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Ask your doubts here

Post by Admin on Wed Jan 13, 2016 9:22 am

You can ask your doubts/queries in this forum

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Re: Ask your doubts here

Post by saidheeraj on Wed Jan 13, 2016 12:43 pm

I have a doubt in Question no : 37. When the switch is on for a long time, We will have the charge in the 1st capacitor(2 uf) to dissipate across the resistor and the second capacitor will not be able to dissipate its charge. Since we have no initial charge/ Voltage specified I had assumed it as zero. When the switch is now closed. The capacitor will act as a short in series with the initial voltage ( which I had taken as zero. And we also have a condition that the capacitors do not allow the voltage across them to change instantaneously. So vc(0-) = Vc(0+). I felt the Voutput should be zero. And the approach specified in the solution would be when you have initially charged and then the switch opened, or the voltage across the capacitor at steady state. Correct me, if I am wrong.

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Query on Question no. 37

Post by Admin on Wed Jan 13, 2016 2:30 pm

saidheeraj wrote:I have a doubt in Question no : 37. When the switch is on for a long time, We will have the charge in the 1st capacitor(2 uf) to dissipate across the resistor and the second capacitor will not be able to dissipate its charge. Since we have no initial charge/ Voltage specified I had assumed it as zero. When the switch is now closed. The capacitor will act as a short in series with the initial voltage ( which I had taken as zero. And we also have a condition that the capacitors do not allow the voltage across them to change instantaneously. So vc(0-) = Vc(0+). I felt the Voutput should be zero. And the approach specified in the solution would be when you have initially charged and then the switch opened, or the voltage across the capacitor at steady state. Correct me, if I am wrong.

Your argument lies on the point that capacitor voltage cannot change instantaneously. Hence you have mentioned vc(0-) = Vc(0+). This is correct. But you should remember one thing. Try to reason out why capacitor voltage cannot change instantaneously. It is because if there is a step change in the capacitor voltage then the current through capacitor which is derivative of capacitor voltage will be an impulse signal. Impulse current means an infinite current for a very short duration.This is the reason why capacitor voltage cannot change instantaneously since it leads to an infinite current.
In the problem the capacitor voltages have to change from 0v to a finite value otherwise KVL will be violated ! This is an exception where capacitor voltage changes instantaneously. There are some previous year gate questions on this concept.

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Re: Ask your doubts here

Post by saidheeraj on Thu Jan 14, 2016 10:52 am

Thank you.

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Re: Ask your doubts here

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