Electronics - Inductors - Discussion

57. 

What is the current rise after 2 time constants if Vs = 12 V dc, R = 60 omega.gif, and the inductor is rated at 24 mH?

[A]. 79.9 mA
[B]. 126.4 mA
[C]. 173.0 mA
[D]. 198.6 mA

Answer: Option C

Explanation:

No answer description available for this question.

Chaitali Acharya said: (Dec 2, 2011)  
Please could anyone give a formula for this question?

Prasanna said: (Aug 20, 2012)  
R + jXL= sqrt( 60^2 + 24^2 )
I = Vs/(R+ jXL)
2t=.862

I at 2t is = I*0.862

Sajid said: (Oct 13, 2012)  
I = It*(e^(-2))

It=Vs/Z

Z=(R^2+Xl^2)^1/2
Xl=2*3.14*f*L /// f=0,Since DC ///

Z = R

It = Vs/R
=12/60
=.2

Now,
I = .2(e^(-2))
= .1729
=173mA

Sanjoy said: (Feb 29, 2016)  
Current rise after 2 time constant.

So, IL = I(1-e^-2).

IL = 0.2 (1-0.13) = 173 mA.

Sivapramod said: (Jul 17, 2016)  
What is the value of 'e' in the given formula?

E.Sae said: (Dec 25, 2016)  
@Sivapramod.

e means exponential

E.Sae said: (Dec 25, 2016)  
@Sivapramod.

e means exponential

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