Electrical Engineering - Inductors - Discussion

Discussion :: Inductors - General Questions (Q.No.19)


A 320 H coil is in series with a 3.3 k resistor. How long does it take for current to build up to its full value?

[A]. 0.48 s
[B]. 0.48 ms
[C]. 0.48 s
[D]. 48 s

Answer: Option C


No answer description available for this question.

Kirthikumar Rao said: (Nov 23, 2011)  
time constant T=(L/R)
to reach max current it will take appr 5T

Feliciano said: (Apr 25, 2012)  
Time constant for an RL circuit is not R*L! It is instead L/R, which makes Time Constant equal to 5* (320u/3. 3K) = 0.4848.

Rrgiri said: (Mar 1, 2014)  
How it s possible ? @Vikas time constant is R/L or R*L ?

Rajat said: (Mar 4, 2014)  
What is time constant?

Ganesh said: (Apr 26, 2014)  
Time constant for RL circuit is L/R SO time constant = (320*10^(-6)/3.3*10^(3)) time taken to reach final value is approximately 5T = 0.48*10^(-6).

Ekanath Methre said: (Apr 30, 2015)  
How did you took 5T? Please explain it.

Tejas Gouda said: (May 6, 2015)  
Why we need to multiply it by 5? How the time taken will be approximately 5T? Why can't it be 10T? Time constant means L/R not LR.

11Kv said: (Jun 12, 2015)  
Time constant for inductor is L/R.

Roopesh D M said: (Jun 18, 2015)  
The time required for the current flowing in the LR series circuit to reach its maximum steady state value is equivalent to about 5 time constants or 5T.

This time constant T, is measured by T = L/R, in seconds, were are is the value of the resistor in ohms and L is the value of the inductor in Henries.

Stuti Kushwaha said: (Feb 15, 2016)  
Time constant of LR circuit is L/R. So here after multiplying by 5 we get answer is 0.484 micros.

Shoaib said: (Oct 28, 2016)  

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.

So, first we know that the maximum possible current through the circuit will be full, means the 100% of max current and the time constant will be 5Ts so,

Time constant => T= (L/R) = (320u/3.3k).
When it reaches max current(100%) it will take 5Ts.
So 5Ts = 5(320u/3.3k),

Dhaval said: (May 4, 2017)  
Thanks @Shoaib.

Rehman said: (Dec 5, 2019)  
Good @Shoaib.

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