AC circuits with R, L, C

Plug a resistor into an AC socket and the current tracks the voltage step for step — Ohm's law, moment by moment. Swap in an inductor or a capacitor and something odd happens: the current still oscillates at the same frequency, but it no longer peaks at the same instant as the voltage. It runs a quarter-cycle late for an inductor, a quarter-cycle early for a capacitor. This lag or lead is not a minor detail — it is why AC power engineering needs an entirely different toolkit from DC circuits, and it is the gateway to reactance, impedance, and eventually resonance in the series LCR circuit. This article works out, from first principles, exactly how much a resistor, an inductor, and a capacitor each oppose alternating current, and precisely how far out of step the current runs with the voltage in each case.

The core insight

In an AC circuit a resistor keeps current and voltage in phase, an inductor makes current lag voltage by 90 degrees with reactance $X_L=\omega L$ growing with frequency, and a capacitor makes current lead voltage by 90 degrees with reactance $X_C=1/\omega C$ shrinking with frequency.

The core idea

A phasor is a rotating arrow of length equal to the peak value ( $V_0$ or $I_0$ ) of a sinusoidal quantity, spinning counter-clockwise at angular frequency $\omega$ ; its shadow on a fixed axis traces out the instantaneous value. Phasors let us read off phase relationships at a glance instead of solving a differential equation every time.

Resistor — no phase shift. Ohm's law holds instant by instant: $i=v/R$ . If $v=V_0\sin\omega t$ , then $i=I_0\sin\omega t$ with $I_0=V_0/R$ . Current and voltage phasors point the same way at every moment — they rise and fall together.

Inductor — current lags by 90 degrees. The inductor's back-EMF condition $v=L\,di/dt$ forces the current to trail the voltage by a quarter cycle:

i = I_0\sin\left(\omega t-\frac{\pi}{2}\right), \qquad I_0=\frac{V_0}{X_L},\qquad X_L=\omega L

where: $X_L$ is the inductive reactance in ohms, $L$ the inductance in henries, and $\omega=2\pi f$ the angular frequency. Because $X_L\propto\omega$ , an inductor barely resists a slow-changing current but strongly resists a fast one.

Capacitor — current leads by 90 degrees. With $q=Cv$ and $i=dq/dt$ , the current runs a quarter cycle ahead of the voltage:

i = I_0\sin\left(\omega t+\frac{\pi}{2}\right), \qquad I_0=\frac{V_0}{X_C},\qquad X_C=\frac{1}{\omega C}

where: $X_C$ is the capacitive reactance in ohms and $C$ the capacitance in farads. Because $X_C\propto1/\omega$ , a capacitor blocks slow-changing current (and blocks DC completely, since $\omega\to0\Rightarrow X_C\to\infty$ ) but passes fast-changing current readily.

Impedance $Z=V_0/I_0$ is the AC generalisation of resistance for any single element or combination; for a pure R, L, or C alone it equals $R$ , $X_L$ , or $X_C$ respectively. Note that $Z$ , like reactance, can be computed equally well from peak values ( $V_0/I_0$ ) or RMS values ( $V_{rms}/I_{rms}$ ), since the ratio is the same either way — but the two must never be mixed within one calculation. Combining R, L, C together into one impedance is the job of the series LCR circuit topic, not this one; there, the phase differences worked out here for each element separately determine how the reactances add.

Why the phase shift matters physically. A resistor converts electrical energy irreversibly into heat at every instant, regardless of how fast the current is changing, which is exactly why $R$ carries no dependence on $\omega$ . An inductor and a capacitor, by contrast, only store and return energy — they oppose the rate of change of current or voltage respectively, and that opposition is what produces both the frequency-dependent reactance and the 90-degree phase shift. This distinction between elements that dissipate energy and elements that merely store and release it will resurface when the series LCR circuit's power factor is introduced.

Figure

Three phasor diagrams side by side showing current in phase with voltage for a resistor, current lagging voltage by ninety degrees for an inductor, and current leading voltage by ninety degrees for a capacitor

Derivation AWhy the inductor's current lags the voltage by 90 degrees

The goal is to show, starting only from the defining relation for an inductor's back-EMF and one rule of calculus, that connecting an inductor across a sinusoidal source $v=V_0\sin\omega t$ produces a current that is also sinusoidal at the same frequency but shifted by exactly a quarter cycle behind the voltage, and to read off the peak current and the reactance $X_L$ along the way.

Step 1. Write the defining law for an inductor. The voltage across an ideal inductor is proportional to the rate of change of the current through it, $v=L\dfrac{di}{dt}$ . This is not Ohm's law — it links $v$ to how fast $i$ is changing, not to $i$ itself, and that difference is the whole reason a phase shift appears.

Step 2. Substitute the source voltage and isolate $di/dt$ . With $v=V_0\sin\omega t$ , the relation gives $\dfrac{di}{dt}=\dfrac{V_0}{L}\sin\omega t$ . To find $i(t)$ we must undo the derivative, i.e. integrate both sides with respect to time.

