AC applied to an inductor

Touch a running motor or a doorbell transformer right after switching it off and you may feel — or hear — a small spark jump across the switch contacts. That spark is a coil of wire objecting to having its current changed too fast. The same objection, playing out smoothly and continuously instead of as one violent jolt, is what happens every instant inside an inductor connected to an AC supply. A resistor simply eats a fraction of the driving voltage in proportion to the current at that instant; an inductor does something stranger — it develops an opposing "back" voltage that depends not on how much current is flowing, but on how fast that current is changing. Because AC current is always changing — rising, peaking, falling, reversing, sixty or fifty times a second — a pure inductor is perpetually fighting the supply. The result is a frequency-dependent opposition called inductive reactance, measured in the same ohms as resistance yet behaving completely differently: it vanishes for steady DC and grows without bound as the frequency climbs. Stranger still, the current does not merely get smaller — it gets shifted in time, lagging a full quarter-cycle ( $90°$ ) behind the driving voltage. Peaks of voltage and peaks of current never coincide. This single phase-lag fact is one of the two hinges (the other belongs to the capacitor) on which the entire phasor treatment of the LCR series circuit swings later in this chapter. Filament bulbs, chokes in old-style tube-light fixtures, and the coils inside transformers and motors are all, to a rough first approximation, inductors sitting in an AC world — and every one of them behaves according to the ideas built up here: reactance that scales with frequency, a $90°$ phase lag, and — most counter-intuitively of all — an average power dissipation of exactly zero, however large the current flowing through the coil.

Key takeaways

A pure inductor $L$ in an AC circuit develops a back-EMF $v = L\,di/dt$ that opposes the rate of change of current, not the current itself.
Solving for the current shows it is sinusoidal at the same frequency as the voltage but shifted: $i = I_0\sin(\omega t - \pi/2)$ — current lags voltage by $90°$ .
The peak-current relation looks like Ohm's law, $I_0 = V_0/X_L$ , with inductive reactance $X_L = \omega L = 2\pi f L$ , measured in ohms.
$X_L$ is frequency-dependent: it is zero at DC ( $f=0$ ) and grows linearly with $f$ — the opposite frequency behaviour from a capacitor.
Average power over a full cycle is exactly zero ( $\cos 90° = 0$ ): the inductor stores energy in its magnetic field on one quarter-cycle and returns all of it on the next — a wattless current.
The mnemonic ELI (in an inductor, EMF E leads current I, so L comes first) fixes the direction of the lag once and for all.
Key idea: a pure inductor's back-EMF depends on the rate of change of current, which forces the current to lag the voltage by exactly 90°, so reactance $X_L=\omega L$ grows with frequency while the average power dissipated stays zero at every frequency.

The core insight

Because a pure inductor's opposing voltage depends on $di/dt$ rather than on $i$ itself, the current it draws from an AC source lags the voltage by $90°$ , giving a frequency-proportional inductive reactance $X_L=\omega L$ and — since voltage and current are never simultaneously nonzero in the same sense — zero average power dissipation.

Where it fits in the syllabus

This section is the second of the three "AC applied to a single element" building blocks in the CBSE Class 12 Alternating Current chapter (Unit VI), sitting between "AC applied to a resistor" and "AC applied to a capacitor." Everything here — reactance, the $90°$ phase relationship, and the wattless-current idea — is explicitly listed in the 2025-26 prescribed scope, and it feeds directly into the series LCR circuit, which CBSE requires to be analysed only through phasor diagrams, never by solving the governing differential equation from first principles. Accordingly this article justifies the phase lag through the phasor/physical picture used in the source Pass-1 node, and quotes rather than formally re-derives the calculus solution of $v = L\,di/dt$ , exactly as the syllabus intends.

