Electronics - Time Response of Reactive Circuits - Discussion

11. 

How long will it take the capacitor in the given circuit to discharge?

[A]. 16.4 mu.gifs
[B]. 32.8 mu.gifs
[C]. 65.6 mu.gifs
[D]. 82 mu.gifs

Answer: Option D

Explanation:

No answer description available for this question.

Phaneendra said: (Feb 14, 2011)  
Solution:
Ton+Toff=T
f=(1/RC)=6090...
1/f=Toff
Toff*5=82microsec.

Vasu said: (Oct 14, 2011)  
I dint get last two steps. Y are you multiplying toff*5 ?

Whats the correct formula ? anyone please explain ?

Priya said: (Nov 22, 2011)  
T=R*C
T=82K*200PF=0.0000164

The capacitor will take 5 time constant for fully discharge.
So, 0.0000164*5=0.000082 Sec is equal to 82 µs.

Maneet said: (Oct 5, 2013)  
How get time constant = 5?

Saheed said: (Dec 4, 2014)  
5 is the Toff period which will be obtained by .0000164/32.8 = 5.

5 times the RC constant = Charging time.

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