Electrical Engineering - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 5)
5.
When a 12 V input pulse with a width equal to one time constant is applied to an RC integrator, the capacitor charges to
Discussion:
10 comments Page 1 of 1.
SANJAY KUMAR said:
5 years ago
Time constant.
1) 68.8%
2) 86.5%
3) 96 %
4) 98 %
5) 100%
1) 68.8%
2) 86.5%
3) 96 %
4) 98 %
5) 100%
Abbas Ramadan said:
6 years ago
c = v * (1_e^-t/Rc)
t = rc,
c = 12 * (1_e^-1)
Then, c = 7.57.
t = rc,
c = 12 * (1_e^-1)
Then, c = 7.57.
MANAPPA said:
7 years ago
How to find the charging time? Please tell me.
Jnan said:
9 years ago
@Dheeraj.
If the option will be 9 point something, is it possible that the answer will be near equal to supply voltage?
If the option will be 9 point something, is it possible that the answer will be near equal to supply voltage?
Waqas habib said:
1 decade ago
@Bimba.
When the input pulse width and time constant are equal, then how capacitor charges to 63.8%?
Please explain.
When the input pulse width and time constant are equal, then how capacitor charges to 63.8%?
Please explain.
Avinash chandra manish said:
1 decade ago
capa volt=v*(1-exp(-t/rc)
v=12*(1-exp(-rc/rc)
v=12*(1-exp(-1))=7.58
v=12*(1-exp(-rc/rc)
v=12*(1-exp(-1))=7.58
Dheeraj said:
1 decade ago
Simply take nearest value of applied voltage.
ANI said:
1 decade ago
It is the charging time.
Amit said:
1 decade ago
Where did you get 63.8%?
Bimba said:
1 decade ago
As we are giving 12V supply,and our integrator is having a time constant of RC. thus, the capacitor will be charged to 63.8% of the applied supply voltage. hence its value will be equal to 12*63.8%=7.56V.
Hope my analysis is right.
Hope my analysis is right.
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