2
\$\begingroup\$

Here is the schematic:

I got a little problem in electronics: it involves simple things but I can't figure it out (and found no other circuit that was the same configuration).

At t < 0 ms, both switches are open.

The problem is the following : when t = 0 ms, the switch I is closed and the switch II is open. Then I need to make an equivalent circuit when one DC source is in series with A and B (from UAB = UTH), to find the value V0 and calculate the exponential equation for the voltage of the capacitor.

Do we remove the branch that is open to calculate UTH or do we keep it? In short, is Uth (when t = 0ms) equal to 0V or 10V?

The question is here: if we close the switch I, and if V1 is equal to 10V, the equation for the capacitor being equal to \$V_1 + (V_0 - V_1) \times e^{(-t/TAU)}\$ will be equal to 10V constant, which doesn't seem logical since we added one resistance in the circuit!

But if V1 is equal to 0V, it would not be logical since the capacitor is in series with the 10V DC...

Can you please guide me in the right direction or tell me what I did wrong?

Improve this question
\$\endgroup\$
0
\$\begingroup\$

For t < 0 ms, the circuit can be re-drawn as shown.

schematic

simulate this circuit – Schematic created using CircuitLab

at t = 0 ms the \$30\Omega\$ resistance comes in parallel with the old resistance. Now the circuit becomes.

schematic

simulate this circuit

You can directly write \$R_{th}\$ and \$U_{th}\$ from this circuit.

The only thing happened here is that you made the capacitor's charging fast. But it won't have any effect on voltage across C if it has reached V1 volts before t = 0 ms.

Improve this answer
\$\endgroup\$
  • \$\begingroup\$ Okay, thanks you a lot. It means when we close the second switch, the voltage would be still 10V ? Then u(t) would be constant 10V from t<0ms to t->+infinite ? And current would be 0A from t<0ms to t->+infinite? \$\endgroup\$ – user2687718 May 21 '14 at 9:22
  • \$\begingroup\$ Nop. When 2nd switch is closed, the capacitor will start to discharge because \$U_{th}\$ in that case will be less than 10V. Hence there will be non zero current whose value will saturate as \$t\rightarrow\infty\$ \$\endgroup\$ – nidhin May 21 '14 at 12:35
  • \$\begingroup\$ I get Rth = 20V and Uth = -8.4V, am I right ? :) \$\endgroup\$ – user2687718 May 21 '14 at 12:48
  • \$\begingroup\$ I think \$R_{th} = 20\Omega\$. But I am not sure about \$U_{th}\$. Are we closing both the switches? or only 2nd switch? \$\endgroup\$ – nidhin May 21 '14 at 12:54
  • \$\begingroup\$ Both of them when t = 10ms. :) \$\endgroup\$ – user2687718 May 21 '14 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.