Well i understand how to perform the mechanics behind mesh analysis, but notice that 800ma does not 'flow' through R2. If 800ma "really" flows through R2 then we could not have 600ma flowing through R3 because the sum of currents entering a node is the same as the sum leaving the node. 800ma through R1 and 800ma Through R2 and 600ma through R3 means we have a deficit of 600ma. 800-800-600=-600, so something there cant be realok, here is the steps to a solution:

Define mesh equations:

View attachment 275003

Simplify:

View attachment 275004

Add 2 times last equation to top equation to eliminate I2 and then solve for I3:

View attachment 275005

Solving for I2:

View attachment 275006

The voltage across R1 is: R1*I2=10*4/5=8V

The voltage across R2 is: R2*(I2-I3)=(4/5-3/5)*10=2V

The voltage across R3 is: R3*I3=20*3/5=12V

No issues becasue:

The voltage around loop 2 is:

-10+8+2=0

The voltage around loop 3 is:

-10+-2+12=0

Verified.

But I2 is 800mA and that is going through R1 and must be coming from V1.

And I3 is 600mA and is going through R3

The current through R2 is 800mA-600mA which is 200mA and thus a 2 volt drop is correct.

The loop currents are what I expect. Not sure what you are trying to say?

The voltage across R3 is not to surprising being when you look at the outer loop the the two voltage sources are in series allowing for up to 20V in the circuit.

View attachment 275007

Yep 800mA is the current through R1 and that happens to be the I2 current that I calculated so it is the I2 current which is 800mA?

So WBahn is correct: The calculated I2 current is 800mA and that is the current going through R1.

Also because both currents have positive values, the assumed direction of current flows (clockwise) is correct.

Resistor R2 is not a calculating machine, it does not know how to add and subtract. If you look inside you do not see 800ma going one way and 600ma going the other way, you just see 200ma going one way. What really happens is at the top central node the 800ma splits into two currents, one 600ma and one 200ma. At the bottom node the 600ma combines with the 200ma and produces 800ma. So there is no addition or subtraction going on in R2.

But as i said before, that's not the only thing we do in circuit analysis that is not 'real', even with other circuits, so this starts to look like a common thing.

Someone mentioned the square root of minus 1: i=sqrt(-1) as being not real but the mesh current is real. But we also use sqrt(-1) in lots of analysis and come out with real values. How is that possible? Because it only has to be 'imaginary' for part of the calculation, and then if it still appears at the end there becomes a different interpretation of that 'quantity'.

We use things that we might call imaginary in lots of calculations so this should not be that big of a surprise.

If you would still like to call it 'real' that's your decision. In the end we still recognize mesh analysis to be a useful tool just like many other types of analysis we do.

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