH, C = 100 µF and f = 50 Hz, |Z| = 74.13 Ω.

Working · rearrange for |Z|, substitute, evaluate

1. Rearrange for |Z|
$|Z|=\sqrt{R^2+X^2}$
2. Substitute values (in SI units)
- L = 300 mH = 0.3 H
- C = 100 µF = 1 · 10⁻⁴ F
$|Z|=\sqrt{40^2+(94.2478-31.831)^2}$
3. Result
|Z| = 74.13 Ω

X > 0 is inductive: current lags. X < 0 is capacitive: current leads. Output φ is voltage phase minus current phase. The complex value is Z = R + jX.

Quick tips

Resonance at f₀ = 1 / (2π · √(L · C)): there XL and XC cancel, Z = R is at its smallest and the current at its largest.
Phasor triangle: R and X are the legs, Z the hypotenuse. So Z is never smaller than R; X > 0 means inductive, X < 0 capacitive.

Assumptions and limits

Ideal linear components R > 0, L > 0 and C > 0 in series; sinusoidal excitation with f > 0 after transients decay. ω = 2πf. Not a parallel circuit, DC, switching transient or a full nonsinusoidal signal. R = 0 is outside this model: free oscillations do not decay, and resonance does not admit the finite steady-current calculation.

Derivation & background

Using the real-part phasor convention, $\widetilde Z=R+\mathrm j\left(\omega L-\frac1{\omega C}\right)$ . Its displayed magnitude loses reactance sign. Positive reactance means voltage leads current; negative reactance means current leads voltage. Assumes ideal linear series components in sinusoidal steady state.

Exam blueprint

Validity range

Magnitude formula for an ideal linear series RLC circuit in sinusoidal steady state; component values are treated as frequency independent.

Derivation steps

Resistance and net reactance are perpendicular in the phasor diagram.

1Choose ideal series RLC in sinusoidal steady state and passive voltage reference.
2The same current passes through all three elements; source voltage is their instantaneous sum.
3With the real-part convention, differentiation becomes $\mathrm j\omega$ , giving $\widetilde Z_L=\mathrm j\omega L$ .
4From $i=C\frac{\mathrm du_C}{\mathrm dt}$ , obtain $\widetilde Z_C=-\frac{\mathrm j}{\omega C}$ .
5Add $\widetilde Z=R+\mathrm j\left(\omega L-\frac1{\omega C}\right)$ .
6Take $Z=|\widetilde Z|$ ; resistance and reactance are perpendicular.
7Check signed phase, current and power using matching RMS or peak values.

Rearrangements

Current amplitude

I_0=\frac{V_0}{Z}

Use peak voltage and peak current together; likewise for RMS values. Z denotes impedance magnitude here and must be positive for a finite-current calculation. With R = 0 at ideal series resonance, division by Z = 0 is invalid.

Task variant

For $R=40\,\Omega$ , $X_L=70\,\Omega$ and $X_C=22\,\Omega$ , find Z.

Approach: $X=X_L-X_C$ , then $Z=\sqrt{R^2+X^2}$ .
Substitute: $X=(70-22)\,\Omega=48\,\Omega$ .
Result: $Z=\sqrt{40^2+48^2}\,\Omega\approx62.5\,\Omega$ .

Check: magnitude exceeds $R=40\,\Omega$ .

Common mistakes

Adding R, XL and XC directly even though reactance and resistance are out of phase.

First form XL−XC, then combine it geometrically with R to obtain impedance.

Exam context

AC circuits, resonance and phase shift in series-RLC problems.

Formula cluster

Alternating current and circuits

Impedance complements resonance frequency with resistance and current amplitude.

T = 2π·√(L·C)

Thomson formula for an LC circuit

U = R·I

Ohm's Law

U_ind = −L·(dI/dt)

Self-Induction of a Coil

C = Q/U

Capacitance of a Capacitor

Worked example

$R=30\,\Omega$ , $X_L=50\,\Omega$ , $X_C=10\,\Omega$ : $Z=\sqrt{30^2+40^2}\,\Omega=50\,\Omega$ . At 10 V peak source voltage, $I_0=\frac{10\,\mathrm V}{50\,\Omega}=0.20\,\mathrm A$ . Voltage leads by about 53.13°; both squared terms have units Ω².

Applications

AC circuits, filters, resonance and frequency response of inductor-capacitor systems.

How do you calculate with Impedance of a series RLC circuit?

Here is how to work through a typical Impedance of a series RLC circuit (Z = √(R² + (ωL − 1/ωC)²)) task step by step:

1
Task
For $R=40\,\Omega$ , $X_L=70\,\Omega$ and $X_C=22\,\Omega$ , find Z.
Solution path
1. Approach: $X=X_L-X_C$ , then $Z=\sqrt{R^2+X^2}$ .
2. Substitute: $X=(70-22)\,\Omega=48\,\Omega$ .
3. Result: $Z=\sqrt{40^2+48^2}\,\Omega\approx62.5\,\Omega$ .
Check: magnitude exceeds $R=40\,\Omega$ .

Understand and apply the equation

An AC source drives the same current through R, L and C, but their voltages do not peak together. We construct impedance from component laws, change the frequency of an original circuit and connect phase, resonance and actual dissipation.