The physical picture

Picture a pure inductor of self-inductance $L$ (no resistance of its own — an idealisation, but a good one for a well-wound coil) connected straight across an AC source supplying $v = V_0\sin\omega t$ . At every instant, Kirchhoff's voltage law for this single-element loop says the source voltage must exactly balance the inductor's self-induced back-EMF:

v = L\frac{di}{dt}

where: $v$ is the instantaneous source voltage (V), $L$ is the self-inductance (henry, H), and $di/dt$ is the instantaneous rate of change of current (A/s). The physical reading is blunt: whatever the source voltage is doing at any instant, that is exactly how fast the current must be changing at that instant — a large $v$ forces a steep slope in $i(t)$ , and $v=0$ forces the current's slope to be momentarily flat (a turning point). Because the source drives the circuit at angular frequency $\omega$ , the current must also be sinusoidal at that same frequency (an inductor cannot invent new frequencies), so we can write $i = I_0\sin(\omega t - \phi)$ and ask what phase angle $\phi$ makes the equation above balance at every instant. Carrying out that calculus — quoted here rather than re-derived, since CBSE's prescribed method for this chapter is the phasor picture, not the differential-equation solution — gives a clean, memorable result:

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

where: $I_0$ is the peak current (A), $X_L$ is the inductive reactance (Ω) — playing the same role $R$ plays in Ohm's law, but frequency-dependent — $\omega = 2\pi f$ is the angular frequency (rad/s), and $f$ is the source frequency (Hz). The physical reading of the phase term is that the current's sine wave is the voltage's sine wave shifted later in time by a quarter-period: the current lags the voltage by $90°$ . Physically, the inductor's opposition is fiercest exactly when the current is changing fastest — which for a sine wave happens as it crosses zero, not at its peak — so the current's own zero-crossing (and hence its peak, a quarter-cycle later) is delayed relative to the voltage's zero-crossing. The frequency dependence of $X_L$ is the other half of the physical picture, and it runs opposite to intuition about ordinary resistance. At $f = 0$ (steady DC), $X_L = 0$ : an ideal inductor offers zero opposition to a constant current, which checks out from $v = L\,di/dt$ itself, since $di/dt = 0$ for constant $i$ . As $f$ rises, $X_L$ rises in direct proportion — an inductor is a "high-frequency blocker," passing DC and low frequencies with ease while throttling rapidly-alternating current. This is precisely why chokes (inductors) are used to smooth ripples in DC supplies and why high-frequency signals are attenuated by coils in a way steady currents never are.

Figure

Voltage and current sine waves for a pure inductor plotted against angular position, with the current curve shifted a quarter cycle to the right of the voltage curve, and a phasor diagram alongside showing the current phasor rotated ninety degrees clockwise from the voltage phasor

Finally, since voltage and current are $90°$ out of phase, the instantaneous power $p = vi$ spends exactly half of every cycle positive (energy flowing from source into the magnetic field) and half negative (energy flowing back out of the field into the source) — the two halves cancel perfectly over one full cycle, so the average power dissipated in a pure inductor is zero. The inductor is not "wasting" the current that flows through it; it is only ever storing energy temporarily and handing every joule of it back. This behaviour is called a wattless current — substantial current can flow, and the ammeter reads a real nonzero value, yet the wattmeter reads zero.

Derivation APeak current and inductive reactance in a pure inductor circuit

The goal is to start from the loop equation for a pure inductor driven by a sinusoidal source, $v = L\,di/dt$ , and arrive at the two working results used throughout this chapter — the $90°$ phase lag of current behind voltage, and the peak-current relation $I_0 = V_0/X_L$ with $X_L = \omega L$ — using the physical reasoning the CBSE phasor approach relies on, quoting the one calculus step rather than working through it in full, exactly as the "phasors only" prescription intends.

Step 1. Write the source voltage as $v = V_0\sin\omega t$ and set up the loop equation for a purely inductive circuit (no resistor, no capacitor in this element's own branch): $V_0\sin\omega t = L\dfrac{di}{dt}$ . This says the instantaneous rate of change of current is proportional to the instantaneous source voltage, scaled by $1/L$ .

