AC Voltage Applied to an Inductor

Physics · Alternating Current · NEET

When AC voltage v = vm sin(wt) is applied to a pure inductor, the current is i = im sin(wt - pi/2), so the current LAGS the voltage by 90 degrees (a quarter cycle). The inductor opposes current with inductive reactance XL = wL = 2 pi f L (in ohms), giving im = vm / XL. Memory hook: in an inductor "voltage comes first, current is late" - the L in inductor reminds you current Lags.
Pure Inductor: current lags voltage by 90 degreeswtquartercycle lagv = vm sin(wt)i = im sin(wt - pi/2)XL = wL = 2 pi f L , im = vm / XL
Voltage (blue) and current (red dashed) in a pure inductor. The current curve is shifted a quarter cycle to the right, showing the current lags the voltage by 90 degrees. Reactance XL = wL sets the current amplitude im = vm / XL.

Your doubts, answered

Does a pure inductor behave like a resistor in an AC circuit?

No. Both an inductor and a resistor limit the current, but they do it differently. A resistor keeps current in phase with voltage and dissipates heat. A pure inductor makes the current lag the voltage by 90 degrees and dissipates NO average power. The opposition of an inductor is called inductive reactance XL = wL (unit: ohm), not resistance.

Why does the current lag the voltage by 90 degrees in an inductor?

Start from Kirchhoff's loop rule: v = L (di/dt), so di/dt = (vm/L) sin(wt). Integrating gives i = -(vm/wL) cos(wt). Since -cos(wt) = sin(wt - pi/2), we get i = im sin(wt - pi/2). The '- pi/2' means the current reaches its peak a quarter cycle AFTER the voltage, so current lags voltage by 90 degrees.

Why is the average power supplied to a pure inductor zero?

Instantaneous power p = i*v = -(im*vm/2) sin(2wt). Over one full cycle the average of sin(2wt) is zero, so the average power is zero. Energy flows into the inductor's magnetic field for a quarter cycle and flows back out the next quarter cycle. This current that carries no net power is called the wattless current.

What does inductive reactance XL depend on?

XL = wL = 2 pi f L. It is directly proportional to both the frequency f and the inductance L. So a higher frequency or a larger inductor gives more opposition. At f = 0 (pure DC steady state) XL = 0, so a pure inductor lets DC pass freely once current is steady.

Is the inductor's opposition to current the same as resistance?

It has the same unit (ohm) and limits current the same way (im = vm/XL), but it is NOT resistance. Resistance is fixed and dissipates energy; reactance XL changes with frequency and dissipates no average power. That is why we call XL reactance, not resistance.

⚠️ The NEET trap
Using XL = L or forgetting the 2 pi factor, so students write XL = f*L instead of XL = 2 pi f L.
XL = wL = 2 pi f L. For L = 25 mH at f = 50 Hz: XL = 2 x 3.14 x 50 x 25 x 10^-3 = 7.85 ohm. Always convert mH to H (x 10^-3) and include 2 pi.
🧠 NTA loves swapping 'current leads' (capacitor) with 'current lags' (inductor). Remember: inductor = current Lags, capacitor = current leads. In L, current is Late.

Solved Alternating Current NEET PYQs

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Frequently asked

In a pure inductor, does current lag or lead the voltage?

The current lags the voltage by 90 degrees (pi/2 radians), i.e. by one quarter of a cycle. Voltage peaks first, current peaks a quarter cycle later.

What is the formula for inductive reactance?

XL = wL = 2 pi f L, measured in ohms. It increases with both frequency and inductance.

What is the current amplitude in a pure inductor?

im = vm / XL = vm / (wL). In rms terms, Irms = Vrms / XL.

Why does a pure inductor consume no average power?

Because current lags voltage by exactly 90 degrees, so cos(phi) = cos(90) = 0. Energy is stored in the magnetic field and returned each cycle, giving zero net average power (wattless current).

What happens to a pure inductor with a DC source in steady state?

For steady DC, f = 0 so XL = wL = 0. The inductor offers no opposition and behaves like a plain wire once the current becomes steady.