Why three voltages do not add like three lengths

An inductor and capacitor can have large individual voltages while the source supplies a much smaller voltage. Their phases differ, so the peaks do not occur together. The phasor picture therefore adds directed contributions first; impedance magnitude follows afterward.[2][1]

We study a series circuit in sinusoidal steady state. The same current flows through all three components. Positive total resistance represents losses; constant ideal L and C store energy. Check this topology before applying the formula.[2][1]

Magnitude alone does not give phase

\begin{aligned}\widetilde Z&=R+\mathrm jX\\ X&=\omega L-\frac{1}{\omega C}\\ Z=|\widetilde Z|&=\sqrt{R^2+X^2}\end{aligned}

We mark the complex phasor ratio with a tilde and call its nonnegative magnitude Z. j is the imaginary unit; R is the total effective series resistance.[2]

A series circuit carries the same current through every component, but its voltages are shifted in time. Source voltage is therefore not the sum of their positive peaks. Inductive and capacitive voltage oppose one another in the phasor picture; their difference is perpendicular to resistor voltage. Magnitude determines current size, while signed reactance also determines whether current leads or lags. Two circuits can have equal impedance magnitudes yet opposite phase relationships.[2]

Which circuit this model describes

Assume ideal linear L and C and a positive constant resistance R, all in series with a sinusoidal source at fixed angular frequency. We consider steady state after the free switching response has decayed. Component voltages use passive reference; source voltage equals the sum of voltage drops. Distributed transmission lines, frequency-dependent losses, nonlinear cores and leakage paths are not automatically represented. For example, real coil resistance belongs within R.[2][5]

The magnitude formula does not apply unchanged to a parallel circuit. There the common voltage is the starting point and complex admittances add. A square wave also contains multiple frequency components: each can be treated separately in a linear model, but a single impedance at the fundamental frequency does not describe the complete current waveform.[5]

Component laws produce the phasor triangle

\begin{aligned}u_R&=Ri\\u_L&=L\frac{\mathrm di}{\mathrm dt}\\u_C&=\frac{q}{C},\qquad \frac{\mathrm dq}{\mathrm dt}=i\end{aligned}

Under passive reference, inductor voltage follows current change and capacitor voltage follows charge.[1]

To read the phasor, use the common current as the horizontal reference. Resistor voltage lies on the same axis. Inductor voltage leads by 90°, capacitor voltage lags by 90°. Subtracting them leaves one perpendicular reactive component. Only now apply Pythagoras, rather than adding three independent positive lengths.[2]

Choose the real part of a complex phasor rotating at positive angular frequency. Differentiation multiplies by jω; integration of the steady AC component divides by jω. This gives positive inductive and negative capacitive reactance. Kirchhoff’s loop rule adds the three complex voltages. Only after that addition do we take the magnitude. This is a consequence of component laws and the circuit model, not a new empirical law.[2][5]

\begin{aligned}\widetilde U&=\left(R+\mathrm j\omega L+\frac{1}{\mathrm j\omega C}\right)\widetilde I\\&=\left[R+\mathrm j\left(\omega L-\frac{1}{\omega C}\right)\right]\widetilde I\\ \varphi&=\operatorname{atan2}(X,R)\end{aligned}

φ is voltage phase minus current phase. For positive X voltage leads; for negative X current leads.[2]

Frequency, units and useful rearrangements

\begin{aligned}\omega&=2\pi f\\I_{\mathrm{rms}}&=\frac{U_{\mathrm{rms}}}{Z},\qquad I_0=\frac{U_0}{Z}\\|X|&=\sqrt{Z^2-R^2}\\ C_{\mathrm{res}}&=\frac{1}{\omega^2L}\end{aligned}

Use RMS and peak values in matching pairs. The magnitude alone cannot recover the sign of X.[2][4]

Hertz is not numerically equal to angular frequency in radians per second. Both ωL and the reciprocal of ωC have units of ohms; the square root contains squared ohms. Convert millihenries and microfarads into henries and farads before substitution. A measured impedance below the assumed positive series resistance contradicts this simple model: inspect measurement, component values or topology. At zero R and exact resonance an ideal finite-voltage source has no bounded steady current; the usual resonance-current ratio is invalid there.[2][5]

Two frequencies, the same circuit

Electrical power distinguishes real power from apparent power in AC. The following series example consistently uses RMS values; the series resistance R determines dissipation.[3]

Inductive operation at 45° phase angle

Given: $R=24\,\Omega$ , $L=80\,\mathrm{mH}$ , $C=125\,\mathrm{\mu F}$ , $\omega=500\,\mathrm{rad/s}$ , source RMS voltage 60 V; ideal steady-state series circuit.
Find: Impedance, current, phase and average resistor power.

Approach: determine signed reactance, then magnitude, current and power.
$\begin{aligned}X&=\omega L-\frac{1}{\omega C}\\Z&=\sqrt{R^2+X^2}\\\overline P&=I_{\mathrm{rms}}^2R\end{aligned}$
Substitution: first evaluate the two reactances separately.
$\begin{aligned}X_L&=(500\,\mathrm{s^{-1}})(0.080\,\mathrm H)=40\,\Omega\\X_C&=\frac{1}{(500\,\mathrm{s^{-1}})(125\times10^{-6}\,\mathrm F)}=16\,\Omega\\X&=24\,\Omega\\Z&=\sqrt{24^2+24^2}\,\Omega=33.9411\,\Omega\end{aligned}$
Current and phase: voltage leads current by 45°.
$\begin{aligned}I_{\mathrm{rms}}&=\frac{60\,\mathrm V}{33.9411\,\Omega}=1.76777\,\mathrm A\\\varphi&=\arctan\frac{24}{24}=45^\circ\\\overline P&=I_{\mathrm{rms}}^2(24\,\Omega)=75.0\,\mathrm W\end{aligned}$

Result: Z ≈ 33.94 Ω; current ≈ 1.768 A; $\varphi=+45^\circ$ ; real power 75 W.

