Thévenin resistance when branch is 0V (RC circuit)

Discussion in 'Homework Help' started by mabtjie, Oct 16, 2011.

1. mabtjie Thread Starter New Member

Oct 16, 2011
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0

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?

2. JoeJester AAC Fanatic!

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

3. crutschow Expert

Mar 14, 2008
18,636
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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.