Is i_0 Zero in This KCL Circuit Analysis Problem?

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SUMMARY

The discussion centers on analyzing a KCL circuit problem with a voltage source of 54V and resistors of 12Ω and 6Ω. The participant calculated a current of 3A in the leftmost loop but questioned the validity of having i_0 equal to zero due to the apparent imbalance of currents at the node. The conclusion reached is that i_0 is indeed zero, as there is no return path for the current, confirming that the currents can be negative, indicating directionality rather than magnitude. The relationship between i_1 and i_2 is established as i_2 = 2i_1, with a current source of 9A being relevant to the analysis.

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ElijahRockers
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Homework Statement



8xi5q.jpg


if [itex]V_g = 54V[/itex]

find [itex]i_0, i_1, i_2[/itex]

The Attempt at a Solution



In the left most closed loop, Since the source voltage is 54, I assumed there must be a voltage drop across the 12Ω and 6Ω resistors equal to 54. This makes VΔ 18V.

Next I tried to KCL the node below it, but it immediately confused me. If the current around the closed loop is 3A (which is what I calculated by doing a KVL of that leftmost loop), then how can there be any more current flowing into that node? Shouldn't i0 be 0?

Then, even if it is, that means in the righthand part of the circuit, one node has three currents leaving it and none coming in, while another node will have three currents entering it and none leaving. The only way I can think of that happening is if all the currents are equal to zero, but surely this is not the answer...

Anyway, the only other work I could do was to note that the 10Ω and 5Ω are in parallel, so i2 = 2i1. Also, the current source is producing 9A, since the supplied current is supposed to be VΔ/2, but that doesn't seem to help my situation much.
 
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Shouldn't i0 be 0?
Yep! There is nowhere for the current to return...

one node has three currents leaving it and none coming in
That's just a sign thing. Currents can be negative.

i2 = 2i1. Also, the current source is producing 9A

You are on the right track -- keep going!
 

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