Check: Check: resistor voltage 42.426 V; inductor voltage 70.711 V; capacitor voltage 28.284 V. Their net reactive voltage 42.426 V and resistor voltage form a right triangle with source voltage 60 V. Also $60\times1.76777\times\cos45^\circ\approx75\,\mathrm W$ .

Halving angular frequency makes the circuit capacitive

Given: Same circuit with $R=24\,\Omega$ , $L=80\,\mathrm{mH}$ , $C=125\,\mathrm{\mu F}$ and RMS source voltage 60 V; now $\omega=250\,\mathrm{rad/s}$ .
Find: New current magnitude and phase direction.

Approach: recalculate both frequency-dependent reactances; do not halve total impedance.
$\begin{aligned}X&=\omega L-\frac1{\omega C}\\I_{\mathrm{rms}}&=\frac{U_{\mathrm{rms}}}{\sqrt{R^2+X^2}}\end{aligned}$
Substitution: inductive reactance halves, capacitive reactance doubles.
$\begin{aligned}X_L&=20\,\Omega,\qquad X_C=32\,\Omega\\X&=-12\,\Omega\\Z&=\sqrt{24^2+(-12)^2}\,\Omega=26.8328\,\Omega\\I_{\mathrm{rms}}&=\frac{60\,\mathrm V}{26.8328\,\Omega}=2.23607\,\mathrm A\\\varphi&=\arctan\frac{-12}{24}=-26.5651^\circ\end{aligned}$
Power: larger current increases dissipation at unchanged R.
$\overline P=(2.23607\,\mathrm A)^2(24\,\Omega)=120\,\mathrm W$

Result: Current leads voltage by about 26.57°; RMS current 2.236 A, power 120 W.

Check: Check: resistor voltage 53.666 V; net reactive voltage −26.833 V; their squared sum is 3600 V². Since Z remains above 24 Ω, current stays below its resonance value 2.5 A.

Small total impedance can hide large component voltages

\begin{aligned}\omega_0&=\frac{1}{\sqrt{LC}}\\Z(\omega_0)&=R,\qquad \varphi(\omega_0)=0\\\overline P&=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi=I_{\mathrm{rms}}^2R\\\cos\varphi&=\frac{R}{Z}\end{aligned}

Resonance cancels total reactance, not the individual voltages or stored energy.[3][4]

For this example current resonance occurs at about 316.23 rad/s or 50.33 Hz. RMS current is then 2.5 A and dissipation 150 W. Both L and C carry about 63.25 V RMS although the source supplies only 60 V: their opposing phasors cancel. With smaller R, voltage magnification can be much larger. Free damped oscillation frequency is a different question from the frequency of maximum steady series current. Capacitor voltage can also reach its maximum at a different frequency.[4][5]

Series current has a resonance maximum

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For

R=24\,\Omega

L=80\,\mathrm{mH}

C=125\,\mathrm{\mu F}

and 60 V RMS. The maximum is 2.50 A; at 200 and 500 rad/s it is 1.768 A. The source holds RMS voltage constant.[2][4]

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The Thomson LC oscillation relation describes the natural frequency of an ideal lossless LC circuit. The same frequency marks current resonance in this ideal series RLC circuit; it does not replace a transient calculation for the damped circuit.[4]

From model to measurement

A filter must be described by the voltage actually taken as output, not Z alone. Output across R is maximal at resonance; L and C have different transfer functions. Real inductance may change at large currents, capacitors have losses and sources have internal resistance. Large resonant voltages may exceed component limits. Calculated real power describes only losses in modeled R; ideal L and C return their stored energy on average.[3][5]

Explore further: half-power bandwidth

Half power is not half current

Given: The same ideal series circuit has $R=24\,\Omega$ , $L=0.0800\,\mathrm H$ , $C=125\,\mathrm{\mu F}$ . The source maintains $U_{\mathrm{rms}}=60.0\,\mathrm V$ as frequency changes. Assume steady state.
Find: Find the two frequencies giving half maximum resistor power.

Approach: halving $I_{\mathrm{rms}}^2R$ divides current by $\sqrt2$ and multiplies Z by $\sqrt2$ . Thus $|X|=R$ .
$\begin{aligned}\overline P&=\frac{U_{\mathrm{rms}}^2R}{R^2+X^2}\\\overline P=\frac{\overline P_{\max}}2&\ \Longrightarrow\ X=\pm R\\\omega_-&=\frac{-R+\sqrt{R^2+4L/C}}{2L}\\\omega_+&=\frac{R+\sqrt{R^2+4L/C}}{2L}\end{aligned}$
Substitution: the lower boundary is capacitive; the upper is inductive.
$\begin{aligned}\omega_-&=200\,\mathrm{rad/s}\\\omega_+&=500\,\mathrm{rad/s}\\f_-&=\frac{200}{2\pi}\,\mathrm{Hz}\approx31.8310\,\mathrm{Hz}\\f_+&=\frac{500}{2\pi}\,\mathrm{Hz}\approx79.5775\,\mathrm{Hz}\\I_{\mathrm{rms}}&=\frac{60}{24\sqrt2}\,\mathrm A\approx1.76777\,\mathrm A\\\overline P&=75.0\,\mathrm W\end{aligned}$

Result: At about 31.83 Hz and 79.58 Hz power is 75.0 W, half its resonance value of 150 W. Phases are −45° and +45°.

Check: Direct reactance calculation gives ±24 Ω at the boundaries. Bandwidth $\omega_+-\omega_-=300\,\mathrm{rad/s}$ equals $R/L$ ; their geometric mean is resonance angular frequency, not their arithmetic mean.

Five exercises on magnitude, phase and resonance

Calculate · Application

An ideal series RLC circuit has $R=18\,\Omega$ , $X_L=42\,\Omega$ , $X_C=18\,\Omega$ at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

Show hint

First take the difference of reactances.

