# 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
0 V
12 V
6.3 V
7.56 V
Explanation:
No answer description is available. Let's discuss.
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%

Abbas Ramadan said:   6 years ago
c = v * (1_e^-t/Rc)
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?

Waqas habib said:   1 decade ago
@Bimba.

When the input pulse width and time constant are equal, then how capacitor charges to 63.8%?

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

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.