Engineering Circuits 1 help with this circuit in a DC state

[sammyqw]

Homework Statement

http://imgur.com/a/4YfkJ
http://imgur.com/a/4YfkJ

Homework Equations

for t<0 I understand that Capacitor acts as an open circuit and goes away but why did the 90v and 6ohm resistor disappeared? is it because current doesn't go that way? and also in that KCL at node V1 wouldn't it be v1-(30)/6 + v1/6 + v1-(-2ix)/8 ?

The Attempt at a Solution

http://imgur.com/a/3jlrz[/B]

http://imgur.com/a/3jlrz

 

[gneill]

for t<0 I understand that Capacitor acts as an open circuit and goes away but why did the 90v and 6ohm resistor disappeared? is it because current doesn't go that way? and also in that KCL at node V1 wouldn't it be v1-(30)/6 + v1/6 + v1-(-2ix)/8 ?

The 90v and 6ohm resistor "disappear" because the switch branch blocks any effects it might have on the rest of the circuit. No potential difference can be developed across a perfect conductor. Any current in the leftmost loop will be confined there.

Why do you assign a double negative to the last term (the "i_x" term")?

As far as I can tell the solution in the image is fine as far as it goes. Of course it doesn't answer the problem's part (a), which wants the capacitor voltage.

 

[sammyqw]

yeah it makes sense now thanks. For part d i got 10e^-2000t Amps ,how can I find part e?

 

[gneill]

yeah it makes sense now thanks. For part d i got 10e^-2000t Amps ,how can I find part e?

I haven't done the math for part (d) so I have no comment to make on your suggested solution. If you want to have it checked, present the details of your work.

Part (e) asks for i_x at times t = 0^- and t = 0⁺. One is during steady state and the other just after the switch opening. For the first you can ignore the capacitor because at steady state it won't impact any currents -- it behaves as an open circuit. At t = 0⁺ things are changing so you need to consider the capacitor. For that instant of time you can replace the capacitor with a voltage source that has the same potential difference that the capacitor had the instant before the switch opened (capacitors can't change their potential difference instantaneously). The resulting circuit can be solved by the usual methods. Naturally the solution only applies to that instant after the switch opens.