University Physics II · Inductance and AC Circuits · 11.6
Reactance, Impedance & Phasors
Drive a resistor, a capacitor and an inductor with one sinusoidal current and their voltages peak at three different moments. Phasors turn that timing into geometry, and the geometry returns one number for the whole circuit: the impedance.
Build the model
Connect the measurement to the mechanism.
One current flows through every element of a series AC circuit, so measure every voltage against it. The resistor's voltage peaks with the current. The inductor's peaks a quarter cycle earlier, because v = L di/dt is largest where the current crosses zero.
The capacitor's peaks a quarter cycle later, because its charge is the accumulated current. Each element still fixes an amplitude ratio in ohms — R, XL = ωL, XC = 1/(ωC) — but two of the three arrive with a fixed 90° shift, so the three voltage amplitudes cannot be added as numbers. Draw each as a phasor, a vector whose length is the amplitude and whose angle is the phase, and the loop rule becomes vector addition: Z = √(R² + (XL − XC)²), with tan φ = (XL − XC)/R.
Because XL rises with frequency while XC falls, one circuit is capacitive at low frequency and inductive at high, and the whole frequency response follows from those two trends.
- Simple definition
- Reactance is the amplitude ratio V/I of an inductor or capacitor at a given frequency, measured in ohms but carrying a 90° phase shift. Impedance combines it with resistance: Z = √(R² + (XL − XC)²), the ratio the source sees.
- Example
- At 60.0 Hz a 0.400 H inductor has XL = ωL = 151 Ω and a 6.00 μF capacitor has XC = 1/(ωC) = 442 Ω. In series with R = 200 Ω that gives Z = 353 Ω, so a 120 V rms supply drives 0.340 A.
Differentiating a cosine advances its phase by 90°; integrating retards it by 90°.
The same i passes through all three in series; only the timing differs
At 60.0 Hz a quarter cycle is 4.17 ms between the two peaks.
Inductor voltage leads the current; capacitor voltage lags it
Grows with ω because a faster current change forces a larger back emf.
Ω for L in H and ω in rad s⁻¹; 0.400 H at 60.0 Hz gives 151 Ω
Falls with ω because less charge accumulates before the polarity reverses.
Ω for C in F; 6.00 μF at 60.0 Hz gives 442 Ω
R = 200 Ω, XL = 151 Ω, XC = 442 Ω → Z = 353 Ω and φ = −55.5°.
V = IZ; φ > 0 is inductive (voltage leads), φ < 0 is capacitive
A capacitor blocks DC and passes fast signals; an inductor does the reverse.
XL = XC at ω₀ = 1/√(LC) = 646 rad s⁻¹ here, or 103 Hz
The series current sets the reference
Once the transients have died away, a series circuit driven at angular frequency ω settles into a steady state in which the current and every voltage oscillate at that same ω. Only the amplitudes and the phases differ. Charge conservation forces one current through the resistor, the inductor and the capacitor in turn, so that current is the natural reference: write i = I cos ωt and quote every voltage as an amplitude plus a phase measured against it. Nothing here weakens Kirchhoff's loop rule — vR + vL + vC equals the source voltage at every instant, without exception. What changes is the bookkeeping. The three terms reach their maxima at different times, so the sum of their peak values is not the peak of their sum, and the arithmetic that worked for DC has to be replaced.
Three elements, three timings
Put i = I cos ωt through each element. The resistor obeys Ohm's law instant by instant, so vR = IR cos ωt: in phase, peaking with the current. The inductor obeys vL = L di/dt = −ωLI sin ωt = ωLI cos(ωt + 90°). The derivative is largest where the current crosses zero, so the inductor's voltage peaks a quarter cycle before the current — the voltage leads. The capacitor holds charge q = ∫i dt = (I/ω) sin ωt, with no constant term in the steady state, so vC = q/C = (I/(ωC)) cos(ωt − 90°). Charge has to accumulate before a voltage appears, so the capacitor's voltage peaks a quarter cycle after the current — the voltage lags. Differentiating a cosine advances its phase by 90° and integrating retards it by 90°; every phase relation in this lesson is that one fact applied to L and C.
