Hello,

this information was provided by our Professor:
-U1(t) had for long time an initial value of 12 V.
-At t = 0.5 ms the U1(t) drops to 0 V. (You can see that in the graph at the bottom of the page).

Parts A and B I solved by myself and I got the same results as my Prof.
A) U2(t) @ 0 ms? Result: 3 V. Method: Simple Voltage divider.
B) What is the Voltage across the capacitor @ 0 ms? Result 9 V. I just subtracted 12 V - 3 V = 9 V. Is that all I have to do?

I am struggling with part C.
I get the same equivalent circuit and it appears to me, that he uses the Voltage divider formula ( 12 V * R1/(R1+R(TH)) ), but I am not certain (at all) why he does that? The next step is equally puzzling, why is he subtracting 1 V - 9 V?
Is there a different equivilant circuit that I could use?

Tobias

 

Since this is an English-only forum, there probably aren't too many people that can read the questions. So you need to provide the translations for (c) and on just like you did for (a) and (b).

 

Excelltent point! I totally forgot to translate the Question! Sorry about that!

Here is the question: What happens to U2(t) right after the drop of U1(t)?

 

Hint: Since R1 and C1 are in series, they can be swapped for analysis purposes (that don't involve the voltage at the junction between them, anyway).

 

Is this the equivalent circuit you were talking about? If so, than I understand why he used the voltage divider formula, BUT why do I have to subtract 9 V from 1 V?

 

No, that's not what I was talking about. I was talking about swapping R1 and C1 to (potentially) make things a bit easier to visualize.

What he has done is replace everything to the right of node 2 with it's Thevenin equivalent.

Also, if the blue 9V is the instructor's answer for (b), then the instructor might need to pay more attention to the specified polarities of the signals. Unless I am mistaken (and it's possible, since I seldom us arrow conventions for voltages instead of + and - labels), the arrow above the capacitor means that the voltage on the right side of the capacitor is greater then the voltage on the left side of the capacitor whenever Uc1 > 0.

 

Unless I am mistaken (and it's possible, since I seldom us arrow conventions for voltages instead of + and - labels), the arrow above the capacitor means that the voltage on the right side of the capacitor is greater then the voltage on the left side of the capacitor whenever Uc1 > 0.

I don't know why but Germans use different convention. The arrow tip (head) point into the lower potential. Notice how Vth is mark.

Tobias Hildebrandt

Can you solve for U2 for this circuit ?

 

The Helmholtz Method seems to give me the right answer, I think this is what you had in mind, aye?
It seems that my Prof used a different method to calculate the value, but both are equally correct, aye?
In what cases can I transform a capacitor into a voltage source?

 

I never heard about "Helmholtz Method" but it looks like you are using a superposition theorem.
Do you know the Kirchhoff's voltage law (KVL) ? Can you write a KVL equation for this circuit (a single loop)?

 

What happens to U2(t) right after the drop of U1(t)?

Easy: the voltage across the capacitor cannot change instantaneously.

Before the voltage drop, the capacitor's right has a potential of 3v; The capacitor's left has a potential of 12v. For a difference of -9v, from right to left.

When V1 changes, that voltage does not change instantaneously.

The rest is easy.

 

I don't know why but Germans use different convention. The arrow tip (head) point into the lower potential. Notice how Vth is mark.

Wonderful -- all the more reason to avoid that notation (but, of course, if that is what you are taught and/or expected to use...).

 

The Helmholtz Method seems to give me the right answer, I think this is what you had in mind, aye?
It seems that my Prof used a different method to calculate the value, but both are equally correct, aye?
In what cases can I transform a capacitor into a voltage source?

Like Jony130, I've never heard of "the Helmoltz Method", but it looks like what we refer to as "superposition".

As for transforming the capacitor into a voltage source, that is done because at the instant the change in the circuit is made, the voltage across the capacitor cannot change, which means that it looks like a voltage source at that instant in time.

 

It seems that my Prof used a different method to calculate the value, but both are equally correct, aye

Simple, your Prof first find a voltage drop across R1 (VR1 = 1V) then he used the KVL to find U2 voltage.
U2 = V1 - Vcap = 1V - 9V = -8V

 

1) Thanks to everyone that helped me with this problem!
2) http://www.amplifier.cd/Tutorial/Grundlagen/Superposition.htm (The pictures give you a good idea what they are doing, even without understanding what is said. I am sorry for posting so much non English texts!)
There is a Helmholtz Method, but I think you guys are right, what I used is just a good old Superposition method.

 

Question D: Provide a function u2(t) for t > 0.5 ms.

So at this point the capacitor starts to discharge, so I use the formula: u(t) = U(max) * e^( - (t / RC )).

The calculation of the time constant is strait forwart. It is just R times C.

The +3 V comes from the voltage divider.

The - 0.5 ms part in he exponent is rather obvious as well, but where does the - 11 V come from?

It seems that I always ask similar questions, is that sth I can solve with the superposition method?

 

The general formula for capacitor charging/discharging phase looks like this
Vc = V∞ + (Vo - V∞)*e^(-t/(τ))
And in this formula Vo includes the initial capacitor voltage (-8V).
And V∞ - is a steady state final voltage (+3V)
(Vo - V∞) = (-8V - 3V) = -11V