Alternating Voltage Applied to an Inductor :
An alternating source is shown connected to an ideal inductor of inductance (L). Such a circuit is known as purely inductive circuit.
The alternating voltage across the inductor is given by
$V = V_0 \sin \omega t \quad \text{…(i)}$
The induced e.m.f. across the inductor = $- L \frac{dI}{dt}$
which opposes the growth of current in the circuit.
As there is no potential drop across the circuit, so,
$V + \left(-L \frac{dI}{dt}\right) = 0$
$L \frac{dI}{dt} = V$
$\frac{dI}{dt} = \frac{V}{L}$
Using eqn. (i), we get,
$\frac{dI}{dt} = \frac{V_0}{L} \sin \omega t$
$dI = \frac{V_0}{L} \sin \omega t \, dt \quad \text{…(ii)}$
Integrating both sides, we get,
$\int dI = \int \frac{V_0}{L} \sin \omega t \, dt = \frac{V_0}{L} \int \sin \omega t \, dt$
or
$I = \frac{V_0}{L} \left(-\frac{\cos \omega t}{\omega}\right) = \frac{V_0}{L \omega} (-\cos \omega t)$
Since
$(-\cos \omega t) = \sin \left(\omega t - \frac{\pi}{2}\right)$
and peak value
$I_0 = \frac{V_0}{L \omega}$
$I = I_0 \sin \left(\omega t - \frac{\pi}{2}\right) \quad \text{…(iii)}$
Comparison of eqns. (i) and (iii) shows that in this case the current lags behind the voltage by an angle of π/2.
Circuit diagram showing an a.c. source \( V = V_0 \sin \omega t \) connected to an inductor \( L \) with induced e.m.f. \( -L \frac{dI}{dt} \).
Time Diagram showing \( V = V_0 \sin \omega t \) and \( I = I_0 \sin (\omega t - \frac{\pi}{2}) \).
Phasor Diagram showing voltage and current vectors with a 90° phase difference.
Inductive Reactance $(X_{L})$ :
The inductive reactance is the effective opposition (resistance) offered by the inductor to the flow of current in the circuit. In other words, inductive reactance plays the same role in a.c. circuit as the resistance plays in d.c. circuit.
The peak value of alternating current flowing through an inductor is given by
$I_0 = \frac{V_0}{L \omega} \quad \text{…(1)}$
As per Ohm’s law,
$I_0 = \frac{V_0}{R}$
From equations (1) and (2), we find that the term \(Lω\) has the dimensions of resistance \(R\). The term \(Lω\) is known as inductive reactance (represented by \(X_L\)).
Thus,
$X_L = Lω = L \cdot 2π \nu$
$\omega = 2π \nu$
Power Supplied to an Inductor :
When alternating voltage,
$V = V_0 \sin ωt$
is applied across the inductor, the alternating current flowing through the inductor is:
$I = I_0 \sin \left(ωt - \frac{π}{2}\right)$
Instantaneous power supplied to inductor:
$P_L = VI = (V_0 \sin ωt) \cdot \left[I_0 \sin \left(ωt - \frac{π}{2}\right)\right]$
Since
$\sin(\omega t - \frac{\pi}{2}) = -\cos \omega t$
$P_L = -V_0 I_0 \cos(\omega t) \sin(\omega t)$ $P_L = -\frac{V_0 I_0}{2} \sin(2\omega t)$
(∵ \(2 \sin \omega t \cos \alpha t = \sin 2\omega t\))
Average power over a full cycle is given by
$\overline{P_L} = -\frac{V_0 I_0}{2} \text{average of } (\sin 2\omega t)$
$= -\frac{V_0 I_0}{2} \langle \sin 2\omega t \rangle$
Now, average of (\sin 2\omega t\) for full cycle is given by,
$\langle \sin 2\omega t \rangle = \frac{1}{T} \int_0^T \sin 2\omega t \, dt$
$= \frac{1}{T} \left[ \frac{\cos 2\omega t}{2\omega} \right]_0^T$
$= \frac{1}{2\omega T} [\cos 2\omega T - \cos 0]$
$= \frac{1}{2\omega T} [\cos \frac{4\pi}{T} \times T - \cos 0]$
$= \frac{1}{2\omega T} [1 - 1] = 0$
$\overline{P_L} = 0$
Thus,
Therefore, average power supplied to an inductor over a complete cycle is zero.
In other words, energy stored by the inductor in one quarter period \(\left(\frac{T}{4}\right)\) is returned to the source in the next quarter period \(\left(\frac{T}{4}\right)\).