mesh analysis to find currents

lemon

Joined Jan 28, 2010
125
Hi:
I need to use mesh analysis to find iA, iB and iC on the attached image. Please tell me whether my approach is correct.
I decided that:
Loop1, on the left is anti-clockwise (iA)
Loop 2 is clockwise (iB)

Mesh1 gave: -21+10iA+6iB=0
Mesh2 gave: -6iA-18iB+28=0

after solving I got:
iA=1.4583 Amps
iB=1.0694 Amps

But I'm confused about iC and the directions of the currents on the diagram.
How do I find iC and why are the directions of currents going in a non-senical direction?

Thank you

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ke5nnt

Joined Mar 1, 2009
384
In mesh currents, all directions are generally assumed to be the same. If your current value for a current comes out as a negative value, then the current just flows the other direction. You have 2 meshes, assume current is clockwise in both. Write 2 loop equations making sure to remember the effect of both loops through the 6K Ohm resistor. The loop equations are derived from Kirchhoff's Voltage Law which states that the sum of all voltage drops in a closed path will equal the applied voltage. Thus, using the unknown currents Ia and Ic and the known resistor values, you can state loop equations in terms of those currents where the resistance times the unknown current is considered a "voltage drop" (remember V=IR).

Loop 1 (for the 21V battery) is... -21V - (3KΩ*Ia) - (6KΩ*Ia) + (6KΩ*Ic) = 0
Loop 2 (for the 28V battery) is... 28V - (12KΩ*Ic) - (6KΩ*Ic) + (6KΩ*Ia) = 0

Simplify the loops:
1: -21V -9KΩIa + 6KΩIc = 0
2: 28V -18KΩIc + 6KΩIa = 0

Solve loop 1 for one of the unknowns. Let's solve Ia.
-21V -9KΩIa + 6KΩIc = 0
-21V +6KΩIc = 9KΩIa (Moved Ia to the other side of the equation using the commutative property of addition, where 9KΩIa is added to each side. Doing this cancels out the term on the left side of the equation to isolate the unknown Ia on one side of the equation.)

Divide both sides of the equation by 9KΩ to get Ia by itself. Remember that 9K divided by 9K equals 0, so by dividing 9KΩIa by 9KΩ, we have completely isolated Ia, which is what we want.

That whole process would look like this:
$$\frac{-21V}{9K\Omega} + \frac{6K\Omega I_C}{9K\Omega} = \frac{9K\Omega I_A}{9K\Omega}$$

The result of that problem is:
-2.333mA + 0.667ΩIc = Ia

Now that we have a value for Ia, substitute (-2.333mA + 0.667ΩIc) for every instance of Ia in loop 2 equation.

2: 28V -18KΩIc + 6KΩ(-2.333mA + 0.667ΩIc) = 0
Simplify by distributing the 6KΩ and combining like terms
28V - 18KΩIc - 14V + 4KΩIc = 0 (remember that when you multiply 6KΩ by 2.333mA it results in a voltage).
Simplify: 14V - 14KΩIc = 0 (now solve for Ic)

14V - 14KΩIc + 14KΩIc = 0 + 14KΩIc
14V = 14KΩIc
Divide both sides by 14KΩ remembering that the 14V divided by 14KΩ equals a current, we get:
1.0mA = Ic (answers may vary slightly depending on how you round). Ic is a positive number, so the current direction we assumed (clockwise) is the correct current direction.

Now we can solve Ia using the known value of Ic:
-21V -9KΩIa + 6KΩIc = 0
-21V -9KΩIa + 6KΩ(1mA) = 0
-21V -9KΩIa + 6V = 0
-15V - 9KΩIa = 0
-15V - 9KΩIa + 9KΩIa = 0 + 9KΩIa
-15V = 9KΩIa
-1.667mA = Ia

Since we got a negative value for Ia, our assumed clockwise direction is incorrect. Current flows counter-clockwise through the 21V loop.

Now use Kirchhoff's Current Law to determine the current through the 6KΩ resistor. If Ia flows counter-clockwise and Ic flows clockwise, then both currents flow the same direction through the 6KΩ resistor, and thus are added together. Ib = Ia + Ic or Ib = 1mA + 1.667mA
Ib = 2.667mA

Ia = 1.667mA
Ib = 2.667mA
Ic = 1.0mA

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DerStrom8

Joined Feb 20, 2011
2,390
Ah, I am guessing you did not solve for iB, but for iC. To find iB, you need to use Kirchoff's current law (current going into a node equals the current coming out). iC is solved using the second mesh, and iB is solved by using the values you found for iA and iC. An arrow pointing in the opposite direction as suggested by the value simply means it is the opposite value. For example, if there is a clockwise current of 5 amps labeled by iA pointing counter-clockwise, iA is simply -5 amps.

Does this make sense?

