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The problem is stated as follows: Find the value of Vs in the circuit below such that the power supplied by the current source is 0. enter image description here

I realize that voltage across the current source has to be zero for the power supplied to be zero. I've tried to solve the problem using nodal analysis by grounding the node in between the 3 and 8-ohm resistors and am getting an answer of -72 V (for Vs) but am not confident in that answer. Would solving the problem using KVL and KCL equations be better? Please help. Thank you.

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2 Answers 2

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Draw the circuit and simplify it in steps, as shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

The blue arrows show you the steps I took here. The first step just reordered the two voltage sources in order to get the resistors into position for easier understanding. The next step combines the resistors (no harm in doing so.) The final step sets a "ground point" (which you are allowed to do in exactly one place.) Then, by definition, you also know that there is a virtual ground up at the top of the current source, of course, as you point out must occur.

I would imagine this would be pretty easy to solve now.

Is this sufficient help?

Or are you looking to just "stamp out" rote mathematics to achieve the same thing?


Noting your comment:

"I'm just a little confused how you can just switch around the voltages and the resistors to get them in series. Doesn't the original problem show the resistors in parallel?"

The \$2\:\Omega\$ and \$3\:\Omega\$ resistors are not parallel. They are in series. Similarly, the \$6\:\Omega\$ and \$8\:\Omega\$ resistors are not parallel. They are also in series. Let's look at the first schematic (upper left corner) again, and then the transformation moving to the right upper left corner (from above):

schematic

simulate this circuit

No difference at all from the earlier examples above. But now they look like they are in series.

You need to break yourself of the idea that if two resistors are drawn parallel to each other on paper, then they must be in parallel as far as a circuit is concerned.

A schematic drawing is just the way a human draws something in order to (hopefully) communicate accurately with another human. But like any other communication between humans, it's possible to lie, dissemble, and be disingenuous. To trick you or to pull the wool over your eyes, so to speak. Though sometimes it's just a matter of having the time to draw out a tight schematic that communicates well. As Bjarne Stroustrup has been quoted,

"I would have preferred this paper to be either much longer or much shorter, but I did not have the time to do either."

Part of learning circuitry is learning how to tear down someone else's "statement" and taking the time needed for rebuilding it back into something that communicates better.

To paraphrase from Dr. Rudolf Flesch in his book, "The Art of Plain Talk," as an example of bad writing:

"An ultimate consumer as used here means a person or group of persons,
 generally constituting a domestic household, who purchase eggs generally
 at the individual stores of retailers or purchase and receive deliveries
 of eggs at the place of abode of the individual or domestic household
 from producers or retail route sellers and who use such eggs for their
 consumption as food."

Clearer is,

Ultimate consumers are those who buy eggs to eat them.

It's the same way with schematics.

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  • \$\begingroup\$ Nice....those guys at uni only told us about shorting out one source and calculate for others and do recursively...never thought of solving these kind of problems in this way...damn I could have better grades :D but \$\endgroup\$
    – Dumbo
    Commented Oct 6, 2017 at 22:36
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    \$\begingroup\$ @Sean87 There are two facets: (1) "How to think about the world around you"; and, (2) "How to deal with a complex world you've decided you cannot apprehend well, but for which you still need to calculate." Computers all have to take #2 because, quite frankly, they cannot think. But humans can. So a person should be taught to do #1, if possible, and then rely on #2 when that fails. The recursive method you mention is #2 -- it gets you there when you have decided the problem cannot be understood in simpler terms on its own. \$\endgroup\$
    – jonk
    Commented Oct 6, 2017 at 22:43
  • \$\begingroup\$ Thanks for your response. I'm just a little confused how you can just switch around the voltages and the resistors to get them in series. Doesn't the original problem show the resistors in parallel? \$\endgroup\$
    – user165063
    Commented Oct 6, 2017 at 23:59
  • \$\begingroup\$ @user165063 No, your eyes say that the resistors are in parallel because someone drew them that way. If only electrons could see what your eyes see. But they don't. Note that all of the current through the 2 ohm resistor must also flow through the 3 ohm resistor. They are in series. \$\endgroup\$
    – jonk
    Commented Oct 7, 2017 at 0:13
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I think the solution is rather simple in this case. The \$V_s\$ source must be adjusted so that it absorbs the current generated by the 18-V source when the 3-A source voltage is 0 V plus the 3 A from the source itself. The current delivered by the 18-V source with 0 V across the current source is simply \$\frac{18}{3+2}=3.6\;A\$ to which you add the 3-A contribution and you obtain a total current flowing in \$V_s\$ of \$6.6\;A\$. The \$V_s\$ source must thus absorb the current to "null" the output. The source voltage is then negative and equal to \$V_s=-6.6\times(8+6)=-92.4\;V\$. A quick sim as below in which you display the operating points confirms the result. The source in the right side auto-adjusts to "null" the voltage across the current source.

enter image description here

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  • \$\begingroup\$ What is the sim software? Doesnt look like ltspice \$\endgroup\$
    – Dumbo
    Commented Oct 7, 2017 at 8:29
  • \$\begingroup\$ This is Intusof's IsSpice (the demo version) but LTspice will do the same. \$\endgroup\$ Commented Oct 7, 2017 at 9:45
  • \$\begingroup\$ Thank you! I got -92.4 V as the answer now and see how the current actually flows through the resistors as if they were in series. \$\endgroup\$
    – user165063
    Commented Oct 7, 2017 at 22:48
  • \$\begingroup\$ Glad if I could help! \$\endgroup\$ Commented Oct 8, 2017 at 7:34

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