Find the voltage at specific time

Thread Starter

crackhead227

Joined Feb 11, 2019
22

Translation: The circuit in the figure closes at t = 0.
What is the voltage, Ux , at t = 1 milliseconds if the capacitor was uncharged at t = 0.


I’m not sure how to solve this problem. I attempted a solution but I think I’m doing something wrong.
 
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crutschow

Joined Mar 14, 2008
24,402
The resistors are in series.
The capacitor is in parallel with R2.
So first you find the equivalent resistance at each of the capacitors terminals (assume the voltage source looks like a short), and the DC voltage at each capacitor terminal.
That gives you a Thevenin equivalent circuit.
From that you have the RC time constant and can calculate the response.
 

Jony130

Joined Feb 17, 2009
5,145
If they are not parallell, and not in series, what is the relationship?
Well I think you got that part right Rth is equal to (R1+R3)||R2 ≈ 762Ω and that the capacitor will charge from 0V to 1.19V at 5*Rth*C
You even got the 0.73 part right. But you made an error in your final step.
 

Thread Starter

crackhead227

Joined Feb 11, 2019
22
My thevenin circuit looks like this:



The voltage for Ux at t = 0.001s should be 3.71.

How can I get that answer with this formula?

Vc = V(1-e^(-t/RC))
 

WBahn

Joined Mar 31, 2012
25,283
How is the voltage going to go above your 1.19 V source?

Where did that answer come from?

What do you get for your time constant?

How many time constants is 1 ms?
 

MrAl

Joined Jun 17, 2014
7,179
Hi,

Good question, and interestingly i got a similar answer i wont spell out just yet.
So how did he get that.
 

Thread Starter

crackhead227

Joined Feb 11, 2019
22
How is the voltage going to go above your 1.19 V source?

Where did that answer come from?

What do you get for your time constant?

How many time constants is 1 ms?
The answer should be correct, got it from the same book as this exercise. 3.71V

Time constant: 762 * 1 * 10^-6 = 7.62 * 10^-4
 

Jony130

Joined Feb 17, 2009
5,145
This question is tricky because the capacitor is not connected to the reference point (GND) but Vx is.
Vx change from Vx =3.4357V (at t = 0) to 3.8V (at t =∞), only by 0.372V.
And the voltage across R3 resistor is changing form 3.4375V to 2.62V, drops by 0.8175V.
Also notice that 0.372V + 0.8175V ≈ 1.19V
But you shoud care only about Vx change ( ΔVx = 0.372V)
 

MrAl

Joined Jun 17, 2014
7,179
Hello again,

Often it helps to transform the circuit into one that looks easier to solve.
Notice R3 has been moved into a position that does not alter the result as long as the result voltage is taken across the same two elements.

RC-Circuit_20190212-2.gif
 
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