Differential equation in simple RC-circuit

Hi guys,

I know this should be obvious, but there's something I am just NOT getting.

Imagine a simple series RC circuit with a DC source as shown in the attachment. As can be seen from the picture, I have solved the differential equation in capacitor current in the time domain. In order to be able to solve the problem, I have assumed that dE/dt = 0 as this is a dc source.

What I wish to know is why can't I take the Laplace transform of both sides of the equation (*)? I know that L(0) = 0, so this would give a bogus equation (i.e. I(s) = 0, which is wrong). But if my equation (*) is right, then why can't I use the laplace transform of both sides at this point?

From textbooks, I read that the DC source is considered as a step input, thus in the Laplace domain, this would be E/s. So again, what is wrong with equation (*)? And why do I get the right answer at the end?

Again, as I said, I think this should be obvious, so please don't hesitate to point out trivialities.

Thanks for your understanding.
e.

The shown differential equation solution is correct for a step function input, but not for a DC source. For the latter, steady state would be already reached at t=0 and the assumption VC(t=0)=0 incorrect. To correct the circuit, you need to add a switch, that is closed at t=0.

FvM said:

The shown differential equation solution is correct for a step function input, but not for a DC source. For the latter, steady state would be already reached at t=0 and the assumption VC(t=0)=0 incorrect. To correct the circuit, you need to add a switch, that is closed at t=0.

Click to expand...

Thanks for your reply. Assuming I add a switch that is closed at t=0, why is it not possible to take the Laplace transform of both sides of the equation labelled (*)? Thanks again.

darkfeffy,

What I wish to know is why can't I take the Laplace transform of both sides of the equation (*)? I know that L(0) = 0, so this would give a bogus equation (i.e. I(s) = 0, which is wrong). But if my equation (*) is right, then why can't I use the laplace transform of both sides at this point?

Click to expand...

What says you can't? From your equation(*), we get (s*I-E/R)+I/RC= 0 . Solving for I we get, I = E/[R(s+1/RC)] . Taking the inverse LaPlace gives the equation you seek.

Ratch

Ratch said:

darkfeffy,



What says you can't? From your equation(*), we get (s*I-E/R)+I/RC= 0 . Solving for I we get, I = E/[R(s+1/RC)] . Taking the inverse LaPlace gives the equation you seek.

Ratch

Click to expand...

Thanks Ratch. I knew I was being stupid somewhere. I totally forgot about the initial capacitor current. Problem solved.:-D