20.4.2 AC Through an Inductor

Introduction

When an alternating current (AC) source is connected to a pure inductor (a coil with negligible resistance), the inductor opposes changes in current through the phenomenon of self-induction. This opposition is characterised by inductive reactance and produces a specific phase relationship between voltage and current.

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Circuit Analysis

Consider a pure inductor of self-inductance $L$ connected to an AC source:

$V=V_o\sin (\omega t)$

By Faraday's law, the back emf of the inductor equals the applied voltage:

$V=L\frac{dI}{dt}$

Solving for the current:

$I=\frac{V_o}{\omega L}\sin \left(\omega t-\frac{\pi }{2}\right)=I_o\sin \left(\omega t-\frac{\pi }{2}\right)$

where $I_o=\frac{V_o}{\omega L}$.

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Phase Relationship

This can be visualised on a phasor diagram: the voltage phasor $\vec{V}$ leads the current phasor $\vec{I}$ by $90^{\circ }$.

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Inductive Reactance ($X_L$)

The quantity $\omega L$ plays the role of resistance in limiting the current. It is called Inductive Reactance:

$\boxed{X_L=\omega L=2\pi fL}$

• SI unit: Ohm ($\Omega$)
• $X_L$ is directly proportional to frequency $f$ and inductance $L$.
• The peak current is: $I_o=\frac{V_o}{X_L}$
• In RMS terms (Ohm's Law analogy): $I_{rms}=\frac{V_{rms}}{X_L}$

Variation of $X_L$ with Frequency

Since $X_L=2\pi fL$, the graph of $X_L$ vs $f$ is a straight line through the origin with slope $2\pi L$.

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Inductor as a Choke

An inductor used to block or reduce AC while allowing DC to pass is called a choke.

Why it works:

• For DC: $f=0\Rightarrow X_L=0$ → inductor offers no opposition; DC passes freely.
• For AC: $X_L=2\pi fL$ → higher frequency means higher opposition.
• A choke does not dissipate energy as heat (unlike a resistor); energy is stored in the magnetic field and returned to the circuit.

Applications of chokes:

• Smoothing pulsating DC in power supplies (in series with the load)
• Tuning circuits in radio receivers
• Fluorescent lamp ballasts

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Power in a Purely Inductive Circuit

The instantaneous power is:

$P=VI=V_o\sin (\omega t)\cdot I_o\sin \left(\omega t-\frac{\pi }{2}\right)$

Using the identity $\sin (\omega t-\pi /2)=-\cos (\omega t)$:

$P=-V_o{I}_o\sin (\omega t)\cos (\omega t)=-\frac{V_o{I}_o}{2}\sin (2\omega t)$

The average power over a complete cycle is:

$\boxed{P_{avg}=V_{rms}{I}_{rms}\cos (90^{\circ })=0}$

The power factor $\cos \theta =\cos (90^{\circ })=0$. Energy is alternately stored in the magnetic field and returned to the source — no net energy is dissipated.

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Summary Table