Given The switch which has been open for a very long time is

Given: The switch, which has been open for a very long time, is closed at t = 0.

Required: Determine the initial inductor current, iLo, the steady-state inductor current, iL?, the time constant, ?, in effect for t > 0 and the value of the inductor current at t = 101.25 ?s.

Solution:

iLo = mA

iL? = mA

? = ?s

iL(101.25 ?s) = mA

450 mH IL t=0 120 V 20 k 20 k2

Solution

t<0

Req = 5+(20/2) = 15k ohm

Supply current = 120/15k

= 8 mA

I(0) = 8x 20/(20+20)

= 4mA.

Steady state current = I (oo)

Two 20k ohm resistors are parallel with short circuit.

I(oo) = 8mA.

Time constant T= L/R

= 450m / 4k

= 112.5 micro sec.

I (t) = I(oo)+ {I(0) - I(oo) } exp (-t/T)

= 8 + ( 4- 8) exp ( -t/ 112.5 micro)

= 8- 4 exp (-t / 112.5 micro)

At t= 101.25 micro sec

I(101.25 micro sec) = 8 - 4 exp ( - 101.25/112.5)

= 6.36 mA