Electronics - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 1)
1.
What voltage will the capacitor charge up to in the given circuit for the single input pulse shown?
Discussion:
17 comments Page 1 of 2.
Syed Shakeeb said:
3 years ago
Vin = 5v.
t = 32.8 * 10^-6.
RC=16.4 * 10^-6.
VC=vin(1-e^-t/RC),
5 * (1 - e(-32.8 * 10-6 ÷ 16.4 *10-6)).
Answer = 4.3V.
t = 32.8 * 10^-6.
RC=16.4 * 10^-6.
VC=vin(1-e^-t/RC),
5 * (1 - e(-32.8 * 10-6 ÷ 16.4 *10-6)).
Answer = 4.3V.
(1)
Arun said:
1 decade ago
@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.
So output voltage is:
Vo = Vin (1-e^(-t/RC)).
T = 32.8 us.
R = 82 k.
C = 200 PF.
Elkaanutil said:
4 years ago
1 trc = (1-e^-1) = 0.6321 or 63.21%,
2 trc = (1-e^-2) = 0.8647 or 86.47 %,
Thus,
n trc = (1-e^-n).
2 trc = (1-e^-2) = 0.8647 or 86.47 %,
Thus,
n trc = (1-e^-n).
Asha said:
8 years ago
Thank you @Gaurav.
Kanha said:
8 years ago
(1-e^(-t/RC))=0.63 for t=RC and it is 0.86 for t = 2RC.
Seema said:
8 years ago
Yeah! Please explain about 1trc and 2trc calculation.
Prabhu said:
9 years ago
@Gaurav.
Please explain about the 1 trc and 2 Trc calculation.
Please explain about the 1 trc and 2 Trc calculation.
Padma said:
9 years ago
Yeah! @Gaurav is absolutely right.
Gaurav said:
10 years ago
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.
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.
Phaneendra said:
1 decade ago
f=1/RC
Toff=1/f
Ton+Toff=T;32.4+16.2=48.6
Ton/T=Von(drop)
beacause we hav calucalate voltge during on state .
V-Von=4.3v
5-0.66=4.3v
Toff=1/f
Ton+Toff=T;32.4+16.2=48.6
Ton/T=Von(drop)
beacause we hav calucalate voltge during on state .
V-Von=4.3v
5-0.66=4.3v
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