RLC Impedance Calculator
Work out the impedance, current, phase angle and resonant frequency of a series RLC circuit driven by an AC source.
What it does
An alternating current circuit behaves differently from a battery circuit. A coil of wire opposes changes in the current, a capacitor opposes changes in the voltage, and the two effects fight each other. The result is an impedance that depends on the frequency, and a current that runs out of step with the voltage driving it.
The OpenStax University Physics text sets out the geometry that makes this simple. The same current flows through every element of a series combination, so each element gets a voltage in step with the current for a resistor, a quarter cycle behind it for a capacitor and a quarter cycle ahead for an inductor. Adding those three voltages as vectors gives the source voltage, and the angle of the resultant is the phase angle between the voltage and the current.
From that diagram come the numbers. The impedance is the magnitude of the resultant, the square root of the resistance squared plus the difference of the two reactances squared, and the current follows the alternating version of Ohm's law, I equals V over Z. The phase angle is the arctangent of the difference of the reactances divided by the resistance, and it says whether the current leads or lags the voltage.
This page runs the whole calculation for a series RLC circuit. Give it the resistance, the inductance, the capacitance, the frequency and the peak source voltage, and it returns the two reactances, the impedance, the peak and rms current, the phase angle in radians and degrees, the resonant frequency, the quality factor and the average power in the resistor.
The limits are worth stating. The formulas assume a single series loop of ideal components, with no winding resistance, no core loss and no stray capacitance, and the average power figure is what the resistor dissipates, which is not the same as the power the source delivers when the circuit is reactive.
How to use
- Enter the resistance, the inductance and the capacitance of the series combination.
- Enter the frequency of the AC source, in hertz.
- Enter the peak amplitude of the source voltage. A mains supply of 230 V rms is a peak of about 325 V.
- Read the two reactances and the impedance, which is the number a design calculation needs.
- Read the peak and rms current, and the phase angle to see whether the current leads or lags the voltage.
- Compare the working frequency with the resonant frequency. Near resonance the impedance is smallest and the current is largest, and the quality factor says how sharp that peak is.
Formula
- XC
- Capacitive reactance, in ohms.
- XL
- Inductive reactance, in ohms.
- Z
- Impedance of the whole series circuit, in ohms.
- f0
- Resonant frequency, the frequency at which the reactances cancel.
- Q
- Quality factor, the sharpness of the resonance peak.
The phase angle is the arctangent of the difference of the reactances divided by the resistance, positive when the circuit behaves inductively. The quality factor is omega0 times L over R, and the average power in the resistor is half the peak current squared times the resistance, which is the same as the rms current squared times the resistance.
Worked example
Example: 4 ohms, 3 mH and 800 uF at 200 Hz
Defaults: a resistance of 4 ohm, an inductance of 3 mH, a capacitance of 800 uF, a frequency of 200 Hz and a peak source voltage of 0.1 V.
- The angular frequency is 2 pi times 200 = 1256.64 radians per second.
- The capacitive reactance is 1 over 1256.64 times 800e-6, which is 0.9947 ohm, and the inductive reactance is 1256.64 times 3e-3, which is 3.7699 ohm.
- The impedance is the square root of 4 squared plus 3.7699 minus 0.9947 squared, which is 4.8684 ohm.
- The peak current is 0.1 over 4.8684 = 0.020540 A, and the rms current is that over the square root of 2, which is 0.014524 A.
- The phase angle is the arctangent of 2.7752 over 4, which is 0.6066 radians, or 34.75 degrees, and it is positive, so the circuit is inductively reactive and the current lags the voltage.
- The resonant frequency is 1 over 2 pi times the square root of 3e-3 times 800e-6, which is 102.734 Hz, and there the impedance would be only the 4 ohms of resistance.
- The tool shows 0.9947 ohm, 3.7699 ohm, 4.8684 ohm, 0.020540 A, 0.014524 A, 0.6066 rad, 102.734 Hz, a quality factor of 0.4841 and 0.000844 W of average power.
