Electronics - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 11)
11.
How long will it take the capacitor in the given circuit to discharge?
16.4 
s
s32.8 s
s65.6 s
s82 s
sDiscussion:
5 comments Page 1 of 1.
Priya said:
1 decade ago
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.
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.
Vasu said:
1 decade ago
I dint get last two steps. Y are you multiplying toff*5 ?
Whats the correct formula ? anyone please explain ?
Whats the correct formula ? anyone please explain ?
(1)
Saheed said:
1 decade ago
5 is the Toff period which will be obtained by .0000164/32.8 = 5.
5 times the RC constant = Charging time.
5 times the RC constant = Charging time.
Phaneendra said:
1 decade ago
Solution:
Ton+Toff=T
f=(1/RC)=6090...
1/f=Toff
Toff*5=82microsec.
Ton+Toff=T
f=(1/RC)=6090...
1/f=Toff
Toff*5=82microsec.
Maneet said:
1 decade ago
How get time constant = 5?
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