Electrical Engineering - Inductors - Discussion

In a RL circuit V=v(1-e^-t*R/L) as t = 2*L/R

So e^-2*L/R*R/L would be e^-2 as a result V = .86*v.

So current i =V/R i.e. 86*20/220 = 78 ma(approx.)

Please explain that how l/r term comes?

TIME CONSTANT AND CURRENT RELATIONSHIP:

0 0% of max current.
1Ts 63.2% of max current.
2Ts 86.5% of max current.
3Ts 95% of max current.
4Ts 98.2% of max current.
5Ts 100% of max current.

First find the maximum possible current through the circuit. Inductor obeys completely a short circuit on DC(ignoring coil resistance). So the impedance of the circuit is only R.

Max current is I = V/R. ie,. 20/220 = 0.090909.

Question is asked for the two time constant that is '2Ts'..

So the Equivalent current is 86.5% of max current(0.090909) from the above data = (86.5/100)x0.090909 =0.0786363 or 78mA.

How comes time constant and current relationship?

RL circuit V=v(1-e^-t*R/L).
as t = 2*L/R.

So e^-2*L/R*R/L.
e^-2.
as a result V = .86*v.

So current I =V/R i.e. 86*20/220 = 78 mA(approx.).

While calculating current, why inductive reactance doesn't come in picture?

@Roopesh.

Because it is battery, a DC source. So Xl won't come into a picture.

Why taking t=2L/R?
This time constant is for series RLC circuit.
And there is no capacitor in the circuit. This circuit is RL circuit.for RL circuit time constant is t=L/R.

Yes, you gave the right solution, Thanks @Giri.

In RL circuit V=v(1-e^R/L t)
Given time constant L/R =2
V=20*(1-e^-2)
=17.29v.

Current through inductor is I = V/R
=17.29/220=78mA.