From Andre LaMothe's course on Udemy, I'm trying to figure out how I can ascertain the voltage drops across node 1-2 and node 2-3.

I came up with 2 equations but they aren't enough to solve the problem for me as there are 3 unknowns.

Voltage across node 1-2 is V1, and node 2-3 is V2. Total current is I.

(1) 12 = V1 + V2


(2) I = V1/24 + V2/17 + (50*10^-3)

Any help on the thought process for working through this one?

Edit: I just noticed my mistake. Equation (2) is incorrect. Total current I is just the same as V1/24. I.e. current going in to node 2 is the same as current going out of node 2. So eqn 2 becomes the following.

V1/24 = V2/17 + (50*10^-3)

Only 2 unknowns, so can be solved.

  • \$\begingroup\$ Use Norton's (or it is Thevenin?) theorem and convert the current source to voltage source. \$\endgroup\$
    – Eugene Sh.
    Mar 15 '21 at 20:48
  • \$\begingroup\$ @EugeneSh ah, I think that was the missing piece of the puzzle! \$\endgroup\$
    – JC123
    Mar 15 '21 at 20:49
  • \$\begingroup\$ Your current equation is not correct. You need to mark the names and directions of the currents, then apply KCL at node 2. \$\endgroup\$ Mar 15 '21 at 21:06

The best way to solve a circuit with multiple sources of voltage and current is by using the superposition principle: Remove the current source from the circuit and replace it with an open circuit.

Calculate currents and voltages.

Then return back to your original circuit and remove the voltage source and replace it with a closed circuit.

Recalculate currents and voltages.

Then for currents which pass through the same node add those currents.

And for voltages between the same nodes add those voltages.

(Don't mix up the signs).

  • \$\begingroup\$ Again, you are saying something that is not true. "The best way to solve a circuit with multiple sources of voltage and current is by using the superposition principle" - this is a really naive thing to say. Solving multiple source problems is usually done best using Millman's theorem but it totally depends on the detail so you shouldn't be making such a generalism in my book. Notice that I'm not downvoting your answer. \$\endgroup\$
    – Andy aka
    Mar 19 '21 at 17:24
  • \$\begingroup\$ I have never heard of "Millman's theorem" and reducing a circuit to less complex circuits and then add them up is quite easy and avoids errors, don't you think? \$\endgroup\$
    – Miss Mulan
    Mar 20 '21 at 17:51
  • \$\begingroup\$ If you have never heard of Millman's theorem then you are in no position to assert what is a best method. That is my point. Your answer is naive and you are also too loose with your downvotes on perfectly good answers. \$\endgroup\$
    – Andy aka
    Mar 20 '21 at 18:50

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