**4.2. A.C. Through Pure Inductance Alone**

Fig. 24 (a) shows the circuit containing a pure inductance of L henry.

Let the alternating voltage applied across the circuit be given by the equation,

v = V max sin rot … (i)

Whenever an alternating voltage is applied to a purely inductive coil, a back e.m.f. is produced due to the self-inductance of the coil. This back e.m.f. opposes the rise or fall of the current through the coil. Since there is no ohmic drop in this case, therefore, the applied voltage has to overcome this induced e.m.f. only. Thus at every step,

Fig. 24 (b), (c). A. C. through pure inductance alone. Resultant power zero.

u = L di / dt

V _{max} sin wt = L di / dt

di = V _{max} / L sin wt dt

Integrating both sides, we get

ʃ di = ʃ V _{max} sin wt dt

i = V _{max} / L ( – cos wt / w) = V _{max} / wL sin [ wt – π / 2 ]

i = V _{max} / X_{L} sin [ wt – π /2 ]

where X_{L} = wL (opposition offered to the flow of alternating current by a pure inductances) and is called Inductive reactance. It is given in ohms if L is in henry and w is in radian/second.

The value of current will be maximum when sin ( wt – π / 2) = 1

I _{max} = V _{max} / X_{L}

** Power.** Refer Fig. 24 (c)

Instantaneous power, P = ui=V _{max} sin wt x I _{max} sin ( wt – π / 2)

= – V _{max} I _{max} sin wt . cos wt

= – V _{max} I _{max} / 2 x 2 sin wt cos wt

= – V _{max} / √ 2 . I _{max} / √ 2 . sin 2 wt

Power for the whole cycle, P = – V _{max} / √ 2 . I _{max} / √ 2 ʃ^{2x} _{0} sin 2wt = 0

Hence average power consumed in a pure inductive circuit is zero.

Hence in a pure inductive circuit, we have :

1. current I = V / XL = V / wL = V / 2πfL amp.

2. Current always lags behind the voltage by 90°.

3. Average power consumed is zero.

** **

**Variation of XL and f:**

Since XL = wL = 2πfL, and here if L is constant, then

Fig. 25, shows the variation. As frequency is increased XL increases and the current taken by the circuit decreases.