Learning goal: Compute current and dissipation from signed reactances.

Show solution and calculation

Approach: impedance is the hypotenuse formed by R and X; R dissipates power.
Substitution gives $X=24\,\Omega$ , $Z=30\,\Omega$ , $I_{\mathrm{rms}}=0.4\,\mathrm A$ and 2.88 W.

Check: power factor 0.6; multiplying the source product by 0.6 also gives 2.88 W.

\begin{aligned}X&=42-18=24\,\Omega\\Z&=\sqrt{18^2+24^2}\,\Omega=30\,\Omega\\I_{\mathrm{rms}}&=\frac{12\,\mathrm V}{30\,\Omega}=0.4\,\mathrm A\\\overline P&=(0.4\,\mathrm A)^2(18\,\Omega)=2.88\,\mathrm W\end{aligned}

[2][3]

Rearrange · Application

At $\omega=400\,\mathrm{rad/s}$ , an ideal series circuit with $L=50\,\mathrm{mH}$ and $R=10\,\Omega$ should resonate. What capacitance is required?

Show hint

Equate the two reactances.

Learning goal: Rearrange resonance condition for capacitance.

Show solution and calculation

Approach: resonance requires zero net reactance.
Rearranging for C and substituting SI values gives 125 μF.

Check: both reactances are 20 Ω and total impedance is 10 Ω.

\begin{aligned}\omega L&=\frac{1}{\omega C}\\C&=\frac{1}{\omega^2L}=\frac{1}{(400\,\mathrm{s^{-1}})^2(0.050\,\mathrm H)}\\&=125\times10^{-6}\,\mathrm F\end{aligned}

[4]

Interpret · Foundation

Two ideal series circuits have $R=24\,\Omega$ and $X=+24\,\Omega$ or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Show hint

Squaring loses the sign; phase does not.

Learning goal: Distinguish equal impedance magnitude from equal phase.

Show solution and calculation

Approach: compare magnitude and argument separately.
Both have Z ≈ 33.94 Ω, equal RMS current and equal resistor dissipation.
In the first circuit voltage leads by 45°; in the second voltage lags current by 45°.

Check: both have the same positive power factor.

\begin{aligned}Z&=24\sqrt2\,\Omega\\\varphi_+&=45^\circ\\\varphi_-&=-45^\circ\end{aligned}

[2]

Error analysis · Application

Someone uses $R=24\,\Omega$ , $X_L=40\,\Omega$ , $X_C=16\,\Omega$ to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

Show hint

Capacitive and inductive phasors oppose; R is perpendicular to them.

Learning goal: Correct scalar addition of phase-shifted voltages.

Show solution and calculation

Approach: add complex impedances, not their positive magnitudes.
Z is not $24+40+16$ .
Net reactance is 24 Ω and magnitude $24\sqrt2\,\Omega$ .
Current is about 1.768 A.

Check: the perpendicular voltage components are each 42.426 V and give a 60 V source.

\begin{aligned}\widetilde Z&=(24+\mathrm j24)\,\Omega\\Z&=24\sqrt2\,\Omega\\I_{\mathrm{rms}}&=\frac{60}{24\sqrt2}\,\mathrm A=1.76777\,\mathrm A\end{aligned}

[2]

Transfer · Application

An ideal series circuit has $R=5\,\Omega$ , $L=0.10\,\mathrm H$ , $C=40\,\mathrm{\mu F}$ and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?

Show hint

At resonance $Z=R$ , but individual reactances are not zero.

Learning goal: Reconcile resonance voltage magnification with energy conservation.

Show solution and calculation

Approach: calculate resonance frequency, then current and component voltages.
$\omega_0=500\,\mathrm{rad/s}$ , both reactances 50 Ω, current 2 A and both component voltages 100 V RMS.
Their phasors oppose.
Source power is 20 W, all dissipated in R; ideal storage elements consume no average power.

\begin{aligned}\omega_0&=\frac{1}{\sqrt{(0.10\,\mathrm H)(40\times10^{-6}\,\mathrm F)}}=500\,\mathrm{s^{-1}}\\I_{\mathrm{rms}}&=\frac{10\,\mathrm V}{5\,\Omega}=2\,\mathrm A\\U_{L,\mathrm{rms}}=U_{C,\mathrm{rms}}&=(2\,\mathrm A)(50\,\Omega)=100\,\mathrm V\\\overline P&=(2\,\mathrm A)^2(5\,\Omega)=20\,\mathrm W\end{aligned}

[4][3]

In brief

For ideal steady-state series RLC, add complex contributions first. Magnitude sets current size; reactance sets phase. At current resonance opposing reactive terms cancel in the total phasor, not in component voltages. Real power is dissipated in the modelled resistance.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Linked real power and LC natural frequency in context; preserved the series model, phase, resonance limits, exercises and frequency diagram.
October 1, 2026Derived series-RLC phasor and signed phase; developed two original frequency examples, resonance limits and five varied exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

Cite this reference page

Quanta. Impedance of a series RLC circuit. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wechselstrom-impedanz-rlc

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Frequently asked questions about Impedance of a series RLC circuit

What does RLC impedance describe physically?+

RLC impedance describes the voltage-to-current phasor ratio of a linear series circuit at a specified sinusoidal frequency. Its magnitude determines the peak or RMS current associated with a correspondingly specified source voltage. Its phase additionally describes whether voltage or current leads. Resistance dissipates energy; ideal inductive and capacitive contributions store and return it. They are therefore not added as three positive resistances. First take the signed difference of the two reactances, then combine that difference geometrically with R. Magnitude alone cannot distinguish an inductive circuit from an equally reactive capacitive one. The series formula also does not describe an arbitrary parallel arrangement.

