Reactance, Impedance & Phasors

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 — v_R + v_L + v_C 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.

Put i = I cos ωt through each element. The resistor obeys Ohm's law instant by instant, so v_R = IR cos ωt: in phase, peaking with the current. The inductor obeys v_L = 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 v_C = 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.

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 X_L = ωL, and for the capacitor the capacitive reactance X_C = 1/(ωC). Both are in ohms and both obey V = IX, but neither is a resistance. X_L 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. X_C 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 X_L = 151 Ω and X_C = 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 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; V_R lies along it, V_L points 90° ahead, and V_C points 90° behind. V_L and V_C are therefore antiparallel, which is why their magnitudes subtract rather than add.

Kirchhoff's loop rule says v = v_R + v_L + v_C at every instant, so the source phasor is the vector sum of the three. The reactive pair collapses to a single arrow of length |V_L − V_C| perpendicular to V_R, leaving a right triangle: V = I√(R² + (X_L − X_C)²) = IZ, with tan φ = (X_L − X_C)/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 X_L = 151 Ω and X_C = 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 V_R = 67.9 V, V_L = 51.2 V and V_C = 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.

Push ω towards zero and X_C = 1/(ωC) diverges while X_L = ωL vanishes: the capacitor blocks, the inductor is just wire, Z → ∞ and the current dies. Push ω towards infinity and the roles swap — X_L diverges, X_C vanishes, Z → ∞ again. Between them sits one frequency where X_L = X_C, 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.