1
\$\begingroup\$

I can't find the rules about how to calculate the initial voltage across the capacitor \$C_1\$. A my friend told me that I can calculate \$C_1\$ following the meshes lines: if I do this, I will come across \$R_3\$ and \$R_1\$ or \$R_2\$ and \$L_1\$. So I am tempted to say \$V_c(0^-) = |V_{R_1}|+|V_{R_3}|=|Ia*R_1|+|Ib*R_3|\$, is this correct? If yes, what are the signs (+ or -) of \$|V_{R_1}|\$ and \$|V_{R_2}|\$?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ The given information are the only I have written on the schematic. What I need to calculate is \$V_c(0^-)\$ (I mistakenly wrote \$V_L\$). \$\endgroup\$ – user128414 Oct 31 '16 at 11:16
  • \$\begingroup\$ Olin's comment that we can't tell you about the initial conditions if we are not given any info about momentary behaviour at t=0 still holds. Possibly the exercise states that the circuit is at steady state until t=0 and then something happens (like changing the current of I1). We need to know that to answer the question. Therefore downvoted, because this question doe not have a unique answer. \$\endgroup\$ – Michael Karcher Oct 31 '16 at 11:31
  • \$\begingroup\$ Actually, at a \$ t < 0\$ correspond I1 = 1, from \$t>=0\$, I1 became equals to 0. Can you help me now? \$\endgroup\$ – user128414 Oct 31 '16 at 13:52
  • \$\begingroup\$ The question is answerable now. You actually look for the steady-state voltage at I1=1A. Downvote reverted. \$\endgroup\$ – Michael Karcher Oct 31 '16 at 14:01
2
\$\begingroup\$

The initial voltage across a capacitor and initial current thru a inductor are state variables that have to be given. You can't calculate them because they depend on previous history. By definition, "initial" conditions are before there is any history. They must be specified to be able to analyze the circuit going forwards.

What is possible is to compute the steady state conditions after everything settles. This is independent of initial conditions in a circuit with all passive linear components like you show. You can determine what the capacitor voltage and inductor current will be eventually without having to know the initial conditions. That's because these settle to the same value regardless of what they started at.

Added

Now that you've specified that the current source is 1 A for all time before t=0, the "initial" conditions (actually conditions at t=0) are the steady state with the current source producing 1 A.

This is easy to solve by inspection. In steady state, capacitors are opens and inductors are shorts. This leaves the resistance across the current source as R1 // (R2 + R3), which can be seen from inspection is 500 mΩ. That means the current source voltage is 500 mV. That applied to R1 is 500 mA, which is the t=0 state of L1. Again from inspection it is obvious that the left end of C1 is at 500 mV and the right end at 250 mV. The t=0 state of C1 is therefore -250 mV.

\$\endgroup\$
  • \$\begingroup\$ I corrected the description of my problem. What I need to calculate is \$V_c(0^-)\$ (I mistakenly wrote \$V_L\$). \$\endgroup\$ – user128414 Oct 31 '16 at 11:18
  • \$\begingroup\$ Olin, in the context of a simulation, the "initial conditions" are by convention the steady-state conditions before any time-varying stimulus has been applied to the circuit. \$\endgroup\$ – Dave Tweed Oct 31 '16 at 12:08
1
\$\begingroup\$

In the steady state, you replace all inductors by short circuits and all capacitors by open circuits (supposing there is no AC applied), so you get 1 ohm in the left half and another ohm parallel to that in the right half. The source is loaded by 0.5 ohms, so it will produce 0.5 volts. Try to find the capacitor voltage yourself with this info.

\$\endgroup\$
  • \$\begingroup\$ (Please check the updated schematic) -- I supposed two current \$ I_1\$, \$I_2\$ and both running anticlockwise. After a Mesh analysis I calculated \$I_1 = -3/4 A\$ and \$I_2 = 1/4 A \$. Now I can calculate \$ V_c(0^-) = I_1*R_1+I_2*R_2\$, assuming that \$R_3\$ has a '+' in the top and \$R_1\$ has a '+' in the bottom. Am I wrong? \$\endgroup\$ – user128414 Oct 31 '16 at 14:54
  • \$\begingroup\$ Olin already posted the correct answer, upvote him if you understand it. Something must have gone wrong with your mesh analysis, though. You should obtain 0.5A in either half of the lower part, with the right half having -0.5A if accounted counterclockwise. \$\endgroup\$ – Michael Karcher Oct 31 '16 at 22:48
  • \$\begingroup\$ You're right, I made some mistakes, now my mesh analysis agrees with yours. Thank you very much. \$\endgroup\$ – user128414 Nov 1 '16 at 15:10

Your Answer

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