Q.Find out the phase relationship between voltage and current in a pure inductive circuit.

AC Through an Inductor: From Intuition to the 90° Lag

The Core Intuition — Why an Inductor "Fights" Change

Imagine you are pushing a heavy box across a floor. If you push steadily, the box moves at a constant speed. But if you try to suddenly jerk it forward, the box resists — its inertia makes it want to stay where it is. The harder you push, the more it pushes back, but only while you are changing the speed.

An inductor does the same thing, but with electric current. It does not resist steady current (DC) — a perfect inductor has zero resistance. What it resists is change in current. The faster you try to change the current, the harder the inductor pushes back with a voltage that opposes that change.

This opposition is called self-induction, and the property that causes it is inductance $L$, measured in henries (H).

The Physics: Faraday's Law in Action

When current flows through a coil, it creates a magnetic field. If the current changes, the magnetic field changes, and that changing field induces a voltage in the same coil — a back emf. Faraday's law gives the magnitude:

The sign matters: the induced voltage always acts to oppose the change that produced it (Lenz's law). So if current is increasing ($di/dt > 0$), the inductor generates a voltage that tries to push current the other way — like a spring that pushes back harder the faster you compress it.

AC Through an Inductor: The Mathematics

Now feed the inductor with an AC voltage source:

where $\omega = 2\pi f$ is the angular frequency. The circuit equation (Kirchhoff's voltage law) gives:

So:

Integrate to find the current:

The constant $C$ is zero for steady-state AC (no DC offset). Using $\cos(\omega t) = \sin(\omega t + 90^\circ)$:

The 90° Lag — What It Means Physically

Compare the voltage and current:

• Voltage: $v(t) = V_m \sin(\omega t)$
• Current: $i(t) = I_m \sin(\omega t - 90^\circ)$

The current reaches its peak exactly one-quarter cycle after the voltage does. We say current lags voltage by 90° (or $\pi/2$ radians).

Why? Look at the derivative relationship. When voltage is at its peak, $di/dt$ is maximum — the current is changing fastest. But the current itself is passing through zero at that instant (think of a sine wave: its steepest slope is at the zero crossing). When voltage passes through zero, $di/dt = 0$, and the current is at its peak (the sine wave is flat at the top).

Inductive Reactance — The "AC Resistance"

The amplitude of the current is:

Define inductive reactance $X_L$:

Then $I_m = V_m / X_L$, which looks like Ohm's law — but $X_L$ is not a resistance. It depends on frequency: …