# RL Transient analysis

#### shilpZM

Joined Apr 28, 2021
9

If the switch is open before t=0 for long time, find value of iL at t=125ms.

#### ericgibbs

Joined Jan 29, 2010
13,585
hi,
E

#### shilpZM

Joined Apr 28, 2021
9
hi,
E
is this right???I am not sure..

#### ericgibbs

Joined Jan 29, 2010
13,585
hi,
That looks like a clip from a electronics text book, why have you posted that and not your actual calculations.?

E

#### shilpZM

Joined Apr 28, 2021
9
hi,
That looks like a clip from a electronics text book, why have you posted that and not your actual calculations.?

E
This is my calculation.I made this in wordpad,so it will be easy to post

#### shilpZM

Joined Apr 28, 2021
9
This is my calculation.I made this in wordpad,so it will be easy to post

Here first I edited in ppt and then pasted in word

#### ericgibbs

Joined Jan 29, 2010
13,585
hi,
OK, understood.
E

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#### shilpZM

Joined Apr 28, 2021
9
so is this wrong???

#### ericgibbs

Joined Jan 29, 2010
13,585
hi ZM,
Is IL(t=0) really 0Amps.?
There is a steady current thru the 40R and 20R||20R resistances, what do you calculate that to be.?
E

Another point: why have you chosen 400mSec for 'T'

Update:
Ref the Rth value.
Consider when the switch is Closed, the 80V is applied across the middle 20 ohm resistor, so how can Rth be 20||20.?

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#### shilpZM

Joined Apr 28, 2021
9
is this right???

#### ericgibbs

Joined Jan 29, 2010
13,585
hi ZM,
The questions asks for the current thru the 4H inductor.

What is the Inductor current before the switch is Closed.??
[Note this moves the Base line of the current Up above 0Amps before the switch is Closed.]

When the switch is Closed what will be the Time Constant and the Transient time value of the series R L circuit.?

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#### MrAl

Joined Jun 17, 2014
8,473
It looks like your answer is 2.3 amps but there is another character between the '3' and the 'A' for amps that i cant quite make out. But 2.3 amps is correct rounded to two significant figures, although we usually strive for a few more digits or at least one more. The more exact result is slightly less than 2.3 amps.

#### ericgibbs

Joined Jan 29, 2010
13,585
hi ZM,
Look at this simpler method.
E

I would solve the initial <t0 IL current this way IL= [80V*(10R/50R)]/20R = 0.8Amps
also IL max = 80R/20R = 4Amps

then @t125m+

IL = 4A +( 0 - (4A-0.8A)*e^ -tc) ; where tc = 0.125 * 5 = 0.625
So
IL = 4A + ( -3.2*e^-0.625)
IL =4A + ( -3.2 *0.5352)
IL = 4A - 1.7126A = 2.28A

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