0
\$\begingroup\$

I am once again stuck on a task. The circuit initially looks like this:

enter image description here

And the question is:

enter image description here

So my attempt:

The first thing that came up in my mind was that the switch had been closed for a long time. That made me think that initially the capacitor behaves like an open circuit. So I redrawed the circuit like this:

enter image description here

Where Vc = Vab

So I thought that I could find Req and find Vab through the voltage divider. However there are two things that makes it difficult form the to continue:

  1. The question states that when t = 1 μs then the switch opens, but the plot of Vc(t) should be from 0

  2. I don't know how I should consider the current source. Most problems that I solved have either had a current source or a voltage source. How should I look on the circuit when I have a current and voltage source?

Thanks in advance!

Edit:

Here is my new circuit after some help, however on the final state I am uncertain if the switch should be open and the capacitor not acting like an open circuit? :

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ You can use superposition. Solve the circuit with the current source opened, then solve the problem with the voltage source shorted, then sum the results for node voltages and branch currents. You can also replace the voltage source with a Norton equivalent current source. \$\endgroup\$ Commented Apr 21, 2020 at 12:40
  • \$\begingroup\$ You can also just solve the circuit using KVL or KCL systems of equations. \$\endgroup\$ Commented Apr 21, 2020 at 12:42
  • \$\begingroup\$ What is the direction of I0? \$\endgroup\$
    – Bart
    Commented Apr 21, 2020 at 13:10
  • \$\begingroup\$ @Bart that is not stated in the question, so I don't know. \$\endgroup\$
    – Vetenskap
    Commented Apr 21, 2020 at 13:18
  • \$\begingroup\$ @Vetenskap: It can't be deduced from the schematic, but it is important information. Failing this, you would need to solve for two cases, current up and current down. \$\endgroup\$
    – Bart
    Commented Apr 21, 2020 at 13:20

1 Answer 1

1
\$\begingroup\$

I'm assuming that the current direction of the current source is as shown below: -

enter image description here

I don't know how I should consider the current source. Most problems that I solved have either had a current source or a voltage source. How should I look on the circuit when I have a current and voltage source?

My preference is to make the voltage source \$V_A\$ into a current source but first...

Your redraw is correct but there's one more thing you can do. R1 (in series with the current source) has no effect on the circuit because it is purely in series with a current source. It's the same for resistors in parallel with a voltage source - they do not affect the voltage source and can be turned into open circuits. In R1's case it becomes a short circuit.

Then, turn \$V_A\$ into a current source of 6 mA in parallel with R3. Do you see what I did here? The effective series combo of 6 volts and 1000 ohms become a current source of 6 mA in parallel with 1000 ohms.

So now you have R2 || R3 being fed by a current source of 2 mA (from the left) and a current source of 6 mA from the right. Total current is 8 mA into R2 || R3 (500 ohm) or, put another way, 4 mA flows into R2.

That sets the initial charged capacitor voltage.

Can you take it from here?

\$\endgroup\$
10
  • \$\begingroup\$ I've redrawn the circuit now, so that I have the current source 4 mA in parallel with R2 (1kΩ) and on the left side of the R2 should my capacitor still be like an open circuit or in this final state should I consider the switch open again? \$\endgroup\$
    – Vetenskap
    Commented Apr 21, 2020 at 13:17
  • \$\begingroup\$ The help you asked for was in order so that you could determine steady state Vc prior to the switch opening. Vc (in my answer) is shown to be 4 volts (4 mA flows into R2). Now you have to start over with the knowledge that Vc is charged to 4 volts and the switch is open so, you don't have R3 and \$V_A\$ any more. \$\endgroup\$
    – Andy aka
    Commented Apr 21, 2020 at 13:24
  • \$\begingroup\$ and on the left side of the R2 - the only thing on the left of R2 from what I can see is the current source and R1 (not needed). I can't see your redrawn circuit but i can tell you that there is no 4 mA current source involved - that original 4 mA was only to calculate the steady state voltage of C. \$\endgroup\$
    – Andy aka
    Commented Apr 21, 2020 at 13:27
  • \$\begingroup\$ Sorry I meant right side. I put a schematic on the original post. \$\endgroup\$
    – Vetenskap
    Commented Apr 21, 2020 at 13:29
  • \$\begingroup\$ OK I see that now. That tells you the steady state value of voltage across Vc i.e. 4 volts. But that isn't what you use to progress the problem after switch opens. That just tells you the steady state voltage i.e. 4 volts prior to the switch opening. \$\endgroup\$
    – Andy aka
    Commented Apr 21, 2020 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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