This question requires you to find the node voltage between nodes 1 & 2 as well as condense the circuit into its Thevenin Equivalent with R2 as the load. The correct answers are 4 V between nodes 1 & 2 and 36.5 V & 7.8 for Thevenin Voltage & Resistance respectively. Could someone please explain this to me? I'm especially confused as to why the Thevenin Voltage is higher than the source voltage. Thanks in advance.


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    \$\begingroup\$ But notice that you also have a constant current source in your circuit. And this current source will also provide a power. The voltage drop across R1 resistor is 5mA x 6.3kΩ = 31.5V \$\endgroup\$ – G36 Dec 3 '18 at 17:42
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    \$\begingroup\$ Step 1 of the nodal method is to label your nodes. If you do that it will make it easier to discuss your circuit. \$\endgroup\$ – The Photon Dec 3 '18 at 18:03
  • \$\begingroup\$ @G36 thanks for the feedback. Does KVL still apply to that circuit or not, and if so how? I'm struggling to reconcile the two ideas and I think that's where I'm confused. \$\endgroup\$ – Qwerty Dec 3 '18 at 20:31

Redraw the schematic so that the whole thing isn't quite so confusing. (Positive at the top, negative at the bottom, signal input at left, signal output at right. Don't bus power rails around -- it's confusing to do that.)


simulate this circuit – Schematic created using CircuitLab

Note that I've used the schematic editor that is available to you when you create a question here? You should use it, too.

You should be able to see where I got the \$-5\:\text{V}\$ node point. I've labeled all of the remaining possible nodes (ground and \$-5\:\text{V}\$ are already labeled, obviously.)

At this point, I can see that you really only need to do nodal analysis on \$V_X\$. The other nodes are almost certainly unimportant.

Also, you should be able to see that \$R_4\$ is completely irrelevant since the current source has \$\infty\$ impedance. You would be perfectly in your rights to simply remove it and everything "behind" the current source. The only thing that matters is what it injects into the \$V_X\$ node.

Some notes here, assuming you were NOT doing nodal analysis:

  1. You should be able to see that \$R_2\$ and \$R_3\$ can be summed up into an equivalent resistor value.
  2. You should be able to see that once (1) is complete, that you are left with a simple voltage divider made up of two resistors. You should be able to Thevenize this almost immediately into an equivalent voltage source and an equivalent series resistance from it.
  3. Given the current source, this then means you can work out the voltage at \$V_X\$, almost trivially.

Of course, if you do the above all you've done is avoid the nodal analysis.

At this point, you really should be able to perform your nodal analysis. You can include a statement for both \$V_X\$ and \$V_Y\$ and solve them simultaneously. Or you can lump \$R_2\$ and \$R_3\$ and only write one equation for \$V_X\$. Which method you apply depends entirely on what you feel is more appropriate to your circumstances.

  • \$\begingroup\$ Thanks! I'm relatively new to circuits in general, so I'm a bit unfamiliar with the schematic you posted above. Once I get home from school I'll take a better look at it. In the meantime, which node is ground? Additionally, is converting circuits to its Thevenin equivalent the best way to analyze circuits, or does each method has its own merit? \$\endgroup\$ – Qwerty Dec 3 '18 at 20:35
  • \$\begingroup\$ @Qwerty Each method has its own merits. And it helps a lot to use all of them because it gives you various ways to "look" at something. The circuit you provided already has a ground shown. All I did was to retain it in my re-drawing of it. The two schematics are identical in every detail. One important thing you should learn to do, soon, is to remove wires that do nothing more than bus voltage rails around. These serve little purpose except to keep you from being able to focus on what is important. They distract. So learn to remove them. And stick with (+) on top, (-) on bottom. \$\endgroup\$ – jonk Dec 3 '18 at 20:51

enter image description here. According to thevenin's theorem we can solve difficult circuit analysis.That is the reason we solve the above problem easily


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