Step 2. Because the source is a single pure sinusoid at frequency $\omega$ , and an ideal inductor cannot generate new frequencies or a DC offset, the current in steady state must also be a pure sinusoid at the same $\omega$ , only possibly shifted in phase: $i = I_0\sin(\omega t - \phi)$ for some peak value $I_0$ and phase lag $\phi$ still to be found.

Step 3. Substituting this trial current into $v = L\,di/dt$ and carrying out the one calculus step — differentiating $\sin(\omega t - \phi)$ , which turns a sine into a cosine of the same argument — shows the equation balances at every instant only if $\phi = \pi/2$ (90 degrees) and $V_0 = I_0\omega L$ . This calculus step is quoted here as the syllabus permits, since the differential-equation method itself is explicitly out of scope; the phasor picture in the physical section above gives the same $90°$ result by reasoning about when the current changes fastest.

Step 4. Read off the two working results. The phase result is $i = I_0\sin(\omega t - \pi/2)$ : current lags voltage by exactly $90°$ , always, for a pure inductor at any frequency. The magnitude result rearranges $V_0 = I_0\omega L$ into $I_0 = V_0/(\omega L) = V_0/X_L$ , defining the inductive reactance $X_L = \omega L$ as the AC "resistance" of the coil, in ohms.

Step 5. Check the limits, which is how a phasor-only derivation is validated without calculus. At $f\to 0$ , $X_L \to 0$ , so $I_0 \to \infty$ for any nonzero $V_0$ — matching the fact that a pure inductor is a short circuit to true DC, since it has no resistance to limit a steady current once $di/dt$ has settled to zero. At high $f$ , $X_L$ grows without bound, so $I_0 \to 0$ — the inductor increasingly blocks rapidly alternating current. Both limits match the physical picture already built up, confirming the result.

Worked examples

Board level - reactance, current and power for a 0.2 H inductor on 220 V, 50 Hz mains

An inductor of $L = 0.2\ \text{H}$ , assumed to have negligible resistance, is connected to an AC source of RMS voltage $V_{rms} = 220\ \text{V}$ and frequency $f = 50\ \text{Hz}$ . Find (a) the inductive reactance, (b) the RMS current drawn, and (c) the average power dissipated.

Step 1. Find the angular frequency: $\omega = 2\pi f = 2\pi(50) = 314\ \text{rad/s}$ .

Step 2. Find the inductive reactance: $X_L = \omega L = 314 \times 0.2 = 62.8\ \Omega$ .

Step 3. Find the RMS current using the AC analogue of Ohm's law: $I_{rms} = \dfrac{V_{rms}}{X_L} = \dfrac{220}{62.8} \approx 3.5\ \text{A}$ .

Step 4. Find the average power. Since current lags voltage by $90°$ in a pure inductor, $\cos\phi = \cos 90° = 0$ , so $P_{avg} = V_{rms}I_{rms}\cos\phi = 220 \times 3.5 \times 0 = 0\ \text{W}$ . Answer: $X_L = 62.8\ \Omega$ , $I_{rms} \approx 3.5\ \text{A}$ , and $P_{avg} = 0\ \text{W}$ — a substantial 3.5 A flows, yet no net energy is dissipated as heat.

Competitive level - working backwards from measured current to L, then to peak current and stored energy

A pure inductor draws an RMS current of $2\ \text{A}$ from a $100\ \text{V}$ (RMS), $50\ \text{Hz}$ AC source. Find (a) the self-inductance $L$ , (b) the peak current $I_0$ , and (c) the maximum energy stored in the inductor's magnetic field during a cycle.

Step 1. Find the reactance from the RMS Ohm's-law relation: $X_L = \dfrac{V_{rms}}{I_{rms}} = \dfrac{100}{2} = 50\ \Omega$ .