Step 3. Integrate to find the current. $i=\displaystyle\int \frac{V_0}{L}\sin\omega t\,dt = -\frac{V_0}{\omega L}\cos\omega t + \text{constant}$ . Taking the constant of integration as zero (no steady DC offset in a driven AC circuit), $i=-\dfrac{V_0}{\omega L}\cos\omega t$ .

Step 4. Rewrite in sine form to compare phase directly with $v$ . Using $-\cos\omega t=\sin(\omega t-\pi/2)$ , we get $i=\dfrac{V_0}{\omega L}\sin\left(\omega t-\dfrac{\pi}{2}\right)$ . Matching this to $i=I_0\sin(\omega t-\pi/2)$ identifies $I_0=V_0/(\omega L)$ , so defining $X_L=\omega L$ gives the compact result $I_0=V_0/X_L$ — exactly the inductive reactance and the 90-degree lag stated above, both falling directly out of integrating $v=L\,di/dt$ .

Worked example

A pure inductor of 0.2 H is connected to a 220 V (RMS), 50 Hz AC source. Find (a) the inductive reactance, (b) the RMS current, (c) the phase relationship between current and voltage.

Step 1. Find the angular frequency: $\omega=2\pi f=2\pi\times50=314.16\ \text{rad/s}$ .

Step 2. Compute the reactance: $X_L=\omega L=314.16\times0.2=62.83\ \Omega$ .

Step 3. Apply the AC form of Ohm's law with RMS values: $I_{rms}=\dfrac{V_{rms}}{X_L}=\dfrac{220}{62.83}=3.50\ \text{A}$ .

Step 4. State the phase relationship: current lags the voltage by $\pi/2$ (90°), since the inductor's back-EMF always delays the current's response to the applied voltage. Answer: $X_L=62.83\ \Omega$ , $I_{rms}=3.50\ \text{A}$ , current lagging voltage by 90°.

Competitive stretch

What is the average power dissipated in this inductor over a full cycle? Instantaneous power is $p=vi$ , and since $v\propto\sin\omega t$ while $i\propto\sin(\omega t-\pi/2)=-\cos\omega t$ , the product $p\propto -\sin\omega t\cos\omega t=-\tfrac12\sin2\omega t$ oscillates symmetrically about zero and averages to zero over one full cycle. The average power delivered to a pure inductor is zero — energy flows into the magnetic field for a quarter cycle and flows straight back out for the next, with no net dissipation, unlike a resistor which converts all delivered power to heat.

Common traps

Mixing up which element leads and which lags — use "ELI the ICE man": in an inductor (L), EMF leads current; in a capacitor (C), current leads EMF.
Treating reactance like resistance and assuming it is frequency-independent — $X_L=\omega L$ rises with frequency, $X_C=1/\omega C$ falls with frequency; only $R$ stays fixed.
Forgetting that average power is zero for a pure L or pure C over a full cycle, even though the instantaneous power is non-zero and changes sign twice per cycle.
Mixing peak and RMS values within one calculation — $Z=V_0/I_0$ using peaks matches $V_{rms}/I_{rms}$ using RMS values, but never combine a peak voltage with an RMS current.

Ask the tutor: Ask the tutor for a phasor-diagram walkthrough showing exactly how differentiating or integrating a sinusoid produces the $\pi/2$ phase shift.

Why does a capacitor block DC but let AC through?

For steady DC, $\omega=0$ , so $X_C=1/\omega C\to\infty$ — the capacitor charges up fully, the voltage across it matches the source, and no current can keep flowing once it is fully charged. For AC, the source keeps reversing before the capacitor can fully charge, so charge is continually flowing on and off the plates, which looks like a continuous current through the circuit even though no charge actually crosses the gap between the plates.

If average power in a pure L or pure C is zero, is any energy used at all?

Energy is exchanged but not consumed. In an inductor, energy is stored in the magnetic field while current builds up and returned to the source as the field collapses; in a capacitor, energy is stored in the electric field while it charges and returned as it discharges. Over one full cycle the energy borrowed equals the energy returned, so the net (average) power is zero, even though instantaneous power swings between positive and negative twice per cycle.

Why is the phase angle exactly 90 degrees and not some other value for a pure L or pure C?

It follows directly from calculus: differentiating or integrating a sine wave always produces a cosine wave, and $\sin$ and $\cos$ are exactly a quarter cycle (90°) apart. The inductor's law $v=L\,di/dt$ involves a derivative, and the capacitor's law $i=C\,dv/dt$ also involves a derivative — in both cases that derivative operation is what shifts the phase by precisely 90°, no more and no less. A resistor's law $v=iR$ involves no derivative at all, which is exactly why it produces zero phase shift.

AC circuits with R, L, C

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads