Discussion :: Time Response of Reactive Circuits - General Questions (Q.No.1)
What voltage will the capacitor charge up to in the given circuit for the single input pulse shown?
Answer: Option B
No answer description available for this question.
|Phaneendra said: (Feb 14, 2011)|
beacause we hav calucalate voltge during on state .
|Elkaa said: (Mar 26, 2011)|
If the timeconstant = RC is 16.4 usec, and the pulse is 32.4 usec, then I would expect that the capacitor charges up to Vin, as it has a lot more time then RC.
|Sreekumar said: (Jul 16, 2011)|
|I agree with Elkaa, it will charge upto Vin.|
|Nitesh said: (Sep 13, 2011)|
|Vc(voltage across capacitor) = V0(1-e-t/to)|
|Vasu said: (Oct 13, 2011)|
|Can anyone explain this in brief please.|
|Ganesh said: (Apr 21, 2012)|
|I agree with elka.|
|Mehta said: (Apr 25, 2012)|
|Elkaa is totally wrong. It takes 5 timeconstant to charge upto Vin.
So it won't charge upto Vin.
Nitesh is right. In that equation substitute t = 32.8us and to = RC = 16.4us.
|Leena said: (Jun 22, 2013)|
|Writing the kirchoff's current law you get Vin-iR-(1/C) inetgral (idt).
On applying the laplace transform we get the equation,
On substituting we get 4.3.
Where to = 1/RC.
Vo = 5.
|Arun said: (Oct 29, 2013)|
|@Mehta is right. Given circuit is a low pass filter circuit.
So output voltage is:
Vo = Vin (1-e^(-t/RC)).
T = 32.8 us.
R = 82 k.
C = 200 PF.
|Gaurav said: (Jan 16, 2016)|
|T constant = rc = 82 k x 200 pf = 16.4 micro sec.
Plus time = 32.8 micro sec.
Trc = 32.8/16.4 = 2.
1 Trc = 63.3%.
2 Trc = 86%.
So, output voltage = 5x(86/100) = 4.3 v.
|Padma said: (Aug 31, 2016)|
|Yeah! @Gaurav is absolutely right.|
|Prabhu said: (Nov 24, 2016)|
Please explain about the 1 trc and 2 Trc calculation.
|Seema said: (May 7, 2017)|
|Yeah! Please explain about 1trc and 2trc calculation.|
|Kanha said: (Jul 7, 2017)|
|(1-e^(-t/RC))=0.63 for t=RC and it is 0.86 for t = 2RC.|
|Asha said: (Jul 20, 2017)|
|Thank you @Gaurav.|
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