How can I check a calculation involving RLC impedance?+

First check that the components really are in series and sinusoidal steady state has been reached. Calculate inductive and capacitive reactance separately, then subtract the capacitive contribution. With resistance 30 ohms, inductive reactance 50 ohms and capacitive reactance 10 ohms, net reactance is 40 ohms and impedance magnitude is 50 ohms. A peak source voltage of 10 volts therefore gives peak current 0.20 amperes. Check that magnitude is not below R. Phase is positive, so voltage leads. Peak and RMS quantities must not be mixed in this ratio. Finally, verify the voltage phasor sum rather than adding three positive voltage magnitudes.

What assumptions and limits apply to RLC impedance?+

The formula assumes an ideal linear series circuit driven sinusoidally at a fixed frequency. R, L and C are treated as constant within the model, and switching transients and free oscillations must already have decayed. Real coil resistance can be included in R, but nonlinear cores or strongly frequency-dependent losses require additional modeling. For a parallel circuit, add complex admittances rather than using this series expression. A nonsinusoidal signal may need decomposition into frequency components. At exact resonance with zero resistance, an ideal finite-voltage source has no bounded steady current. That limiting case is not suitable for the usual current calculation by division.

How do I check units and scale for RLC impedance?+

Use inductance in henries and capacitance in farads, converting millihenries and microfarads first. Use angular frequency rather than directly substituting frequency in hertz: they differ by a factor of two pi. Both inductive and capacitive reactance have units of ohms, so the magnitude square root contains only squared ohms. For a given positive series resistance, impedance magnitude is at least R. Resonance reaches that lower bound. Well below resonance the capacitive contribution usually dominates; well above it the inductive contribution dominates. These limits belong to the ideal model. Real parasitic components can change high-frequency behavior and must be included when interpreting relevant measurements.

When should I combine RLC impedance with another relation?+

Combine impedance with the ratio of RMS voltage to RMS current when calculating current magnitude. Average real power additionally requires the power factor or the resistive series contribution. The product of RMS voltage and RMS current alone is not real power when phase shift is present. For resonance, equate the two reactances; the resulting frequency describes maximum steady series current for fixed components and fixed source voltage. Large individual voltages can occur across L and C even though their phasors cancel in the total. A filter additionally requires specifying which component supplies the output voltage. Free damped oscillation frequency is a separate question.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §15.3: RLC Series Circuits with AC

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Symbol

Meaning

Unit

Z

Magnitude of total impedance

R

Ohmic resistance

ω

Angular frequency

rad/s

L

Inductance

C

Capacitance

Derivation & background

Exam blueprint

Validity range

Magnitude formula for an ideal linear series RLC circuit in sinusoidal steady state; component values are treated as frequency independent.

Derivation steps

Resistance and net reactance are perpendicular in the phasor diagram.

1Choose ideal series RLC in sinusoidal steady state and passive voltage reference.
2The same current passes through all three elements; source voltage is their instantaneous sum.
3With the real-part convention, differentiation becomes $\mathrm j\omega$ , giving $\widetilde Z_L=\mathrm j\omega L$ .
4From $i=C\frac{\mathrm du_C}{\mathrm dt}$ , obtain $\widetilde Z_C=-\frac{\mathrm j}{\omega C}$ .
5Add $\widetilde Z=R+\mathrm j\left(\omega L-\frac1{\omega C}\right)$ .
6Take $Z=|\widetilde Z|$ ; resistance and reactance are perpendicular.
7Check signed phase, current and power using matching RMS or peak values.

How do you calculate with Impedance of a series RLC circuit?

Here is how to work through a typical Impedance of a series RLC circuit (Z = √(R² + (ωL − 1/ωC)²)) task step by step:

Task

For $R=40\,\Omega$ , $X_L=70\,\Omega$ and $X_C=22\,\Omega$ , find Z.

Solution path

Approach: $X=X_L-X_C$ , then $Z=\sqrt{R^2+X^2}$ .
Substitute: $X=(70-22)\,\Omega=48\,\Omega$ .
Result: $Z=\sqrt{40^2+48^2}\,\Omega\approx62.5\,\Omega$ .

Check: magnitude exceeds $R=40\,\Omega$ .

Why three voltages do not add like three lengths

Magnitude alone does not give phase

\begin{aligned}\widetilde Z&=R+\mathrm jX\\ X&=\omega L-\frac{1}{\omega C}\\ Z=|\widetilde Z|&=\sqrt{R^2+X^2}\end{aligned}

We mark the complex phasor ratio with a tilde and call its nonnegative magnitude Z. j is the imaginary unit; R is the total effective series resistance.[2]

Which circuit this model describes

Component laws produce the phasor triangle

\begin{aligned}u_R&=Ri\\u_L&=L\frac{\mathrm di}{\mathrm dt}\\u_C&=\frac{q}{C},\qquad \frac{\mathrm dq}{\mathrm dt}=i\end{aligned}

Under passive reference, inductor voltage follows current change and capacitor voltage follows charge.[1]

\begin{aligned}\widetilde U&=\left(R+\mathrm j\omega L+\frac{1}{\mathrm j\omega C}\right)\widetilde I\\&=\left[R+\mathrm j\left(\omega L-\frac{1}{\omega C}\right)\right]\widetilde I\\ \varphi&=\operatorname{atan2}(X,R)\end{aligned}

φ is voltage phase minus current phase. For positive X voltage leads; for negative X current leads.[2]