Reactance is an amplitude ratio, not a resistance
Each result has the form (voltage amplitude) = (current amplitude) × (something in ohms). For the resistor that something is R. For the inductor it is the inductive reactance XL = ωL, and for the capacitor the capacitive reactance XC = 1/(ωC). Both are in ohms and both obey V = IX, but neither is a resistance. XL climbs with frequency, because a faster current change means a larger back emf, and at ω = 0 an ideal inductor is a piece of wire. XC falls with frequency, because at high ω the polarity reverses before much charge accumulates, while at ω = 0 no steady current passes at all. With L = 0.400 H and C = 6.00 μF at 60.0 Hz, ω = 377 rad s⁻¹ gives XL = 151 Ω and XC = 442 Ω. The second difference matters as much as the first: with the voltage 90° out of step with the current, the instantaneous power iv varies as sin 2ωt about zero — positive for one quarter cycle, negative for the next — so a reactance stores energy and returns it, dissipating nothing on average.
A phasor turns timing into geometry
A sinusoid of fixed frequency carries exactly two numbers, an amplitude and a phase, and a vector in a plane carries exactly two numbers as well. That is the whole of the phasor idea: draw an arrow of length equal to the amplitude at an angle equal to the phase, let the diagram rotate at ω, and the instantaneous value is the arrow's projection on the horizontal axis. Since every quantity in the circuit shares one ω, the arrows keep their relative angles forever, and the rotation can be ignored — the diagram is rigid. Adding two sinusoids of the same frequency then becomes adding two vectors, because projection is linear: the projection of a sum is the sum of the projections. The 90° shifts fall straight out of the geometry. Place the current phasor along the reference axis; VR lies along it, VL points 90° ahead, and VC points 90° behind. VL and VC are therefore antiparallel, which is why their magnitudes subtract rather than add.
Adding the loop: the series impedance
Kirchhoff's loop rule says v = vR + vL + vC at every instant, so the source phasor is the vector sum of the three. The reactive pair collapses to a single arrow of length |VL − VC| perpendicular to VR, leaving a right triangle: V = I√(R² + (XL − XC)²) = IZ, with tan φ = (XL − XC)/R. Positive φ means the source voltage leads the current and the circuit is inductive; negative φ means the current leads and it is capacitive. Take R = 200 Ω, L = 0.400 H and C = 6.00 μF at 60.0 Hz, so XL = 151 Ω and XC = 442 Ω. Then Z = √(200² + (−291)²) = 353 Ω and φ = arctan(−291/200) = −55.5°, a current leading by 2.6 ms. A 120 V rms supply drives I = 120/353 = 0.340 A, and the rms readings are VR = 67.9 V, VL = 51.2 V and VC = 150 V. Those add to 269 V, which no meter will show across the supply; added as phasors they give √(67.9² + (51.2 − 150.1)²) = 120 V, as they must.
The two frequency limits, and where the model stops
Push ω towards zero and XC = 1/(ωC) diverges while XL = ωL vanishes: the capacitor blocks, the inductor is just wire, Z → ∞ and the current dies. Push ω towards infinity and the roles swap — XL diverges, XC vanishes, Z → ∞ again. Between them sits one frequency where XL = XC, at ω₀ = 1/√(LC); there the reactive phasors cancel, Z falls to R and φ = 0. For the components above ω₀ = 646 rad s⁻¹, or 103 Hz, so driving at 60.0 Hz puts the circuit on the capacitive side — which is why φ came out negative. What happens at that crossover is the resonance topic's business. The model is narrow in three ways. It assumes ideal linear components, so a real inductor's winding resistance and a real capacitor's losses have to be folded into R. It assumes one driving frequency, since Z is defined frequency by frequency and a non-sinusoidal drive must be decomposed into sinusoids first. And it describes the steady state only: the phasor diagram says nothing about the transient that precedes it.
Change one variable at a time
Make the relationship visible.
Drag the frequency up through 103 Hz and watch the reactance leg flip from pointing down (capacitive, current leading) to pointing up (inductive), passing through zero where XL = XC and Z collapses to R — then hold f and drag R to see the same reactance produce a completely different φ.
INDUCTIVE XL151 Ω
CAPACITIVE XC442 Ω
IMPEDANCE Z353 Ω
PHASE φ-55.5 °
Live interpretationINDUCTIVE XL: 151 Ω. CAPACITIVE XC: 442 Ω. IMPEDANCE Z: 353 Ω. PHASE φ: −55.5 °
Catch the common trap
Explain before calculating.