EDIT: Apparently, ke5nnt beat me to the response. @ke5nnt, though, since this is homework, perhaps it would have been better not to give out the answer? An explanation of how to get the answer would have been sufficient. Just my opinion, no disrespect intended.

lemon

Joined Jan 28, 2010
125
Thank you both. And while DerStrom8 is right, of course. I am a very mature student who isn't in this to acquire good grades only, or to just beat the system without doing the ground work. I will use the information given as a guide only and work my way through it to understand fully. But to be totally honest, being Xmas holiday it is nice to speed things along a little so I can enjoy that mulled wine

DerStrom8

Joined Feb 20, 2011
2,390
Thank you both. And while DerStrom8 is right, of course. I am a very mature student who isn't in this to acquire good grades only, or to just beat the system without doing the ground work. I will use the information given as a guide only and work my way through it to understand fully. But to be totally honest, being Xmas holiday it is nice to speed things along a little so I can enjoy that mulled wine
I trust you are honest, but I still think it would have been better for you to do the work. It may be a good idea to go ahead and go over it again yourself anyway, without looking, and see if you get the same answer. School is there for you to learn, not simply to be given the answers.

Best wishes,
Der Strom

ke5nnt

Joined Mar 1, 2009
384
My intention wasn't to simply give the answer without the benefit of learning, but more of a complete walk-through of the process similar to examples that might be found in a text book, except explained more thoroughly. I have to assume that similar to my introductory circuits course, a homework assignment would contain more than 1 of these types of questions. I'm sorry if you feel I went overboard by actually solving the problem, but my hope was that by seeing the entire process of working through the problem, the OP might have gained a better understanding of how to solve the rest of them.

Didn't mean to spoil your homework assignment, I'll try to be a little more vague next time.

DerStrom8

Joined Feb 20, 2011
2,390
My intention wasn't to simply give the answer without the benefit of learning, but more of a complete walk-through of the process similar to examples that might be found in a text book, except explained more thoroughly. I have to assume that similar to my introductory circuits course, a homework assignment would contain more than 1 of these types of questions. I'm sorry if you feel I went overboard by actually solving the problem, but my hope was that by seeing the entire process of working through the problem, the OP might have gained a better understanding of how to solve the rest of them.

Didn't mean to spoil your homework assignment, I'll try to be a little more vague next time.
It's nothing too serious, and there's really no need to apologize. Just a suggestion for next time. I understand your intentions, I would just recommend going about it in a slightly different way next time. No harm done

Regards

lemon

Joined Jan 28, 2010
125
@ke5nnt
You said:
Solve loop 1 for one of the unknowns. Let's solve Ia.
-21V -9KΩIa + 6KΩIc = 0
-21V +6KΩIc = -9KΩIa Moved Ia to the other side of the equation using the commutative property of addition.
Divide both sides of the equation by -9KΩ get Ia by itself
2.333mA - 0.667ΩIc = Ia

should that be 2.333V and not mA?
And why did you use the commutative property of addition. Would it not have been simpler just to take the -9 across to the other side to become +9?

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lemon

Joined Jan 28, 2010
125
sorry - i got it
Volts divided by Ohms = current
Silly me

ke5nnt

Joined Mar 1, 2009
384
@ke5nnt
You said:
Solve loop 1 for one of the unknowns. Let's solve Ia.
-21V -9KΩIa + 6KΩIc = 0
-21V +6KΩIc = -9KΩIa Moved Ia to the other side of the equation using the commutative property of addition.
Divide both sides of the equation by -9KΩ get Ia by itself
2.333mA - 0.667ΩIc = Ia

should that be 2.333V and not mA?
And why did you use the commutative property of addition. Would it not have been simpler just to take the -9 across to the other side to become +9?
We are trying to isolate the term Ia and get it by itself on one side of the equation. The commutative property of addition (and multiplication for that matter) allows us to do this. The process basically states that in an equation, which is a problem that contains and equals sign as ours does, we can perform the same operation on both sides of the equals sign without changing the result. Thus, our full process looks like this:
$$-21V - 9K\Omega I_a + 6K\Omega I_c = 0$$ Our initial equation.
Now I'm going to move the ENTIRE term containing $$I_a$$ to the other side of the equation in an effort to isolate the SINGLE factor $$I_a$$ That is done by adding the term $$9K\Omega I_a$$ to both sides of the equal sign. If I have a simple equation such like $$8 - X = 0$$ I can find the value of X by adding X to both sides $$8 - X + X = 0 + X$$ where the result is that $$X = 8$$ The same principal is applied to:
$$-21V - 9K\Omega I_a + 6K\Omega I_c = 0$$
$$-21V - 9K\Omega I_a + 9K\Omega I_a + 6K\Omega I_c = 0 + 9K\Omega I_a$$
$$-21V + 6K\Omega I_c = 9K\Omega I_a$$ Now that we have the term that contains Ia by itself on one side of the equation, and remember that the term that contains Ia is still a voltage drop that we have expressed in terms of a resistance times a current (V=IR), we can use the commutative property of multiplication to clear the resistance from Ia. In order to do this, we must divide each side of the equation by the 9KΩ resistance. That looks like this:
$$\frac{-21V}{9K\Omega} + \frac{6K\Omega I_c}{9K\Omega} = \frac{9K\Omega I_a}{9K\Omega}$$

When we take Voltage Divided by Resistance, we get current, thus $$\frac{-21V}{9K\Omega} = -2.333mA$$

Well I'm glad you got it, I hope my explanation helps clear up any confusion on the other parts.
Regards

EDIT: Oops, I just realized my typo. When I moved -9 to the other side, it should have been a positive 9 once it got moved. That was a mistake on my part, my apologies. Matter of fact, that one little typo screws up the entire solution. I'm going to go back through it and fix my error so that anyone that comes upon this in the future wont be horrendously confused by my math. Sorry about that.