Frequency, units and useful rearrangements

\begin{aligned}\omega&=2\pi f\\I_{\mathrm{rms}}&=\frac{U_{\mathrm{rms}}}{Z},\qquad I_0=\frac{U_0}{Z}\\|X|&=\sqrt{Z^2-R^2}\\ C_{\mathrm{res}}&=\frac{1}{\omega^2L}\end{aligned}

Use RMS and peak values in matching pairs. The magnitude alone cannot recover the sign of X.[2][4]

Two frequencies, the same circuit

Electrical power distinguishes real power from apparent power in AC. The following series example consistently uses RMS values; the series resistance R determines dissipation.[3]

Inductive operation at 45° phase angle

Given: $R=24\,\Omega$ , $L=80\,\mathrm{mH}$ , $C=125\,\mathrm{\mu F}$ , $\omega=500\,\mathrm{rad/s}$ , source RMS voltage 60 V; ideal steady-state series circuit.
Find: Impedance, current, phase and average resistor power.

Approach: determine signed reactance, then magnitude, current and power.
$\begin{aligned}X&=\omega L-\frac{1}{\omega C}\\Z&=\sqrt{R^2+X^2}\\\overline P&=I_{\mathrm{rms}}^2R\end{aligned}$
Substitution: first evaluate the two reactances separately.
$\begin{aligned}X_L&=(500\,\mathrm{s^{-1}})(0.080\,\mathrm H)=40\,\Omega\\X_C&=\frac{1}{(500\,\mathrm{s^{-1}})(125\times10^{-6}\,\mathrm F)}=16\,\Omega\\X&=24\,\Omega\\Z&=\sqrt{24^2+24^2}\,\Omega=33.9411\,\Omega\end{aligned}$
Current and phase: voltage leads current by 45°.
$\begin{aligned}I_{\mathrm{rms}}&=\frac{60\,\mathrm V}{33.9411\,\Omega}=1.76777\,\mathrm A\\\varphi&=\arctan\frac{24}{24}=45^\circ\\\overline P&=I_{\mathrm{rms}}^2(24\,\Omega)=75.0\,\mathrm W\end{aligned}$

Result: Z ≈ 33.94 Ω; current ≈ 1.768 A; $\varphi=+45^\circ$ ; real power 75 W.

Halving angular frequency makes the circuit capacitive

Given: Same circuit with $R=24\,\Omega$ , $L=80\,\mathrm{mH}$ , $C=125\,\mathrm{\mu F}$ and RMS source voltage 60 V; now $\omega=250\,\mathrm{rad/s}$ .
Find: New current magnitude and phase direction.

Approach: recalculate both frequency-dependent reactances; do not halve total impedance.
$\begin{aligned}X&=\omega L-\frac1{\omega C}\\I_{\mathrm{rms}}&=\frac{U_{\mathrm{rms}}}{\sqrt{R^2+X^2}}\end{aligned}$
Substitution: inductive reactance halves, capacitive reactance doubles.
$\begin{aligned}X_L&=20\,\Omega,\qquad X_C=32\,\Omega\\X&=-12\,\Omega\\Z&=\sqrt{24^2+(-12)^2}\,\Omega=26.8328\,\Omega\\I_{\mathrm{rms}}&=\frac{60\,\mathrm V}{26.8328\,\Omega}=2.23607\,\mathrm A\\\varphi&=\arctan\frac{-12}{24}=-26.5651^\circ\end{aligned}$
Power: larger current increases dissipation at unchanged R.
$\overline P=(2.23607\,\mathrm A)^2(24\,\Omega)=120\,\mathrm W$

Result: Current leads voltage by about 26.57°; RMS current 2.236 A, power 120 W.

Check: Check: resistor voltage 53.666 V; net reactive voltage −26.833 V; their squared sum is 3600 V². Since Z remains above 24 Ω, current stays below its resonance value 2.5 A.

Small total impedance can hide large component voltages

\begin{aligned}\omega_0&=\frac{1}{\sqrt{LC}}\\Z(\omega_0)&=R,\qquad \varphi(\omega_0)=0\\\overline P&=U_{\mathrm{rms}}I_{\mathrm{rms}}\cos\varphi=I_{\mathrm{rms}}^2R\\\cos\varphi&=\frac{R}{Z}\end{aligned}

Resonance cancels total reactance, not the individual voltages or stored energy.[3][4]

Series current has a resonance maximum

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For

R=24\,\Omega

L=80\,\mathrm{mH}

C=125\,\mathrm{\mu F}

and 60 V RMS. The maximum is 2.50 A; at 200 and 500 rad/s it is 1.768 A. The source holds RMS voltage constant.[2][4]

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From model to measurement

Explore further: half-power bandwidth

Half power is not half current

Given: The same ideal series circuit has $R=24\,\Omega$ , $L=0.0800\,\mathrm H$ , $C=125\,\mathrm{\mu F}$ . The source maintains $U_{\mathrm{rms}}=60.0\,\mathrm V$ as frequency changes. Assume steady state.
Find: Find the two frequencies giving half maximum resistor power.

Approach: halving $I_{\mathrm{rms}}^2R$ divides current by $\sqrt2$ and multiplies Z by $\sqrt2$ . Thus $|X|=R$ .
$\begin{aligned}\overline P&=\frac{U_{\mathrm{rms}}^2R}{R^2+X^2}\\\overline P=\frac{\overline P_{\max}}2&\ \Longrightarrow\ X=\pm R\\\omega_-&=\frac{-R+\sqrt{R^2+4L/C}}{2L}\\\omega_+&=\frac{R+\sqrt{R^2+4L/C}}{2L}\end{aligned}$
Substitution: the lower boundary is capacitive; the upper is inductive.
$\begin{aligned}\omega_-&=200\,\mathrm{rad/s}\\\omega_+&=500\,\mathrm{rad/s}\\f_-&=\frac{200}{2\pi}\,\mathrm{Hz}\approx31.8310\,\mathrm{Hz}\\f_+&=\frac{500}{2\pi}\,\mathrm{Hz}\approx79.5775\,\mathrm{Hz}\\I_{\mathrm{rms}}&=\frac{60}{24\sqrt2}\,\mathrm A\approx1.76777\,\mathrm A\\\overline P&=75.0\,\mathrm W\end{aligned}$

Result: At about 31.83 Hz and 79.58 Hz power is 75.0 W, half its resonance value of 150 W. Phases are −45° and +45°.