A series RLC circuit is driven at a fixed frequency. Rms voltmeters read 30 V across the resistor, 80 V across the inductor and 40 V across the capacitor. What is the rms source voltage, and how does the current sit relative to it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 25.0 mH inductor and a 40.0 μF capacitor each carry the same 400 Hz sinusoidal current of amplitude 60.0 mA. Find the reactance of each and the voltage amplitude across each.
- ω = 2πf = 2π × 400 Hz = 2513 rad s⁻¹ — one ω for both, because they share one current.
- XL = ωL = 2513 rad s⁻¹ × 25.0 × 10⁻³ H = 62.8 Ω.
- XC = 1/(ωC) = 1/(2513 rad s⁻¹ × 40.0 × 10⁻⁶ F) = 1/0.1005 S = 9.95 Ω.
- Each reactance is an amplitude ratio V = IX: VL = 0.0600 A × 62.8 Ω = 3.77 V, and VC = 0.0600 A × 9.95 Ω = 0.597 V.
- Only the timing separates them. With T = 1/400 Hz = 2.50 ms, the inductor's peak arrives 0.625 ms before the current's and the capacitor's 0.625 ms after.
AnswerXL = 62.8 Ω and XC = 9.95 Ω; VL = 3.77 V leading the current, VC = 0.597 V lagging it.
MediumA 120 Ω resistor, a 90.0 mH inductor and a 15.0 μF capacitor sit in series across a 24.0 V rms, 200 Hz supply. Find Z, the rms current, the phase angle, and the rms voltage across each element.
- ω = 2π × 200 Hz = 1257 rad s⁻¹, so XL = ωL = 1257 × 0.0900 H = 113.1 Ω and XC = 1/(ωC) = 1/(1257 × 15.0 × 10⁻⁶ F) = 53.05 Ω.
- Subtract, do not add: VL and VC are antiparallel phasors, so the net reactance is XL − XC = 113.1 − 53.05 = +60.05 Ω, inductive.
- Z = √(R² + (XL − XC)²) = √(120² + 60.05²) Ω = √18006 Ω = 134.2 Ω.
- I = V/Z = 24.0 V ÷ 134.2 Ω = 0.1789 A, and tan φ = (XL − XC)/R = 60.05/120 = 0.500, so φ = +26.6°: the source voltage leads the current.
- VR = IR = 0.1789 A × 120 Ω = 21.5 V; VL = IXL = 0.1789 A × 113.1 Ω = 20.2 V; VC = IXC = 0.1789 A × 53.05 Ω = 9.49 V.
- Those three meter readings total 51.2 V, which no supply shows; added as phasors, √(21.5² + (20.2 − 9.49)²) V = 24.0 V — the supply reading, as it must be.
AnswerZ = 134 Ω, I = 0.179 A, φ = +26.6° (inductive); VR = 21.5 V, VL = 20.2 V, VC = 9.49 V.
HardKeep the same 120 Ω resistor, 90.0 mH inductor and 15.0 μF capacitor on a 24.0 V rms supply. At what drive frequency does the current lead the source voltage by exactly 45.0°, and what is Z there?
- A leading current means a capacitive net reactance, so φ = −45.0° and XL − XC = R tan φ = 120 Ω × (−1.000) = −120 Ω.
- Write that in ω: ωL − 1/(ωC) = −120, i.e. 0.0900ω − 66 667/ω = −120. Multiply through by ω to clear the reciprocal: 0.0900ω² + 120ω − 66 667 = 0.
- Take the positive root: ω = [−120 + √(120² + 4 × 0.0900 × 66 667)]/(2 × 0.0900) = (−120 + √38 400)/0.180 = 75.96/0.180 = 422 rad s⁻¹, so f = ω/2π = 67.2 Hz.
- Check it: XL = 0.0900 H × 422 rad s⁻¹ = 38.0 Ω and XC = 1/(422 × 15.0 × 10⁻⁶) = 158.0 Ω, a difference of −120 Ω as required — and 67.2 Hz sits below this circuit's crossover at 137 Hz, which is why it is capacitive.
- At 45° the two legs of the impedance triangle are equal, so Z = √(120² + 120²) Ω = 120√2 Ω = 170 Ω, and I = 24.0 V ÷ 169.7 Ω = 0.141 A.
Answerf = 67.2 Hz; Z = 170 Ω (= 120√2 Ω), giving I = 0.141 A with the current leading by 45.0°.