# Need help in solving DC transients question (Electrical Engg)

1. Sep 19, 2014

### thesidjway

http://i.imgur.com/nhBN1RJ.jpg << Huge image replaced with URL by Moderator >>

The attempt at a solution

I have been able to find
V(0+)=12
V(0-)=12
I(0-)=4

Not able to find a suitable way to get a general solution for Vc and Ix

Last edited by a moderator: Sep 19, 2014
2. Sep 19, 2014

### psparky

This is a tough problem!!!

First off, like all circuits, redraw it so it doesn't look so intimidating.

Slide the switch a little to the left so it lines up vertically with R1.

Take R3, take it's vertical branches and make them horizontal so they are just above the switch and the inductor.
Clearly R3 and the capacitor are in parallel.

I would also slide the inductor to the bottom horizontal branch. When the switch closes, it is then clear that R3, the capacitor and the inductor are in paralell.

Perhaps that will get you started a bit...

3. Sep 19, 2014

### rude man

1. determine the initial conditions on L and C.
2. sum currents to zero at the right-hand node (only) and you're off.

One unknown node, one equation.
Do you know Laplace transforms? You don't have to, but it helps.

4. Sep 19, 2014

### psparky

Laplace transforms would definitely be the easy way....but I get the feeling he is in the differential equation part of his studies....also known as the hard way.

5. Sep 20, 2014

### rude man

Still only one equation & one unknown node.

6. Sep 21, 2014

I hope the answer to differentiation of Vc at t=0+ would be -2V/sec

7. Sep 21, 2014

### rude man

I got infinities for all the t = 0+ derivatives. I'll check it some more later.

Last edited: Sep 21, 2014
8. Sep 24, 2014

### Staff: Mentor

You hope?

The circuit can be resolved into one comprising 3 parallel elements, R, C, and L, then look up or work out the D.E. for that 2nd order system's step response. That is the only way I can see (or the Laplace equivalent).

9. Sep 24, 2014

### rude man

,,, plus a fourth element.

10. Sep 24, 2014

### rude man

Your V(0+) and V(0-) are wrong. Hint: check sign on Vc(0+). Vc(0-) should be obvious. I assume by Vc is meant the right-hand side of C. If by "V(0+)" and "V(0-)" you just mean the voltage across C then you're OK.

Last edited: Sep 24, 2014
11. Sep 27, 2014

I am sorry. I should have used "I think". I solved the problem roughly using general equation of voltage across capacitor Vc(t)=Vc(final)+[Vc(initial) - Vc(final)]e-t/RC. But after your reduced circuit i solved it in the following manner as per given in figures attached.

#### Attached Files:

File size:
46 KB
Views:
74
File size:
40.6 KB
Views:
85
File size:
45.7 KB
Views:
77
• ###### photo 1(2).JPG
File size:
34 KB
Views:
86
12. Sep 27, 2014

### Staff: Mentor

Vc is marked on the schematic as the voltage across the capacitor plates, by essentially a left-pointing arrow.

Vc(0-) = Vc(0+) = +12V

Immediately the switch is thrown, the current into the capacitor changes from 0 to 2A to the RHS plate, this equates to

dVc(0) /dt = -2 V/s

13. Sep 27, 2014

### rude man

Yes. I was computing the voltge at the right-hand node. Either didn't see the vc as marked or it was added later.

Will look at the rest later, altho' the OP seems to have disappeared as usual.

14. Sep 27, 2014

### rude man

Are you doing the rest? The OP seems to have gone AWOL.