Electrical Engineering - Time Response of Reactive Circuits - Discussion
Discussion Forum : Time Response of Reactive Circuits - General Questions (Q.No. 1)
1.
When a 15 V input pulse with a width equal to two time constants is applied to an RC integrator, the capacitor charges to
Discussion:
10 comments Page 1 of 1.
Arvind said:
7 years ago
@Yogesh you are correct.
There have another method. But it's little difficult,
1st time constant = 15*63.2/100=9.48
Remaining voltage is 15-9.48 = 5.52
So, 5.52*63.2/100=3.48.
1st time constant+next time constant=9.48+3.488=12.96.
There have another method. But it's little difficult,
1st time constant = 15*63.2/100=9.48
Remaining voltage is 15-9.48 = 5.52
So, 5.52*63.2/100=3.48.
1st time constant+next time constant=9.48+3.488=12.96.
(3)
Salma hisham said:
1 decade ago
Voltage at Capacitor when charging = V(total) * [1-exp(-t/RC)].
Width = two time constant ====> t = 2RC.
V = V(total) * [ 1 - exp (-2) ].
V = 15 * ( 1 - 0.135 ).
V = 15 * 0.865 = 12.97 V.
Width = two time constant ====> t = 2RC.
V = V(total) * [ 1 - exp (-2) ].
V = 15 * ( 1 - 0.135 ).
V = 15 * 0.865 = 12.97 V.
Yogesh S said:
9 years ago
We know,
1 time constant, its 0.632 of max.
2 time constants, its 0.865 of max.
3 time constants, its 95 of max, so on.
Thus, here @ V = 15.
Its 0.865 * 15 = 12.9 (Answer).
1 time constant, its 0.632 of max.
2 time constants, its 0.865 of max.
3 time constants, its 95 of max, so on.
Thus, here @ V = 15.
Its 0.865 * 15 = 12.9 (Answer).
(2)
Omkar L Karande said:
1 decade ago
Simple thing is for 1 times of TAU it should charge to 63%.
Obviously cap charge to 15V and beyond it.
So A D are neglected..
8.6 is less, not even 63% of 15.
So ans is B.
Obviously cap charge to 15V and beyond it.
So A D are neglected..
8.6 is less, not even 63% of 15.
So ans is B.
Vinay said:
1 decade ago
The capacitor does not charges full voltage that will appox 14. 1.
That all due to some losses the actual input -15 so below 15 is 12. 9v is the answer.
That all due to some losses the actual input -15 so below 15 is 12. 9v is the answer.
Dhanunjay said:
1 decade ago
RC integrator response os given by V(1-exp(-t/RC))
Here t= 2RC so answer is 15(1-exp2) = 12.9
Here t= 2RC so answer is 15(1-exp2) = 12.9
(2)
Netram meena said:
9 years ago
V = Vo(1-exp (-t/RC)).
So here,
Vo = 15.
And,
T = 2 RC.
Then we get,
V = 12.9 V.
So here,
Vo = 15.
And,
T = 2 RC.
Then we get,
V = 12.9 V.
Ujjwal said:
1 decade ago
Vc(t) = V(infinity)-[V(infinity)-V(0)]e^(-t/RC).
Vc(t) = 15(1-e^(-2)) = 12.9.
Vc(t) = 15(1-e^(-2)) = 12.9.
(1)
Mohan bhosale said:
8 years ago
V = Vo(1-exp (-t/RC)).
Given
Vo = 15.
T = 2 RC.
Therefore, V = 12.9 V.
Given
Vo = 15.
T = 2 RC.
Therefore, V = 12.9 V.
(2)
Tarun said:
1 decade ago
Just apply the formula v=1/r integral C dt.
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