# Help in understanding basic network analysis (mesh)

Discussion in 'General Electronics Chat' started by Loonie, Aug 30, 2011.

1. ### Loonie Thread Starter New Member

Aug 30, 2011
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0
Hi all, I have recently enrolled in an Electrical engineering course in a university after 2 years of mandatory military service. In that period of time, my brain have gotten rusty and I have forgotten many of the mathematical and physic concepts taught in high school/college. To my horror, I find myself struggling with questions for hours while the China scholars around me can finish them in minutes. (They are also quite friendly, but when I asked them for them for help, I couldn't understand though their accent, and they just pointed here and there and wrote down some formulas which I couldn't understand.) I did seek help from tutors and the net and things got clearer, but I do have some uncertainties.

For example,

For the Mesh loop l2, seeing that the direction of the mesh current is counterclock wise, why isn't the equation of the mesh:
R2(l2+11) + R4(l2+l3)+R3(l2)=0
I do know both of their sums = 0 but why is the negative convention taken?

Could anyone also enlighten me on this question?

Here are the workings I did (spent many days brute forcing through)

I do have many more pages of workings (tried over many hours and days) but I hope these examples will be able to tell you what I am missing (in conceptual knowledge) =(

This is my final attempt to solve the problem.

As IL is obviously not near the value of 4.44A, (answer is 4.44 + 0.09), I have failed yet again...

Could you guys please help (and apologies if I posted in the wrong forum). I am spending thousands for this university education, and find myself depressed not to be able to understand anything at the start.
Thank you

Last edited: Aug 30, 2011
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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You deserve some credit for perseverance!

I've attached a solution method ...

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3. ### Loonie Thread Starter New Member

Aug 30, 2011
17
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Thank you t_n_k, although the values of the 2 resistors were mixed up, I still got the correct answer using your method.

However, why isn't the load current, IL = I1 + I2 according to KCL? I see that there I1 and I2 are both going into the node on top of Battery 1, and IL coming out from it. But you took I2 = IL for the equations, and it was the correct choice.

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4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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It's good practice that you reversed the sources to find the same answer. In fact it doesn't matter in which orientation the parallel sources are arranged - as the result confirms.

The load current is in fact i2. The mesh currents are notional currents which flow in a user determined path, which can be completely arbitrary as long as the loop is complete - starting and ending at the same point. For the assigned mesh currents (i1 & i2) on our schematic, only mesh current i2 flows in the load. Mesh current i1 is "constrained" by our arbitrary choice to flow only in the left hand loop.

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5. ### hgmjr Moderator

Jan 28, 2005
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Greetings loonie,

In the sample problem you referenced in your original post, one thing to bear in mind is that the equation

$-R_2\left(I_2\ +\ I_1)\ -\ R_4\left(I_2\ +\ I_3)\ -\ I_2R_3\ =\ 0$

is the same as the equation

$R_2\left(I_2\ +\ I_1)\ +\ R_4\left(I_2\ +\ I_3)\ +\ I_2R_3\ =\ 0$

The only difference between these two equation is that the lower equation is nothing more than the upper equation with both sides of the equation mulitplied by -1. In algebra, when you mulitply both sides of any equation by the same value then you do not change the equality expressed by the equation.

In this example, it is somewhat confusing to those unfamiliar with the KVL circuit analysis technique to throw in this switch in the approach being applied to the forming of the loop equations.

For the first and third loop equation shown, it can be seen that the approach used to determine the sign of each of the terms was to assign a negative sign to those currents entering the positive terminal of the element in the loop and a positive sign to those currents entering the negative sign of the element in the the loop. For some reason, the solver elected to change horses in the middle of the stream. While perfectly legal, it is still poor teaching form when the audience for this tutorial text is expected to be encountering these concepts for the first time.

hgmjr

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