Help in figuring out what i did wrong

Thread Starter

enemias

Joined Sep 21, 2019
10
upload_2019-9-21_14-15-15.png

I cant seem to find my mistake. the answer is 0.0233 A. Do i use node voltage to solve it? Or is my mesh currents wrong? Or do i need a supermesh?
 
Last edited:

WBahn

Joined Mar 31, 2012
24,854
Your notations on the diagram have made it impossible to tell.

Your current direction arrows for I1 and I3 are obscuring the notations underneath them, but we can make them out with some level of confidence. However, you scratched out the value of the right-hand voltage source and so there's no way for anyone to solve the problem since they can't tell what the problem is.

Then the first thing that you show in your work is "V: 1190 V"

What voltage is that referring to? Where does the information that it is 1190 V come from?

Where does your equation

2000 (I1 - I3) + 2750·I3 = 0

come from?

(2000 Ω)(I1 - I3) would be the voltage across the middle-left 2 kΩ resistor (left side relative to right side), so somehow the 2750·I3 must be the voltage gain between those same points along some other path. But there is no combination of the resistor values than can cough up 2750 Ω, so where did that magic value come from?
 

Thread Starter

enemias

Joined Sep 21, 2019
10
upload_2019-9-21_15-39-41.png
sorry about that.

what i determined was Vb is 1100I3. and when i did the KVL on the since at the top its 1100ohms with I3 current. When i did my first loop i multiplied 2.5 by 1100I3. and thats what i got
 

WBahn

Joined Mar 31, 2012
24,854
View attachment 186673
sorry about that.

what i determined was Vb is 1100I3. and when i did the KVL on the since at the top its 1100ohms with I3 current. When i did my first loop i multiplied 2.5 by 1100I3. and thats what i got
I have NO idea what "and when i did the KVL on the since at the top its 1100ohms with I3 current" means.

What path are you doing KVL around? I'm guessing I1?

You should make your work flow very clear. Imagine you are doing this for pay (which, in essence, you are -- you are getting paid with a grade) and you only get paid if an auditor (known as a grader in this case) can readily follow your work and determine its correctness.

One way you could make this pretty clear would be to start out:

Circuit with 90 V source set to 0 V.

KVL around I1:

(2 kΩ)(I1 - I3) + (2.5 vb) - (0 V) = 0
vb = R·I3 = (1.1 kΩ)(I3)
(2 kΩ)(I1 - I3) + (2.5)(1.1 kΩ)(I3) = 0
(2 kΩ)(I1 - I3) + (2.75 kΩ)(I3) = 0

Do you see how that makes your line of reasoning evident so that anyone evaluating your work can follow it?

You also need to start tracking your units properly throughout your work -- and that does NOT mean just taking whatever units you want the answer to have onto the end, either.
 

The Electrician

Joined Oct 9, 2007
2,750
In the bottom (black) portion of post #3, you have two cases. For the first case, KVL without 90V, you got two currents I1 = 0.002985 and I2 = 0.015622. Next, for KVL without 40V, you got I1 = 0.034925 and I2 = -0.030224. Next I would have expected you to add these results and get values for I1 total and I2 total. In the upper (white) portion of post #3, I see a couple of values for I1 and I2; where did these values come from? Where is your summation of the currents from the two cases?
 
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