Five exercises on magnitude, phase and resonance

Calculate · Application

An ideal series RLC circuit has $R=18\,\Omega$ , $X_L=42\,\Omega$ , $X_C=18\,\Omega$ at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

Show hint

First take the difference of reactances.

Learning goal: Compute current and dissipation from signed reactances.

Show solution and calculation

Approach: impedance is the hypotenuse formed by R and X; R dissipates power.
Substitution gives $X=24\,\Omega$ , $Z=30\,\Omega$ , $I_{\mathrm{rms}}=0.4\,\mathrm A$ and 2.88 W.

Check: power factor 0.6; multiplying the source product by 0.6 also gives 2.88 W.

\begin{aligned}X&=42-18=24\,\Omega\\Z&=\sqrt{18^2+24^2}\,\Omega=30\,\Omega\\I_{\mathrm{rms}}&=\frac{12\,\mathrm V}{30\,\Omega}=0.4\,\mathrm A\\\overline P&=(0.4\,\mathrm A)^2(18\,\Omega)=2.88\,\mathrm W\end{aligned}

[2][3]

Rearrange · Application

At $\omega=400\,\mathrm{rad/s}$ , an ideal series circuit with $L=50\,\mathrm{mH}$ and $R=10\,\Omega$ should resonate. What capacitance is required?

Show hint

Equate the two reactances.

Learning goal: Rearrange resonance condition for capacitance.

Show solution and calculation

Approach: resonance requires zero net reactance.
Rearranging for C and substituting SI values gives 125 μF.

Check: both reactances are 20 Ω and total impedance is 10 Ω.

\begin{aligned}\omega L&=\frac{1}{\omega C}\\C&=\frac{1}{\omega^2L}=\frac{1}{(400\,\mathrm{s^{-1}})^2(0.050\,\mathrm H)}\\&=125\times10^{-6}\,\mathrm F\end{aligned}

[4]

Interpret · Foundation

Two ideal series circuits have $R=24\,\Omega$ and $X=+24\,\Omega$ or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Show hint

Squaring loses the sign; phase does not.

Learning goal: Distinguish equal impedance magnitude from equal phase.

Show solution and calculation

Approach: compare magnitude and argument separately.
Both have Z ≈ 33.94 Ω, equal RMS current and equal resistor dissipation.
In the first circuit voltage leads by 45°; in the second voltage lags current by 45°.

Check: both have the same positive power factor.

\begin{aligned}Z&=24\sqrt2\,\Omega\\\varphi_+&=45^\circ\\\varphi_-&=-45^\circ\end{aligned}

[2]

Error analysis · Application

Someone uses $R=24\,\Omega$ , $X_L=40\,\Omega$ , $X_C=16\,\Omega$ to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

Show hint

Capacitive and inductive phasors oppose; R is perpendicular to them.

Learning goal: Correct scalar addition of phase-shifted voltages.

Show solution and calculation

Approach: add complex impedances, not their positive magnitudes.
Z is not $24+40+16$ .
Net reactance is 24 Ω and magnitude $24\sqrt2\,\Omega$ .
Current is about 1.768 A.

Check: the perpendicular voltage components are each 42.426 V and give a 60 V source.

\begin{aligned}\widetilde Z&=(24+\mathrm j24)\,\Omega\\Z&=24\sqrt2\,\Omega\\I_{\mathrm{rms}}&=\frac{60}{24\sqrt2}\,\mathrm A=1.76777\,\mathrm A\end{aligned}

[2]

Transfer · Application

An ideal series circuit has $R=5\,\Omega$ , $L=0.10\,\mathrm H$ , $C=40\,\mathrm{\mu F}$ and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?

Show hint

At resonance $Z=R$ , but individual reactances are not zero.

Learning goal: Reconcile resonance voltage magnification with energy conservation.

Show solution and calculation

Approach: calculate resonance frequency, then current and component voltages.
$\omega_0=500\,\mathrm{rad/s}$ , both reactances 50 Ω, current 2 A and both component voltages 100 V RMS.
Their phasors oppose.
Source power is 20 W, all dissipated in R; ideal storage elements consume no average power.

\begin{aligned}\omega_0&=\frac{1}{\sqrt{(0.10\,\mathrm H)(40\times10^{-6}\,\mathrm F)}}=500\,\mathrm{s^{-1}}\\I_{\mathrm{rms}}&=\frac{10\,\mathrm V}{5\,\Omega}=2\,\mathrm A\\U_{L,\mathrm{rms}}=U_{C,\mathrm{rms}}&=(2\,\mathrm A)(50\,\Omega)=100\,\mathrm V\\\overline P&=(2\,\mathrm A)^2(5\,\Omega)=20\,\mathrm W\end{aligned}

[4][3]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Linked real power and LC natural frequency in context; preserved the series model, phase, resonance limits, exercises and frequency diagram.
October 1, 2026Derived series-RLC phasor and signed phase; developed two original frequency examples, resonance limits and five varied exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

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Quanta. Impedance of a series RLC circuit. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/wechselstrom-impedanz-rlc

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Create a curated FSRS exam set for Z = √(R² + (ωL − 1/ωC)²): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Impedance of a series RLC circuit

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Alternating current and circuits

Worked example

Applications

How do you calculate with Impedance of a series RLC circuit?

