Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 17 (Q.No. 49)
49.
For the circuit shown in the figure, the capacitor C is initially uncharged. At t = 0, the switch S is closed. The voltage Vc across the capacitor at t = 1 millisecond is (In the figure shown above, the op-amp is supplied with ± 15 V and the ground has been shown by the symbol)
Discussion:
4 comments Page 1 of 1.
Usha Rani Dash said:
3 years ago
If we use Vc as 10 and find out the value of Vo becomes 20v which is more than the supplied voltage.
Due to this, the output becomes saturated and the virtual short circuit becomes invalid. That's why, the transition method should be used and the answer ll be none of the above (Vc= 8.94).
So, here; All options are incorrect.
Due to this, the output becomes saturated and the virtual short circuit becomes invalid. That's why, the transition method should be used and the answer ll be none of the above (Vc= 8.94).
So, here; All options are incorrect.
Junjun said:
4 years ago
I think the correct answer is D. Derive the formula for the non-inverting op-amp. It will be Ir=Ic. Ir = C*(dVc/dt). Arrange them and you will get, dVc = (1/C)*Ir*dt. Integrate both sides and use the limit.
Also, take note that the current is constant. So, the solution is Vc = (1/C) * Ir * t. Vc=10V.
Also, take note that the current is constant. So, the solution is Vc = (1/C) * Ir * t. Vc=10V.
MAY JOY SIDAYON said:
7 years ago
I think it should be option B.
10(1-e^(-(1us/(1k_ohms)*(0.1uF)))).
10(1-e^(-(1us/(1k_ohms)*(0.1uF)))).
(1)
CARLRIC COLLADO JR said:
7 years ago
Yes! B should be the answer.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers