# Having trouble with this Circuit

Discussion in 'Homework Help' started by gerj91, Jul 24, 2009.

1. ### gerj91 Thread Starter New Member

Jul 24, 2009
3
0
Hi! I'm new in the forums. The thing is I'm having my final exam of circuits next Wednesday and I'm trying to practice and to solve circuits from my book and others, but I'm REALLY stuck with this one. Actually I'm doing poorly on RL and RC circuits in general.

Could you help me with this one?

Determine the value of V(t) and I(t) when t > 0 in the next circuit. (When t > 0 the switch changes of position).

I know that when t < 0 There is no current in the left part of the circuit and when t > 0 I think there is no current in the right part of the citcuit.

I also happen to know that after a long time (t >>>>> 0), the inductor forms a shortcircuit and the capacitor forms an open circuit. (If I'm not using my words correctly, my excuse is that English is not my native language).

What else? Mmmm... with capacitors according to my notes, the voltage is the same before and after 0 (Vc(0-) = Vc(0) = Vc(O+), and with the inductor same goes for current.

Thanks for helping me, it means A LOT.

P.S.: I know I didn't get very far with this one, but this is the one I understand the least.

2. ### millwood Guest

sounds like at t<0, the switch is on the left: no current going through that loop;

at t=0, the switch bangs to the right. no current going through that loop either since the current source is shorted by the inductor.

so your analysis is right.

3. ### gerj91 Thread Starter New Member

Jul 24, 2009
3
0
I think it's the other way around, in t < 0 the switch is on the right. at t = 0 it changes its position to the left so when t > 0 the switch is on the left. Isn't it?

4. ### millwood Guest

i have no idea - it is your circuit.

but in that case, the voltage across the capacitor is zero, and the 10v is resistively divided among a 10k and 12k resistor, at the moment the switch is turned.

5. ### gerj91 Thread Starter New Member

Jul 24, 2009
3
0
I tried to represent the circuit with a simulator (www.falstad.com/circuit/).

When the switch is on the right, the inductor slowly takes over all the current from the source. When the switch is on the left the capacitor slowly charges until it reaches the 10v.

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,227
I view the switch to the left ... charging the capacitor till t = 0. At t=0, the switch is thrown to the right.