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Physics Hypertextbook writes that "The current through the 1 Ω resistor most certainly runs from right to left."

Why is that true?

My approach:

I arbitrarily assigned a counterclockwise direction to the current running through the 1 $\Omega$ resistor. I then applied the junction rule at the node connecting the 2 $\Omega$ Resistor the 20V emf and the wire leading to the 1 $\Omega$ Resistor:

I$_{1\Omega}$ + I$_{2\Omega}$ = I$_{emf}$

where I$_{2\Omega}$ = 4. If the labeled 3A current loops down and to the right.

Also, I$_{emf}$ = 3

It can be shown that the current running through the 2 $\Omega$ resistor is 4A: By the loop rule (for the smaller right-hand circuit): 20V - 2I - 4*3 = 0; I = 4.

If the I$_{emf}$ < 0 (which, by my reasoning, it is), the current actually runs clockwise.

  • $\begingroup$ The voltage on the center node of the three resistors has to be I*R=3A*4Ohm=12V. That leaves a voltage drop of 8V on the 2Ohm resistor, leading to a current flow of 4A and 4A-3A=1A flowing trough the 3Ohm resistor into the unknown voltage source. So that's U=1A*3Ohm=3V of drop, leaving us with 12V-3V=9V for the unknown voltage. So, yes there would be 11A flowing from right to left trough the 1Ohm resistor. $\endgroup$ – CuriousOne Jan 6 '15 at 10:14

The earlier part of the solution (which can be found here) finds that $I_2$ and $I_3$ both have current from right to left. So the left side must be at lower potential. The passive path on the top loop must likewise carry current from right to left.

Yes, $I_2$ is $4A$, but since you haven't calculated the current through the $20V$ source (it isn't $3A$!), you can't use that node to find $I_1$ yet.


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