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A pure inductance is connected through alternating voltage circuit. Find out the value of current, phase difference, reactance and average consumed energy rate in the circuit. Draw phasor diagram also.

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A.C. Voltage Applied on an Inductor:

Let a source of alternating emf be connected across a inductor. The instantaneous value of alternating emf is given by
V = Vm sin ωt …………….. (1)
Here we have assumed that this inductor has negligible resistance though practically its windings have appreciable resistance.
If I is the instantaneous current through L at any time t, then self-induced Faraday emf in the inductor is
ε = \(-\frac{L d I}{d t}\)
Here L is the self inductance of the inductor and the -ve sign shows that induced emf opposes the applied emf (Lenz’s law).

The applied voltage V must be equal and opposite to the induced voltage ε to maintain the flow of current in the circuit.

This equation tells us that the current as a function of time, must be such that the quantity \(\frac{d I}{d t}\) (slope) is a sinusoidally varying quantity, having an amplitude \(\frac{V_{m}}{L}\)and in phase with the source voltage.
The total current flow in the circuit, equation 2 is integrate both the sides

The integration constant here has dimensions of current and does not depend on time. As the source has an emf oscillating symmetrically about zero, the current it sustains should also oscillate symmetrically about zero, so constant exists. Hence, the integration constant is zero.

Comparing equations (1) and (2), it is clear that in an a.c. circuit containing inductance only, current lags behind the voltage by a phase angle of π/ 2 or one-quarter (1/ 4) cycle. This is shown in figure and the corresponding phasor diagram is shown in figure.

As it is clear from the figure that current phasor I is \(\frac{\pi}{2}\) behind the voltage phasor V. When these phasors are rotated in the counter-clockwise direction, they generate the curves given in figure and by

equations (1) and (3). Comparing equation Im = \(\frac{V_{m}}{\omega L}\) with Ohm’s law, we find out that the quantity ωL plays the same role here in the circuit as is done by the resistance in the purely resistive circuit. Thus, it is the effective resistance offered by the inductance L and is called inductive reactance. It is denoted by XL.

Thus, XL = ωL = 2πfL …………… (4)
Where f is the frequency of the a.c. supply. It is obvious from equation (4) that inductive reactance is not constant but increases with increase in the frequency of the a.c. supply. In a d.c. circuit, f = 0 and as such XL = 0, thus a pure inductance offers zero resistance in a d.c. circuit. The inductive reactance is measured in ohms (ω).

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