The Pure Inductive Circuit
Consider an ac source connected to an inductor of self-inductance $L$. Real inductors have appreciable resistance in their windings, but here the winding resistance is taken as negligible, so the circuit is purely inductive. Let the source voltage be $v = v_m \sin\omega t$.
Applying Kirchhoff's loop rule, the only voltage drop is the self-induced Faraday emf $-L\,\dfrac{di}{dt}$, where the negative sign follows from Lenz's law. Setting the algebraic sum of emfs to zero gives the governing equation of the circuit:
$$ v_m \sin\omega t = L\,\frac{di}{dt} $$This says the slope $di/dt$ is a sinusoid in phase with the source voltage, with amplitude $v_m/L$. Integrating with respect to time, and noting that a symmetric source cannot sustain any constant current component (so the integration constant is zero), yields the current.
From $\dfrac{di}{dt} = \dfrac{v_m}{L}\sin\omega t$, integrate to find $i(t)$.
$i = -\dfrac{v_m}{\omega L}\cos\omega t$. Using $-\cos\omega t = \sin\!\left(\omega t - \dfrac{\pi}{2}\right)$, this becomes
$$ i = i_m \sin\!\left(\omega t - \frac{\pi}{2}\right), \qquad i_m = \frac{v_m}{\omega L} $$The current amplitude is $i_m = v_m/(\omega L)$, and the current is shifted $\pi/2$ behind the voltage.
Inductive Reactance $X_L$
The quantity $\omega L$ in the current amplitude plays exactly the role that resistance plays in a resistive circuit. It is called the inductive reactance and is denoted $X_L$:
$$ X_L = \omega L = 2\pi f L, \qquad i_m = \frac{v_m}{X_L} $$Inductive reactance has the same dimensions as resistance, and its SI unit is the ohm ($\Omega$). It limits the current in a purely inductive circuit just as resistance limits the current in a purely resistive circuit. According to NCERT §7.4, the inductive reactance is directly proportional both to the inductance $L$ and to the frequency $f$ of the source.
| Quantity | Symbol | Relation | SI Unit |
|---|---|---|---|
| Inductive reactance | X_L | ωL = 2πfL | ohm (Ω) |
| Current amplitude | i_m | v_m / X_L | ampere (A) |
| Phase of current | φ | lags voltage by π/2 | radian |
| Average power | P | 0 (over one cycle) | watt (W) |
Why Current Lags Voltage
Comparing $v = v_m \sin\omega t$ with $i = i_m \sin(\omega t - \pi/2)$ shows that the current is $\pi/2$ radians (one-quarter cycle) behind the voltage. The current reaches its maximum value later than the voltage by one-fourth of a period, $T/4 = (\pi/2)/\omega$.
The physical cause is Lenz's law. The self-induced back-emf opposes the change in current, so the current cannot rise instantaneously when the voltage rises; it is dragged behind by 90°. NIOS §19.3 reaches the same conclusion, noting that the potential difference peaks one-quarter cycle before the current. This is the exact opposite of a capacitor, in which the current leads the voltage.
"ELI" — and the three things examiners flip
Remember the mnemonic ELI: the voltage E leads the current I in an inductor L. Equivalently, current lags voltage by 90°. Distractor options routinely flip three facts about a pure inductor.
Phase: current LAGS voltage by $\pi/2$ (it does not lead, and is not in phase).
Frequency: $X_L = 2\pi f L$ INCREASES with frequency (capacitive reactance decreases).
Power: a pure inductor consumes ZERO average power over a cycle.
The Phasor Diagram
Voltage and current can be represented as phasors rotating counter-clockwise with angular speed $\omega$; their vertical projections trace the instantaneous values. For an inductor, the current phasor $I$ sits $\pi/2$ behind the voltage phasor $V$.
Plotted against time, the two sinusoids confirm the lag: the voltage curve peaks first, and the current curve reaches the same crest a quarter period later.
New to rotating-vector representation? Start with Phasors to see how amplitude and phase combine before tackling reactance.
Reactance Versus Frequency
Because $X_L = 2\pi f L$, inductive reactance is a straight line through the origin when plotted against frequency: it increases as frequency increases. As $f \to 0$ (a dc source), $X_L \to 0$, so a pure inductor offers no opposition to dc and behaves like a plain conducting wire. NIOS §19.3 stresses that this is consistent with the behaviour of an inductor connected to a battery, and that it is the exact opposite of capacitive reactance, which blows up at low frequency.
Average Power Over a Cycle
The instantaneous power delivered to the inductor is the product of instantaneous voltage and current:
$$ p_L = i\,v = i_m\sin\!\left(\omega t - \tfrac{\pi}{2}\right)\, v_m\sin\omega t = -\frac{i_m v_m}{2}\,\sin 2\omega t $$Averaging over one complete cycle, the mean of $\sin 2\omega t$ is zero, so the average power is zero. Energy flows from the source into the inductor's magnetic field during one quarter-cycle and returns fully to the source during the next. A pure inductor therefore stores and releases energy but dissipates none.
$$ \langle P \rangle = -\frac{i_m v_m}{2}\,\overline{\sin 2\omega t} = 0 $$Worked Example
A pure inductor of $25.0\,\text{mH}$ is connected to a source of $220\,\text{V}$. Find the inductive reactance and the rms current if the source frequency is $50\,\text{Hz}$.
Inductive reactance:
$$ X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 25.0\times10^{-3} = 7.85\ \Omega $$rms current:
$$ I = \frac{V}{X_L} = \frac{220}{7.85} = 28.0\ \text{A} $$The reactance limits the current to $28.0\,\text{A}$, and despite this large current the inductor draws no net power from the source.
One-glance summary
- For $v = v_m\sin\omega t$, the current is $i = i_m\sin(\omega t - \pi/2)$ — current LAGS voltage by $\pi/2$.
- Inductive reactance $X_L = \omega L = 2\pi f L$, measured in ohm; current amplitude $i_m = v_m/X_L$.
- $X_L$ is proportional to both $L$ and $f$; it rises linearly with frequency and is zero for dc.
- Average power over one full cycle is exactly zero — a pure inductor is a non-dissipative element.
- Mnemonic ELI: voltage E leads current I in an inductor L.