# RLC Circuit #2

Discussion in 'Homework Help' started by ihaveaquestion, May 9, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
This one is essentially the same problem as the #1 circuit thread I made...

For some reason though it looks like she solved it differently.

Question:
http://img217.imageshack.us/img217/6278/8888889.jpg

Circuit:
http://img217.imageshack.us/img217/2301/19965838.jpg

http://img217.imageshack.us/img217/6107/40369259.jpg

Again what I don't understand is the circled part

The asterisk I put because I see that since i = Cdv/dt then just differentiating that V = A1e^s1t + A2e^s2t term you get the i equation... Where did she pull that V equation that I circled from?

Is it easier to setup the differential equation for one variable instead of the other? i.e. in the RLC #1 circuit thread she sets it up in terms of i, in this one in terms of v...

2. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Essentially both of these RLC circuits are the same problem, so whatever way is easier to solve I will go with on the test.

This appears to be the one with the differential equation setup in terms of V.

It seems that my answer will be in the form of

x = A1e^(s1t) + A2e^(s2t) depending on what variable I set my second-order differential equation with..

i.e. looking at the two different examples, one has i = A1e^(s1t) + A2e^(s2t) because the differential equation is setup in terms of i and the other has v = A1e^(s1t) + A2e^(s2t) because the differential equation is setup in terms of v.

So I'll just go with the way solved in this #2 example for the test, because now I understand what was done.

I do still however need help with the RL first-order problem on the other thread please

3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
I have a question:

http://img217.imageshack.us/img217/2301/19965838.jpg

What if we switched places of the inductor and capacitor. With the case linked above, we find the intial value of the current in the left-hand circuit and know the voltage across the capacitor is zero.

If we switched the capacitor to where the inductor is and vice versa how would this change our initial conditions? I would think that after a long amount of time the voltage across the capacitor is 50V, so v(0)=50. Likewise the inductor isn't part of the circuit yet so the current is 0A, i.e. i(0) = 0. Correct?

Also, what is the difference between v(0) and v(0+)?

For example what if it asks for i(0+) in the example linked above or v(0+) in the switched scenario described above? These aren't just i(0) and v(0)?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Yes that's correct

None really - the underlying notion is that the changeover switch has reached it's final position. Having now established the initial conditions for the purpose of analysis, the prior "history" of the circuit plays no further part in the analysis.

For practical purposes - yes.