Find v(t) for t > 0. RC Circuit Problem

[bob29]

Homework Statement

The switch in the circuit has been closed for a long time, and it opens at t = 0. Find v(t) for T [tex]\geq[/tex] 0.
[PLAIN]http://img834.imageshack.us/img834/9107/tutorialq2.jpg

Homework Equations

V(t) = V_oe^-t/RC

[tex]\tau[/tex] = RC

The Attempt at a Solution

@ T<0
Short Circuit due to capacitor
V = V_2k
By VDR:
V = 2/(10+2) * 24 = 4V

@ T [tex]\geq[/tex] 0
24V is open therefore
Req = (10//2) = 5/3 ohm
C = 40microFarads

[tex]\tau[/tex] = (5/3)k * 40*10^-6
RC = 1/15

V(t) = 4e^-15000t
Or
[PLAIN]http://img713.imageshack.us/img713/5351/28102010077.jpg

Forgot to add in the extra 1000 due to the K
therefore answer should be 1000 less or V(t) = 4e^-15t

 

[bob29]

Problem solved.
Having trouble with a similar but more complex problem.

1. Homework Statement .
Question is in the image including diagram.
http://img163.imageshack.us/i/electricalproblem.jpg/
[PLAIN]http://img163.imageshack.us/img163/2886/electricalproblem.jpg

Homework Equations

tau = τ = R_eq * Capacitor
v(t) = V(∞) + [v(0) - v(∞)]e^-t/τ

The Attempt at a Solution

[PLAIN]http://img100.imageshack.us/img100/4793/attempt1t0a.jpg[PLAIN]http://img100.imageshack.us/img100/1591/attempt1t0.jpg

 

[CEL]

Capacitor is an open circuit only in steady state (t = infinity).
To solve for the transient you can use superposition.

 

[bob29]

In the working out of the answer.
V(infinity) = (6/2+6)*10 - (2/6+2)*50 = -5
I don't understand how it's subtracting since superposition you have to add.

 

[CEL]

In the working out of the answer.
V(infinity) = (6/2+6)*10 - (2/6+2)*50 = -5
I don't understand how it's subtracting since superposition you have to add.

Because the 50V source has the - sign at the top and the + sign connected to ground.