Reference table
| Frequency | Capacitive reactance | Inductive reactance | Impedance | Peak current |
|---|---|---|---|---|
| 50 Hz | 3.9789 ohm | 0.9425 ohm | 5.0219 ohm | 0.019913 A |
| 100 Hz | 1.9894 ohm | 1.8850 ohm | 4.0014 ohm | 0.024991 A |
| 200 Hz | 0.9947 ohm | 3.7699 ohm | 4.8684 ohm | 0.020540 A |
| 500 Hz | 0.3979 ohm | 9.4248 ohm | 9.8734 ohm | 0.010128 A |
Limits
- The formulas are for a single series loop of ideal components, with no winding resistance, no core loss and no stray capacitance between the parts.
- The average power quoted is the power dissipated in the resistor, and the source delivers more than that whenever the circuit is reactive.
- The resonant frequency assumes the inductance and the capacitance do not change with frequency, which real components do.
- There is no allowance for a source with an internal resistance or for the impedance of the wires.
Common mistakes
- Using the rms voltage in the current formula when the impedance was worked out from the peak. The impedance is the same either way, but the current comes out larger by the square root of two.
- Forgetting that the reactances subtract, so a circuit close to resonance has a much smaller impedance than the resistance alone would suggest.
- Mixing up henries and millihenries, or farads and microfarads, which shifts the resonant frequency by a factor of a hundred.
- Reading the phase angle as the power factor. The power factor is the cosine of that angle.
- Expecting the current to be largest at the working frequency rather than at the resonant frequency, which for most real circuits is somewhere else.
Try it yourself
Quick checks. Answers sit tucked under each question.
What happens to the current at the resonant frequency?
It reaches its maximum, because the two reactances cancel and the impedance falls to the resistance alone. At 102.734 Hz the 0.1 V peak source would drive a peak current of 0.025 A, which is larger than the 0.020540 A at 200 Hz.
Why is the phase angle positive at 200 Hz?
Because the inductive reactance of 3.7699 ohm is larger than the capacitive reactance of 0.9947 ohm, so the circuit behaves like an inductor and the current lags the voltage. The angle is 0.6066 radians, which is 34.75 degrees, and the power factor is its cosine, 0.8216.
What does the quality factor tell you?
It says how sharp the resonance peak is. Here it is 0.4841, which is low, so the peak is broad and the circuit is heavily damped by the resistance. A high Q, which needs a small resistance, gives a narrow peak and a large voltage magnification across the capacitor.
Questions
What is impedance?
The alternating current version of resistance. It combines the resistance, the inductive reactance and the capacitive reactance into one number in ohms, and dividing the voltage by it gives the current.
What is the phase angle telling me?
How far the current is out of step with the voltage. A positive angle means the current lags, which is what an inductive circuit does, and a negative angle means it leads, which is what a capacitive one does.
What is resonance?
The frequency at which the inductive and capacitive reactances are equal, so they cancel and the impedance is as small as it can be. The source notes that at that frequency the phase angle is zero and the current and the voltage are in step.
How is the quality factor defined?
It is the resonant frequency times the inductance divided by the resistance, and it is also the ratio of the resonant frequency to the bandwidth of the peak. A high Q means a sharp peak and a high voltage across the reactive components.
Why does the source have to give a peak voltage?
Because the phasor geometry and the impedance work with the amplitude of the sinusoid. A 230 V rms supply has a peak of 325 V, and the current follows from the peak value before the rms value is taken.
What is the average power in the resistor?
Half the peak current squared times the resistance, which is the same as the rms current squared times the resistance. The rest of the power the source delivers goes back and forth between the coil and the capacitor rather than being used up.
Sources
- OpenStax University Physics Volume 2, 15.3: RLC series circuits with AC, with the impedance and phase angle
- OpenStax University Physics Volume 2, 15.5: resonance in an AC circuit
- OpenStax University Physics Volume 2: the key equations for alternating current circuits