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t_n_k

Joined Mar 6, 2009
5,455
The important lesson to take away from this, is that when using mesh currents, CURRENT ALWAYS IS ASSUMED TO FLOW THE SAME DIRECTION THROUGH EVERY MESH or else the mesh current method will NOT work.
Sorry ke5nnt but I must disagree with you on that point. It doesn't matter which direction one chooses for any mesh current. The choices can be completely arbitrary.

ke5nnt

Joined Mar 1, 2009
384
I am not pretending or claiming to be an expert in the field by any means sir, only repeating what I was taught. I pulled out my textbook on the subject and reread the section on mesh current analysis, since my original answer was pulled from memory. I agree with you that my statement regarding assumed current absolutely has to be the same direction for each mesh or it wont work was incorrect, but it certainly seems to make things easier, which I suppose is why I seem to have recalled it as being a rule. Thanks for pointing out my error, helps us all learn.

Regards

lemon

Joined Jan 28, 2010
125
Well... I just finished working my way through that, which was actually quite a task and has certainly 'learnt' me a lot. Now I will go through it considering your mistake which will give me some fluency practice
But yes, better you post the correct workings as they are certain to be more reliable than my own.

P.S. So, what is the important lesson to be taken away from this, anyone?

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lemon

Joined Jan 28, 2010
125
I have a question regarding the voltage source. If the assumed current goes through the positive terminal of the source, does that mean we write a positive voltage or a negative one?

ke5nnt

Joined Mar 1, 2009
384
I'll try not to muddle this one up, and I would post a graphic to make it easier but again... I am at work and don't have all the cool tools I have at home here.

Okay, so basically, you know that current flows OUT of the negative battery terminal and when we assign a polarity to resistors in a loop, the current enters the "negative" side of the resistor and exits the "positive" side. If we had a loop and we "assume" that current flows clockwise, follow the convention above to get your loop equations.

For instance, imagine a square in your head, on the left side of the square is a 12V battery with the positive terminal facing up. You have 2 resistors in your loop, one on the top of the square and the second on the right side of the square. If we assume that the current through this square is going clockwise, trace your finger around the loop, you would enter the battery's negative terminal, so you'd get -12V, then you would hit resistor 1, we assumed current flows into this side of the resistor so we'd get a negative value here, let's say -3KΩ*I, same thing for the second resistor, -6K*I. Your loop equation would be: -12V - 3K*I - 6K*I = 0. This would result undoubtedly in a negative value for I, and we would know that current actually flows counter-clockwise.

For the same square as above, just turn the battery around so that the positive terminal is facing down. Then the loop equation would be 12V - 3K*I - 6K*I = 0. Basically, once you assume a direction of current flow and assign polarities to resistors based on this assumption, whichever side of each component you hit first will determine whether it's a +value or a -value.

Man I hope that makes sense, lol.

Remember that we don't assign positive or negative to voltage sources, they're always what they're supposed to be. The positive terminal is always positive, negative is always negative.

The short answer to your actual question is, if the assumed current goes through the positive terminal of the source, and by "through" I mean it enters the positive terminal, then you write a positive voltage.

ke5nnt

Joined Mar 1, 2009
384
I went back through my original response and corrected my errors and removed my misstatement that t_n_k pointed out. Hopefully this thread will be in better shape to help someone else out in the future. Thanks for all the patience guys, those little typos or errors mess everything else up down the line.

lemon

Joined Jan 28, 2010
125
What if you have a common component between two loops and the currents in each loop were going through that component in different directions, how would you write the positive or negative for that component/resistor? Would you just not give them initially and wait until your calculations had solved for one current direction?

lemon

Joined Jan 28, 2010
125
ahh - i think i got that. You just do both for the same loop values but in opposite directions based on your earlier statement of current direction enters into a negative component side.
I'm going to try this assuming the left current direction is anti-clockwise and the second is clockwise.

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lemon

Joined Jan 28, 2010
125
ok. I have assumed that the currents are:
left circuit = anti-clockwise
right circuit = clockwise

I got:
ia = 1.0mA
ib = 2.0mA
ic = 1.0mA

ke5nnt

Joined Mar 1, 2009
384
I guess the attached pdf file proves that t_n_k was correct, it can be solved when the currents oppose each other. The .pdf shows your originally posted schematic with the current directions indicated there, and the steps used to solve it. I tried to be thorough.

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