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Why are the currents not adding up to the current going through the R1 resistor? Why is it the same current in loop 1 going through the R1 resistor? What is preventing the current in loop 2 from adding up with the current in loop 1 in the the R1 resistor?

Loops

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  • \$\begingroup\$ The two loops are only connected at one point, assuming that nothing is connected to the two wires going off to the right. \$\endgroup\$ – Peter Bennett Sep 14 '16 at 0:14
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    \$\begingroup\$ R1 is not part of loop 2. Why do you think current from loop 2 should go through R1? \$\endgroup\$ – The Photon Sep 14 '16 at 0:14
  • \$\begingroup\$ I thought the current in loop 2 would split at the junction -- some of it would go through the R1 resistor, and the rest would go because through the current source branch. \$\endgroup\$ – Carl Sep 14 '16 at 0:25
  • \$\begingroup\$ 1. If you're working with loops, you should use KVL, not KCL. 2. To understand why the two loops don't "share" current, remember the other form of KCL: If you draw any line that divides your circuit in two parts, then the net current crossing that line must be 0. Be careful of ground connections when using this rule, but since your circuit only has one ground symbol, it's not an issue here. \$\endgroup\$ – The Photon Sep 14 '16 at 1:16
  • \$\begingroup\$ "If you draw any line that divides your circuit in two parts, then the net current crossing that line must be 0." Would you mind further illustrating this? (preferably with an image) \$\endgroup\$ – Carl Sep 14 '16 at 3:47
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All current from a current source must go back to the same current source. Since your current source doesn't have any connection to the other side of R1, none of its current will go through it.

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  • \$\begingroup\$ Why? I thought the current in loop 2 would split at the junction -- some of it would go through the R1 resistor, and the rest would go because through the current source branch. \$\endgroup\$ – Carl Sep 14 '16 at 0:27
  • \$\begingroup\$ Technically some does split there, but so does the current through the other loop. In the end you can't violate KCL. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 14 '16 at 0:29
  • \$\begingroup\$ I don't understand. Why are the currents trapped in their loops? I did a KCL on paper and I just calculated that the current source's current contributes to the current going through the resistor, but according to a solution, and the simulation, my calculation is wrong. I applied KCL correctly. I don't understand. \$\endgroup\$ – Carl Sep 14 '16 at 0:36
  • \$\begingroup\$ Your calculation is wrong beause you applied KCL incorrectly. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 14 '16 at 0:40
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    \$\begingroup\$ You haven't shown your math, so I don't understand how you expect anyone to explain that to you. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 14 '16 at 0:56
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The current from current source I1 wants to get from positive end of current source to the negative end as quickly as possible with the smallest effort possible. So, current travels to R2, then goes to that node/junction and decides: Where I am going now? To the right, where I see my destination directly, or down to R1? Why should I go to R1? I am lazy, I want to go to my destination directly. And so current flows from that node to negative end of current source.
There is nothing that would attract current to go through R1.

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  • \$\begingroup\$ So in this case: i.stack.imgur.com/VDxGt.jpg, is it the same situation? The current is stuck in the loop. Look at the solution to g11. That is what I was trying to solve. Does the same principle work here? Following your principle, a dependent current source's current shouldn't ever have an escape wherever there's a resistor. Why is the current stuck in a loop in this situation? \$\endgroup\$ – Carl Sep 14 '16 at 11:04
  • \$\begingroup\$ @Spingspoon, circuit on this picture is different than in original question! There are voltage source applied to the right side. That is a BIG difference. \$\endgroup\$ – Chupacabras Sep 14 '16 at 11:07

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