# NOt sure what stage I'm suppose to look at this circuit, finding RC!

1. Apr 1, 2006

### mr_coffee

Hello everyone! I was able to get parts, a-c of this problem, but i'm confused on what stage i'm suppose to use to find T = RC, let t = time.

The problem says: For the cirucit find Vc(t) at t equal to (a) 0-; (b) 0+; (c) infinity; (d) .08s

a: 20v
b: 20v
c: 28v
d: 24.4

Note: that is a unit step fucntion, at time = 0, u have no voltage of 10V, at time t > 0 you will have 10V added to the circuit.
so if this is the circuit:

I know the equation is going to look like this:
Vc(t) = 20*e^-tT;
so all i need is T = RC;
I need to remove all the power sources and look through the conductor right? but thats the part that confusese me, because the conductor is right in the middle, there is a resistor on both sides, so which side do i "look" from? left or right? Or am i not doing this right at all?

THanks!

2. Apr 2, 2006

### andrevdh

If the $1\ mA$ refers to an ammeter then it is short circuiting the capacitor and the two resistors on the right. Which does'nt make sense since the current throught it will then be $0.4\ A$?

3. Apr 2, 2006

### mr_coffee

Yes the 1mA refers to the current source, i don' tknow if its also called a ammeter, its just a source that supplies 1 mA. But i don't understand what part doesn't make sense to you.

Last edited: Apr 2, 2006
4. Apr 2, 2006

### nrqed

It's not an ammeter, it's a *source* of constant current.

5. Apr 2, 2006

### nrqed

You are forgetting what I taught you
You need a function that will give 20 volts at t=0+ and 28 volts at t=infinity. The fucntion you gave give zero at t= infinity!

You need $C_1 + C_2 e^{-t/\tau}$. Imposing that it gives 20 at t=0 gives C_1 + C_2 = 20 and imposing that you get 28 at infinity gives C_1 = 28 volts therefore C_2 = -8 volts.

For the question of determining the R_eq for the calculation of $\tau$, I have to be careful because I *think* it is given by the following trick but I don't remember off hand how to prove this (it has been a very long time since I have done circuits) so I can't be 100% sure that it works for all circuits but it sure works for simple ones (at leastthe ones you have shown me so far). To find R_eq, remove the branches containing currents sources (or if you will, replace the current sources by open switch so that these branches become irrelevant) and replace the voltage sources by ideal wires (i.e. short the voltaghe sources). Then combine all the resistors until you have only one loop with a single capacitor (or inductor) connected with one resistor. That's your equivalent resistor.

In the example above, if you do that you end up with the capacitor connected to the 25 kiloOhms on the left and the 20 and 80 on the right. The equivalent one loop is the capacitore with a 25 connected in parallel with a 100 which gives an equivalent resistance of 20 kiloohms. Your time constant is therefore .100 second.

Then the final result (at 0.08 second) is
$28 - 8 e^{-0.08/0.1} = 24.4 volts$

Hope this helps.

Patrick

6. Apr 2, 2006

### BobG

Patrick is right. For the equivalent resistance, the 25k is in parallel with a 100k (80k + 20k). Use the equivalent resistance to find your time constant.

Patrick's final equation is right, but doesn't really explain how you got there. If you use superposition, it might be clearer.

Your equation was correct for the current source, if the current source were swithed on at time 0. Since it's been on, you just have 20 V at the top of the capacitor as your initial condition.

The equation for the voltage source is $$8-8e^{\frac{-t}{\tau}}$$

Add the two together to get Patrick's final equation.

7. Apr 2, 2006

### mr_coffee

ahh i c! hah thanks for the explanation guys it worked out!! I think i'm studying too much everything is running together! Thanks again!

8. Apr 2, 2006

### mr_coffee

Incase anyone was interested I posted the solution with the help of everyone!

9. Apr 2, 2006