Your equation is wrong. And you do not see it because you do not track the units properly.Smaller circuit for i1: 11 - 62 -19I1 = 0
i1 = -2.68
Several issues.I0 = I1 + I3
Smaller circuit for i1: 11 - 62 -19I1 = 0
i1 = -2.68
i3 = 11 - 15I - 2.68 - 9 = 0
i3 = -.045
Several issues.
First, your thread title indicates that you are supposed to use KVL, yet you're working with KCL equations.
Second, your equations use currents that aren't defined in your diagram. What is I0? I have no idea, because you haven't defined it.
Third, you are making mistakes that would be trivially easy to catch if you had tracked your units properly, but because you aren't do so, you don't catch it and then are left wondering why you aren't getting the correct result.
"Smaller circuit for i1: 11 - 62 -19I1 = 0"
should be
(11 V) - (62 Ω) - (19 Ω)(I1) = 0
So you are subtracting a resistance from a voltage. Not physically possible to do that, so you KNOW the equation is wrong right at this point. No need to waste your time going any further until you address this.
The "62" is from the resistance of R1, which is 62 Ω. Similar for the "19" and R2.So why is it not:
- (62) ohms - (19 ohms) (i1) = 0, I thought that 62 and 19 were voltage drops.
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by Jake Hertz
by Jake Hertz
by Jake Hertz
by Jake Hertz