Step 2. Relate $X_L$ to $L$ using $X_L = \omega L = 2\pi f L$ , so $L = \dfrac{X_L}{2\pi f} = \dfrac{50}{2\pi(50)} = \dfrac{50}{314} \approx 0.159\ \text{H}$ , roughly $160\ \text{mH}$ .

Step 3. Find the peak current from the RMS value: $I_0 = I_{rms}\sqrt2 = 2 \times 1.414 \approx 2.83\ \text{A}$ .

Step 4. Find the maximum stored energy. The energy stored in an inductor at any instant is $U = \tfrac12 L i^2$ , which is largest when $i$ is at its peak $I_0$ : $U_{max} = \tfrac12 L I_0^2 = \tfrac12 (0.159)(2.83)^2 \approx \tfrac12(0.159)(8.01) \approx 0.64\ \text{J}$ . Answer: $L \approx 0.159\ \text{H}$ , $I_0 \approx 2.83\ \text{A}$ , and up to about $0.64\ \text{J}$ is stored (and fully returned) every quarter-cycle. Timing note: this problem chains three separate one-line formulae — $X_L=V_{rms}/I_{rms}$ , then $L=X_L/\omega$ , then $I_0=I_{rms}\sqrt2$ and $U=\tfrac12LI_0^2$ — so the time cost is in not skipping a conversion, not in any single calculation. Budget under two minutes: write $\omega=314\ \text{rad/s}$ once at the top so it is not recomputed by mistake, and keep $I_{rms}$ and $I_0$ clearly labelled since mixing them into the energy formula is the single most common slip on this pattern.

Common traps

Treating an inductor as if it "resists" current the way a resistor does. Reactance $X_L$ is frequency-dependent and carries a $90°$ phase shift with it — ordinary resistance $R$ is frequency-independent and produces no phase shift at all. They are not interchangeable ideas even though both are measured in ohms.
Getting the direction of the phase lag backwards. In a pure inductor, current lags voltage — never the other way round. The mnemonic ELI fixes this: in an inductor (L), EMF (E) comes before current (I) in time.
Assuming that because current flows, energy must be dissipated. For an ideal inductor the average power is exactly zero at every frequency — energy is stored in the magnetic field and returned to the source each half-cycle, never converted to heat.
Confusing $f$ and $\omega$ in $X_L = \omega L$ . Forgetting the $2\pi$ when going from frequency to angular frequency is the single most common arithmetic slip in this topic — always write $\omega = 2\pi f$ explicitly as a first step.
Mixing up peak and RMS values inside a formula, especially in energy calculations where $U = \tfrac12 L I_0^2$ needs the peak current $I_0 = I_{rms}\sqrt2$ , not $I_{rms}$ itself.
Forgetting that this whole $90°$ /reactance picture assumes an ideal, purely inductive coil. A real coil also has some winding resistance, which introduces its own in-phase component and a nonzero (if usually small) power dissipation — a detail that becomes central once resistance is added back in for the full LCR circuit.

Common trapThe biggest misconception: believing current flowing through a coil must mean energy is being lost

Students who have just met $P = I^2R$ for resistors often carry that intuition straight into inductors and assume "current flowing equals heat generated." For a pure inductor this is false at every single frequency: the $90°$ phase lag means the instantaneous power $p = vi$ is positive for exactly one quarter of the cycle (energy flowing into the field), negative for the next quarter (energy flowing back out), and the pattern repeats — the two contributions cancel exactly over a full cycle, giving $P_{avg} = V_{rms}I_{rms}\cos 90° = 0$ precisely, not approximately. A 3.5 A current through a 220 V inductor (as in the board-level example) genuinely dissipates zero watts as heat in the ideal case; the current is real, the power is not. This is exactly what "wattless current" means, and it is the property later exploited when engineers deliberately use reactive elements to shift phase without burning energy.

