My bad... I keep putting the signs down wrong... The confusion in the first place is because ALL example on the net have the batteries ( voltage sources ) opposite so you can reference 0v at the bottom node... This is why I keep writing it down wrong.... I find the only way to get this right is to do a clockwise loop for i1 and an anticlockwise loop for i2 Then I seem to write the signs down correctly....( I did this over 25 years ago.... My brain hurts..)Those last two equations don't match. 0 = +6 +8*i2 + 12(i2 - i1) is not the same as -12*i1 + 20*i2 = 6
The latter one is correct.
.... Is it just guess work
Yes!!Wait,
I just realized I should ask and make sure. Were you trying to use mesh current analysis?
steveB,
"Seriously though, you can work out a procedure, and each person might do it in their own way, but there are basic rules to follow. In another forum, I once wrote out my way. I'll attach it here."
I am confused by your example. Why are there three currents for two loops and two equations? What is the difference between the loop2 current and I3? I would never assign polarity to passive conponents. I would let the assumed currents define what the voltage across them should be.
Ratch
Wrong! You are going to get wrapped around the axle doing that. You will get the same answer doing a KVL either CW or CCW, but be consistent with each loop. You will also find it is much easier to figure out the voltages across components common to two loops than splitting the currents between two or more loops (it's bad to do I3 = I1 - I2).
MrAl,
I believe the term (v-V2)/R3 your node equation should be (v+V2)/R3, because V2 is increasing the current out of the node.
Ratch
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