Why three voltages do not add like three lengths

Magnitude alone does not give phase

Which circuit this model describes

Component laws produce the phasor triangle

Frequency, units and useful rearrangements

Two frequencies, the same circuit

Inductive operation at 45° phase angle

Halving angular frequency makes the circuit capacitive

Small total impedance can hide large component voltages

Series current has a resonance maximum

From model to measurement

Explore further: half-power bandwidth

Half power is not half current

Five exercises on magnitude, phase and resonance

An ideal series RLC circuit has R=18 ΩR=18\,\OmegaR=18Ω, XL=42 ΩX_L=42\,\OmegaXL​=42Ω, XC=18 ΩX_C=18\,\OmegaXC​=18Ω at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

At ω=400 rad/s\omega=400\,\mathrm{rad/s}ω=400rad/s, an ideal series circuit with L=50 mHL=50\,\mathrm{mH}L=50mH and R=10 ΩR=10\,\OmegaR=10Ω should resonate. What capacitance is required?

Two ideal series circuits have R=24 ΩR=24\,\OmegaR=24Ω and X=+24 ΩX=+24\,\OmegaX=+24Ω or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Someone uses R=24 ΩR=24\,\OmegaR=24Ω, XL=40 ΩX_L=40\,\OmegaXL​=40Ω, XC=16 ΩX_C=16\,\OmegaXC​=16Ω to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

An ideal series circuit has R=5 ΩR=5\,\OmegaR=5Ω, L=0.10 HL=0.10\,\mathrm HL=0.10H, C=40 μFC=40\,\mathrm{\mu F}C=40μF and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?

In brief

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Frequently asked questions about Impedance of a series RLC circuit

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Retain Impedance of a series RLC circuit for exams

Impedance of a series RLC circuit

Variables & units

Series RLC impedance calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Alternating current and circuits

Worked example

Applications

How do you calculate with Impedance of a series RLC circuit?

Why three voltages do not add like three lengths

Magnitude alone does not give phase

Which circuit this model describes

Component laws produce the phasor triangle

Frequency, units and useful rearrangements

Two frequencies, the same circuit

Inductive operation at 45° phase angle

Halving angular frequency makes the circuit capacitive

Small total impedance can hide large component voltages

Series current has a resonance maximum

From model to measurement

Explore further: half-power bandwidth

Half power is not half current

Five exercises on magnitude, phase and resonance

An ideal series RLC circuit has R=18 ΩR=18\,\OmegaR=18Ω, XL=42 ΩX_L=42\,\OmegaXL​=42Ω, XC=18 ΩX_C=18\,\OmegaXC​=18Ω at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

At ω=400 rad/s\omega=400\,\mathrm{rad/s}ω=400rad/s, an ideal series circuit with L=50 mHL=50\,\mathrm{mH}L=50mH and R=10 ΩR=10\,\OmegaR=10Ω should resonate. What capacitance is required?

Two ideal series circuits have R=24 ΩR=24\,\OmegaR=24Ω and X=+24 ΩX=+24\,\OmegaX=+24Ω or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Someone uses R=24 ΩR=24\,\OmegaR=24Ω, XL=40 ΩX_L=40\,\OmegaXL​=40Ω, XC=16 ΩX_C=16\,\OmegaXC​=16Ω to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

An ideal series circuit has R=5 ΩR=5\,\OmegaR=5Ω, L=0.10 HL=0.10\,\mathrm HL=0.10H, C=40 μFC=40\,\mathrm{\mu F}C=40μF and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?

In brief

Sources for this explanation

Frequently asked questions about Impedance of a series RLC circuit

Subject references and further reading

An ideal series RLC circuit has $R=18\,\Omega$ , $X_L=42\,\Omega$ , $X_C=18\,\Omega$ at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

At $\omega=400\,\mathrm{rad/s}$ , an ideal series circuit with $L=50\,\mathrm{mH}$ and $R=10\,\Omega$ should resonate. What capacitance is required?

Two ideal series circuits have $R=24\,\Omega$ and $X=+24\,\Omega$ or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Someone uses $R=24\,\Omega$ , $X_L=40\,\Omega$ , $X_C=16\,\Omega$ to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

An ideal series circuit has $R=5\,\Omega$ , $L=0.10\,\mathrm H$ , $C=40\,\mathrm{\mu F}$ and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?

An ideal series RLC circuit has $R=18\,\Omega$ , $X_L=42\,\Omega$ , $X_C=18\,\Omega$ at the stated frequency. Source voltage is 12 V RMS. Find Z, current and average power.

At $\omega=400\,\mathrm{rad/s}$ , an ideal series circuit with $L=50\,\mathrm{mH}$ and $R=10\,\Omega$ should resonate. What capacitance is required?

Two ideal series circuits have $R=24\,\Omega$ and $X=+24\,\Omega$ or −24 Ω at the same frequency. With equal source RMS voltage, which quantities agree and which phase differs?

Someone uses $R=24\,\Omega$ , $X_L=40\,\Omega$ , $X_C=16\,\Omega$ to calculate impedance as 80 Ω and current at 60 V RMS as 0.75 A. Locate and correct the error.

An ideal series circuit has $R=5\,\Omega$ , $L=0.10\,\mathrm H$ , $C=40\,\mathrm{\mu F}$ and a 10 V RMS source at resonance. Find current and voltages across L and C. Do their magnitudes imply free additional real power?