Home | | Physics 12th Std | AC circuit containing only an inductor

Phasor diagram, Circuit Diagram, Formula, Solved Example Problems | Alternating Current (AC) - AC circuit containing only an inductor | 12th Physics : Electromagnetic Induction and Alternating Current

Chapter: 12th Physics : Electromagnetic Induction and Alternating Current

AC circuit containing only an inductor

Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source .

AC circuit containing only an inductor

Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source (Figure 4.47). The alternating voltage is given by the equation.


The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by


By applying Kirchoff’s loop rule to the purely inductive circuit, we get




The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.


where Vm / ωL =Im , the peak value of the  alternating current in the circuit. From equation (4.46) and (4.47), it is evident that current lags behind the applied voltage by Ï€/2 in an inductive circuit. This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90º (Figure 4.48).



Inductive reactance XL

The peak value of current Im is given by Im = Vm / Ï‰. Let us compare this equation with Im = Vm / R from resistive circuit. The quantity Ï‰L plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (XL). It is measured in ohm.

L = Ï‰L

The inductive reactance (XL) varies directly as the frequency.


where f is the frequency of the alternating current. For a steady current, f = 0. Therefore, XL = 0. Thus an ideal inductor offers no resistance to steady DC current.


EXAMPLE 4.20

A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.

Solution

L = 400 x 10-3 H; Ieff = 6 x 10-3A

f = 1000 Hz



Tags : Phasor diagram, Circuit Diagram, Formula, Solved Example Problems | Alternating Current (AC) , 12th Physics : Electromagnetic Induction and Alternating Current
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Physics : Electromagnetic Induction and Alternating Current : AC circuit containing only an inductor | Phasor diagram, Circuit Diagram, Formula, Solved Example Problems | Alternating Current (AC)

Related Topics

12th Physics : Electromagnetic Induction and Alternating Current


Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.