You should be able to…

Explain, using the phasor/physical picture (not the full differential-equation solution), why current lags voltage by exactly $90°$ in a pure inductor.
Define inductive reactance $X_L = \omega L = 2\pi f L$ , state its unit, and describe how it behaves as $f \to 0$ and as $f \to \infty$ .
Use $I_0 = V_0/X_L$ (or the RMS version $I_{rms}=V_{rms}/X_L$ ) to move between peak/RMS voltage and peak/RMS current given $L$ and $f$ , and invert the relation to find $L$ from a measured current.
Justify, from the $90°$ phase angle, why the average power dissipated in a pure inductor is exactly zero over any complete cycle — the wattless-current idea.
Compute the instantaneous magnetic energy stored in an inductor, $U = \tfrac12 L i^2$ , and identify its maximum value in terms of the peak current $I_0$ .
Distinguish the ideal pure-inductor result from a real coil, which also carries winding resistance and therefore does dissipate some genuine heat.

Ask the tutor: if the ELI phase-lag rule or the "current flows but zero power is dissipated" result still feels contradictory, ask the AI tutor to walk through the instantaneous power graph $p=vi$ over one full cycle and show exactly where the positive and negative halves cancel.

If no power is dissipated in an inductor, why do transformers and motors get warm during use?

Real coils are never perfectly ideal. Every practical winding has some resistance in its copper wire, so a real inductor is closer to an inductor and a small resistor in series. The resistive part dissipates ordinary $I^2R$ heat exactly as any resistor would, while the purely inductive part still contributes zero average power on its own. There are also secondary heating effects in real devices — eddy currents induced in the iron core and hysteresis losses as the core's magnetisation reverses each cycle — but those are qualitative, device-level effects, not part of the ideal pure-inductor result derived here. The $P_{avg}=0$ result is an idealisation that isolates the inductive effect cleanly; real components always add some resistive loss on top of it.

Why does the current lag instead of leading in an inductor - what is physically special about the zero-crossing of current?

The back-EMF of an inductor is proportional to $di/dt$ , the slope of the current graph, not to the current's own value. For a sine wave, the slope is steepest exactly where the curve crosses zero and is completely flat exactly at the peaks. So the inductor's opposing voltage is largest when current is crossing zero and smallest (zero) when current is at a peak. Since the source voltage must always match this back-EMF, the source voltage itself is largest near the current's zero-crossings — which means the voltage's own peak arrives a quarter-cycle before the current gets around to its peak. That is exactly the statement that current lags voltage by 90 degrees, read straight off the shapes of the two curves.

Is the inductive reactance formula Xsub L equals omega L on my board exam formula sheet, or do I need to memorise it?

$X_L = \omega L = 2\pi f L$ is one of the small set of formulae CBSE expects you to know and apply directly, along with $X_C = 1/\omega C$ for the capacitor and $Z$ for the full LCR circuit that come later in this chapter. It is short enough that most students simply memorise it, but the safest way to keep it straight under exam pressure is to remember the two limiting behaviours: $X_L \to 0$ as $f \to 0$ (a coil is a short circuit to steady DC) and $X_L$ grows as $f$ grows (a coil chokes off high frequencies). If you can reconstruct those two facts, the formula falls out naturally rather than needing to be recalled cold.

Does the 90 degree phase lag mean the current physically starts a quarter cycle later than the voltage, like a delayed switch?

Not quite — it is not a one-time delay but a permanent, repeating time offset in the steady, ongoing sinusoidal oscillation. Once the AC source has been running for a while and everything has settled into its repeating pattern, every single peak of the current wave arrives one quarter-period (90 degrees of the cycle) after the corresponding peak of the voltage wave, and this keeps happening forever, cycle after cycle, for as long as the source keeps running at that frequency. It is best pictured on the phasor diagram: the current phasor is permanently frozen 90 degrees behind the voltage phasor, and the whole pair spins together at the same angular speed $\omega$ — so the 90 degree gap between them never closes or grows, it just keeps rotating.