# Mesh Analysis

Discussion in 'Homework Help' started by hamid-reza, Oct 31, 2013.

1. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
0
I just don't understand the third equation in page:

First of all I3 is counter clockwise (as shown in pic) so why the equation shows as if it is clockwise???

Next I have no idea why in equation there is 150(I3+I1) & 300(I3-I2) since I3&I1 go through completely different direction.
The same thing for I2&I3 although theyre in the same direction we have 300(I3-I2)?
Thanks for helping.

2. ### anhnha Active Member

Apr 19, 2012
776
48
Hi,

What is the direction of I3 flowing through R1, from A to B or B to A?
What is the direction of I1 flowing through R1, from A to B or B to A?

Now do you see that they are in the same direction?
What is the direction of I3 flowing through R4, from A to D or D to A?
What is the direction of I2 flowing through R4, from A to D or D to A?

Now do you see that they are in the different direction?

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3. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
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Yes you are correct in picture you are right. but look at the equation number 3 if we chose the direction of I3 counter clockwise why the battery is +24- ....?
I think it would be -24+300(I3-I2)+150(I3+I1).

4. ### anhnha Active Member

Apr 19, 2012
776
48
Ah, I see your problem now.

Let's apply KVL for the loop on the left hand side:

$V_{AB} +V_{BD} + V_{DA} = 0$

Where:

$V_{AB} = V_{B} - V_{A}$

$V_{BD} = V_{D} - V_{B}$

$V_{DA} = V_{A} - V_{D}$

And you can check if this equation is correct:

$V_{AB} +V_{BD} + V_{DA} = 0$

OR:

$( V_{B} - V_{A}) + (V_{D} - V_{B}) + ( V_{A} - V_{D}) = 0$

$V_{AB} = 150(I3+I1)$

$V_{DA} = 300(I3-I2)$

What is VBD?

$V_{BD} = V_{BD} = V_{D} - V_{B} = -24$

And we have:

$V_{AB} +V_{BD} + V_{DA} = 0$

OR:

150(I3+I1) -24 + 300(I3-I2) = 0

Can you simplify this?

Last edited: Oct 31, 2013
5. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
0
So my answer was correct but I still can't understand why the third equation is this:
Look at the 'Original form of equation' :

Why do we have 150(I3+I1)? I3 goes through R1 in completely different direction from I1 (Assuming the direction I1&I3 clockwise)

So, why don't we have 150(I3-I1)?

Last edited: Oct 31, 2013
6. ### anhnha Active Member

Apr 19, 2012
776
48

What is the direction of I3 flowing through R1, from A to B or B to A?
What is the direction of I1 flowing through R1, from A to B or B to A?

Now do you see that they are in the same direction?

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7. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
0
Yes both flow through R1 from A to B but look at the battery polarity it makes a clockwise current so if we assume I3 counter clockwise we have:
-24+300(I3-I2)+150(I3+I1)

Correct me if I'm wrong.
Sorry for asking so many questions.

8. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
0
I think I answered my own question sorry for taking your time dear anhnha.
Thank you so much.

9. ### anhnha Active Member

Apr 19, 2012
776
48
I think that may work.
Let's look at it this way. Sorry, I can't explain it more clearly. Hope someone will help you understand it better.

Let's apply KVL to the loop on the left hand side:

$V_{AB} +V_{BD} + V_{DA} = 0$

Where:

$V_{AB} = V_{B} - V_{A}= 150(I3+I1)$

$V_{BD} = V_{D} - V_{B}$

$V_{DA} = V_{A} - V_{D}= 300(I3-I2)$

Now let's consider this:

$V_{BD} = V_{D} - V_{B}$

On the other hand, the battery is 24V this means that the voltage between positive and negative is 24V. This means:

$V_{B} - V_{D} = 24V$

Therefore:

$V_{BD} = V_{D} - V_{B} = -24V$

Finally we have:

$V_{AB} +V_{BD} + V_{DA} = 0$

OR:

150(I3+I1) -24 + 300(I3-I2) = 0

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10. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I think a big part of your problem is that whoever set things up didn't use a consistent direction for their current loops. Let that be a lesson. Get in the habit of ALWAYS using the same direction for all loops. Personally, I run all of my loops counter-clockwise. Yes, there are times that choosing a loop going the other way would be more convenient for a particular problem, but I have long since found that using a technique consistently will reduce the errors I make overall enough to more than compensate for any inconvenience on a problem here or there.

Another thing that this problem does is use carrier flow instead of charge flow. It seems to use it consistently, so it's valid.

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11. ### bwd111 Member

Jul 24, 2013
117
1
Thats a tough one

12. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Not really, IF you are systematic about it.

You first have to decide if you are going to sum up the voltage gains or the voltage drops as you go around the circuit.

You then have to decide if your "currents" represent charge flow or carrier flow. If it represents charge flow, then it is a true current and if you pass through a resistor in the direction of charge flow you always drop in voltage by an amount that is equal to I*R.

If you are using carrier flow, then your "current" may not represent a true current, which is CHARGE per unit TIME. If your carriers are electrons, then they are negatively charged and you have to account for the fact that the true current is the negative of the carrier current. This is done by noting that if you pass through a resistor in the direction of carrier flow, you gain in voltage an amount that is equal to I*R or, putting it another way, you still drop in voltage by and amount that is equal to the charge current times the resistance, but the charge current is -I and so you drop by -I*R.

In either case, as you go through a voltage source you drop in voltage if you are going from the positive to the negative terminal and gain in voltage is you are going the other way.

In this problem, whoever set up the solution has chosen to use carrier flow and to sum up the voltage drops as you go around the loop in the direction of the loop flow. So:

Mesh I1:
24V + (-I3+I2)R4 + (-I3-I1)R1 = 0
24V - I3(R1+R4) + I2(R4) - I1(R1) = 0
- I1(R1) + I2(R4) - I3(R1+R4) = -24V
- I1(150Ω) + I2(300Ω) - I3(450Ω) = -24V

IF a uniform direction had been chosen for all of the mesh currents, then the three mesh equations can be written down by inspections. Using charge flow (a.k.a., conventional current), each equation will have a coefficient for each mesh current that is the sum of the resistors that that current flows through that are in common with the mesh. The mesh current in question will be positive and the bordering currents will be negative. That's the left hand side. The right hand side will then be the sum of the voltage gains due to the sources.

Hence, for mesh currents going counterclockwise, written by inspection:

+ I1(300Ω) - I2(100Ω) - I3(150Ω) = 0
- I1(100Ω) + I2(650Ω) - I3(300Ω) = 0
- I1(150Ω) - I2(300Ω) + I3(450Ω) = -24V

A couple of easy checks can then be made. First, the coefficients are symmetric about the diagonal. If they aren't, then something is wrong. Second, the diagonal coefficients will count every resistor exactly twice unless the resistor is on the periphery, so the total of all the resistors in the circuit should come to 50%*(total of diagonal coefficients+total of periphery resistors). In this case, thats 50%(1400Ω+300Ω)=850Ω, which indeed is the sum of the resistors in the circuit. Another way of saying this is that the sum of the peripheral resistances must be equal to the sum of the diagonal coefficients minus the sum of all of the off-diagonal coefficients, which it is.

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13. ### hamid-reza Thread Starter New Member

Oct 31, 2013
6
0

This is really helping but I don't know why?
Every time I choose loops counter clockwise I get the right answer.
Thanks WBahn.