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A "singular" case~~ circuit problem

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Member level 2
Jun 5, 2005
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Hello ...... all
**broken link removed**

As the figure shows, I can find the current on the resistor. Can I find the current
througt the voltage sourse??
If cannot ~~~~why???
If using the ammeter in series with the voltage source to measure the current flow through the source~~~ what will I measure ????
Thanks a lot

Here's an observation, if you take any 2 of the 3 voltage sources out of the circuit, the current through the resistors would remain the same. When I say take the sources out, I mean to remove them (open circuit). What I am trying to say is that one of the voltages sources is redundant (the outer voltage loop only consists of voltage sources).

With the way that the circuit stands, the current in the voltage sources is arbitrary. However, if you assume a current in either one of the three sources, the rest can be determined. Also, if you take one of the sources out you can find all of the currents.

Best regards,

Added after 16 minutes:

Just to add a little more to the discussion ... the issue is with the voltage sources -- not with the resistors (I guess that is obvious). Consider, for example, the case where you connect two voltage sources (let's say each are 2V) directly across each other. Now note that the "+" and the "-" terminals must match; otherwise, you will violate Kirchoff's voltage law. So if the two voltages sources are and somebody asks you "What is the current in the voltage source?" -- That question cannot be answered. Just like your original problem.

Now, in reality, if you set up the circuit with the three sources, you would be able to measure (and calculate) the currents. Why? That's because the voltage sources in the problem are ideal (zero series resistance) where real source will have some series resistance. And the existence of these resistances will allow us to calculate the currents. So if you go and add three small resistors (say 0.01 Ohm each) it is possible to calculate all of the currents.

Added after 27 minutes:

If you add 10mΩ of series resistance to each of the sources, then you can solve this circuit easily using repeated applications of Thevenin/Norton. I was able to reduce the circuit down to a 29.9mΩ resistor that is being driven by a 10mV source. That resulted in the current of 0.334A (which is the current through the top 2V source). Try it out!

At the end you will find that there is positive current leaving the "+" terminal of each voltage source. All of the sources are delivering power. Current of left 2V source is 1.328A, top 2V source is 0.334A and the 4V source is 1.656A.

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