Thévenin resistance when branch is 0V (RC circuit)

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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?
 

crutschow

Joined Mar 14, 2008
23,508
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.
 
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