Thévenin resistance when branch is 0V (RC circuit)

mabtjie

Joined Oct 16, 2011
1

Prior to t=0, switch is in position A (for long time).
Afterward, switch is in position B.

We're using $$v(t)=v(\infty)+[v(0)-v(\infty)]e^{\frac{-t}{\tau}}$$ where $$\tau=RC$$.

My problem is in the calculation of the thévenin resistance from the perspective of the capacitor.

My answer: $$R=R_\mathrm{th}=(15+10)\parallel 100=20\,\mathrm{k\Omega}$$
Homework answer: $$R=100\,\mathrm{k\Omega}$$.

I was told that it has to do with the entire bottom of the circuit being at 0 V relative to the 100 V, so the left hand loop of the circuit is disabled. I'm not really satisfied with that answer though. I was hoping that someone could shed some light on this?

JoeJester

Joined Apr 26, 2005
4,390
The switch disables the left hand side when in the "b" position. Follow the current flow.

crutschow

Joined Mar 14, 2008
27,725
The Thevenin equivalent from the capacitor's point-of-view is whatever the capacitor sees when going from one capacitor terminal until it reaches the other capacitor terminal. As Joe said, follow the current.