Problem in Thevenin voltage

Question

^{simulate this circuit – Schematic created using CircuitLab}

I want to find out the thevenin equivalent of the circuit across 20 micro farad capacitor as a load. I calculated the thevenin resistance which came out to be 1.5 k-ohm. Is it correct? Actually I am confused with the thevenin voltage across the capacitor. I forgot the concept of supernode which I think will be used in this circuit.

I need the thevenin circuit of the this circuit across the capacitor. Please help and please do confirm my thevenin resistance. If someone can post the full solution, it would be of great help.

Thanks in advance!!

EDIT: Some people are saying that I have done no efforts in solving this circuit, below are my calculations which show how I got the thevenin resistance of 1.5 k.

So, if I short the voltage sources and open the current sources the circuit reduces to,

^{simulate this circuit}

Where node A and B are where the capacitor was connected.

Since node C and node D are same nodes, the 12 k resistor is redundant. Or you can say that 12 k is in parallel with an ideal wire and thus the net resistance is 0, and thus nodes C and D are joined. Hence, the circuit further reduces to;

^{simulate this circuit}

The thevenin resistance thus came out to be 1.5 k. Is it correct?

Answer 1

I've redrawn your schematic a little. I made a choice about which node to call "ground" (you get to do that for exactly one node.)

^{simulate this circuit – Schematic created using CircuitLab}

If I followed your question, the dashed box surrounds the part where you need an Thevenin equivalent.

Let's continue the redrawing process just a little further. The next step is on the left and then a sort-of giant step upon moving towards the right side:

^{simulate this circuit}

You should be able to see how I got to the left side and should also be able to see why \$R_5\$ is irrelevant.

What is perhaps just a tiny bit more subtle is that \$R_1\$ and \$R_2\$ have also become irrelevant. They don't affect the current source in any way, at all. Neither does \$R_3\$, because the current flows through \$R_3\$ regardless of its value.

So the right side now shows you a reduced circuit that is quite a bit easier to apply yourself towards.

Answer 2

The simulator in the circuit editor can be used. Remove the capacitor and solve the DC voltage over R6. That's your Vth. Then replace R6 with a current meter (=short circuit). Solve the meter reading (=Ix). Your Rth = Vth/Ix.

The simulations are done in the following screenshots:

Vth=30,73V and Ix=20,49mA => Rth=1,5kOhm

NOTE: This isn't the solution of your homework, it's only the result.

A hint: When there's no controlled sources, you can find Rth also by replacing all voltage sources with wires, disconnecting the current source and calculating the resistance seen by C1. It's 6kOhm and 2kOhm in parallel.

How to find Vth manually

General circuit analyzing methods work. Just in this case the superposition method seems to be especially useful. Draw three versions of your circuit without C1 and in all versions draw an arrow for Vth = the voltage over R6 considering the plus side to be at right.

Version1: Bat1 active, Bat2 replaced with wire, I1 disconnected

Version2: Bat1 replaced with wire, Bat2 active, I1 disconnected

Version3: Bat1 and 2 replaced with wires, I1 active

Vth can be solved in all versions easily by combining serial and parallel resistors and using ohms law + voltage divider formula

Finally add all solved Thevenin voltages. They are the parts of the total Vth caused by different sources. Be sure to every time assume Vth to be directed from right to left. You'll get minus voltage if the actual direction was different.

Version 3 as an example:

If you give the papers to someone for inspection, you should use different symbols for Vth in different versions, for example Vth1, Vth2 and Vth3 to avoid a direct stamping as "FAILED"