Finding the function of the cap.

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The discussion focuses on calculating the voltage across a capacitor, Vc(t), after a switch is opened in a circuit where the capacitor is initially charged to 10V. The user attempts to apply Thevenin's theorem to find the equivalent resistance and derive the voltage function using the formula Vc(t) = final + (initial - final) e^(t/RC). The correct time constant is identified as RC = 1.67 x 10^-7 seconds, leading to the expression Vc(t) = 10e^(-t/(1.67 x 10^-7)). However, there are concerns regarding the physical validity of the results and the accuracy of the Thevenin equivalent resistance calculation.

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3113203781_233b4fbdd4_o.jpg


This is one of the questions I got in my finals, and I couldn't do it back then. I want to find Vc(t) , given the cap is initially charged up to 10V, the switch is then opened. My approach is to find the thevein eq of the circuit, which I would came up with a resistor in parallel with the cap, and then find the function using the expression Vc(t) = final + (initial - final) e^(t/RC).
sounds good?
 
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johndoe said:
3113203781_233b4fbdd4_o.jpg


This is one of the questions I got in my finals, and I couldn't do it back then. I want to find Vc(t) , given the cap is initially charged up to 10V, the switch is then opened. My approach is to find the thevein eq of the circuit, which I would came up with a resistor in parallel with the cap, and then find the function using the expression Vc(t) = final + (initial - final) e^(t/RC).
sounds good?

After the switch is opened, the cap disappears from the circuit, no?
 
berkeman said:
After the switch is opened, the cap disappears from the circuit, no?

ops the switch should be opened initially and then closed(drawing error)
 
What is the source on the right? It can't be a current source of value 5ix, since there is only one ix flowing through the resistor before the switch is closed...

I think you need to re-draw the circuit correctly, and then show us your work on the solution. Then we may be able to offer some help.
 
Question:
Finding the function of the 0.1uF cap Vc(t) when the switch is closed at t=0 given that the cap is initial charged up to 10V.

3117971925_4c2bc42cf6_o.jpg


ok my attempt at this I find the thevenin eq of the right hand side circuit and come up with :
3117972721_0e2f97c1db_o.jpg


Vc(infinite) = 0, Vc(0)=10 , time constant = RC = 1.67x10^-7

Vc(t) = 0 + ( 10-0) e^(t/(1.67x10^-7 ))

Appreciate if u could have a look at it.
 
How did you get 1.67 ohms for your thevenin equivalent circuit? I got an indeterminate answer for the thevenin equivalent circuit. You could do it without that. Furthermore, your final answer doesn't make much physical sense; Vc(t) would increase exponentially without limit as time passes.
 
Defennder said:
How did you get 1.67 ohms for your thevenin equivalent circuit? I got an indeterminate answer for the thevenin equivalent circuit. You could do it without that. Furthermore, your final answer doesn't make much physical sense; Vc(t) would increase exponentially without limit as time passes.

This is what I did :
3120459633_644b262141_o.jpg


I hook up a 1V source and solve for I1 and thus the Rth.

And I left out a - sign in my final answer
Vc(t) = 10e^-t/RC
 

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