devilstormz

Joined Aug 18, 2011
10
Hey,

I have attached the circuit diagram and my attempt. Please note that the markings on the diagram were given.

I have gone through the notes on the website and have manage to solve the question using the same current directions in notes and applied it to the question attached and it worked. I must be doing something wrong because it shouldnt matter which way the current flows.

There must be a problem with my understanding of applying kvl perhaps?

The correct ans is -1.68 A

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t_n_k

Joined Mar 6, 2009
5,448
Your problems are arising due to the sign notations and inconsistency with relating these to mesh current direction.

You'll note this in your very first equation.

Traversing loop 1 you'll see that (according to your equation) the same current I1 appears to produce voltage drops of opposite sign in the 6Ω and the 1Ω. This doesn't make sense. You need to maintain consistency on that matter. Curiously the "given" resistor sign notations don't match your adopted (conventional flow?) mesh current directions.

devilstormz

Joined Aug 18, 2011
10
Your problems are arising due to the sign notations and inconsistency with relating these to mesh current direction.

You'll note this in your very first equation.

Traversing loop 1 you'll see that (according to your equation) the same current I1 appears to produce voltage drops of opposite sign in the 6Ω and the 1Ω. This doesn't make sense. You need to maintain consistency on that matter. Curiously the "given" resistor sign notations don't match your adopted (conventional flow?) mesh current directions.
Ye it was strange which is what confused me. Is it possible for the 1Ω resistor to have different polarities +/- in each loop because considering the I1 and I2 directions the current appears to follow in the opp directions so how do u account for that?

EDIT:

I tried solving that question by switching the -/+ sign on the first loop but I still keep getting the wrong answer.

Jony130

Joined Feb 17, 2009
4,930
So we have this circuit For the first loop I write

10V - 6Ω*I1 - 1Ω*(I1 - I2) = 0

Loop II

-1Ω*(I2 - I1) - 2Ω*I2 - 3Ω*(I2 + I3) = 0

And the last loop

- 3Ω*(I3 + I2) + 20V- 10Ω*I3 = 0

And if we solve this we get this result
http://www.wolframalpha.com/input
/?i=10+-+6*X1+-+1*%28X1+-+X2%29+%3D+0++-1*%28X2+-+X1%29+-+2*X2+-
+3+%28X2+%2B+X3%29+%3D+0+++-+10*X3+%2B+20+-+3+%28X3+%2B+X2%29+%3D+0+

I1 = 1.34043A; I2 = -0.617021A; I3 = 1.68085A

devilstormz

Joined Aug 18, 2011
10
How have u labelled the polarities on the resistors? When the current enters the resistor a +ve polarity?

So for the 1ohm resistor shared between loop one and two should u use the polarity assigned using the first loop regardless of the direction of the current I2?

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the upstream end of a resistor will always be negative, and the downstream end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not agree with the resistor polarities (assumed current directions)

Jony130

Joined Feb 17, 2009
4,930
How have u labelled the polarities on the resistors? When the current enters the resistor a +ve polarity?
I assume Conventional current flow . Current flow from "+" to "-" From the higher potential to the lower potential. And AAC book uses Electron Flow.
So for the 1ohm resistor shared between loop one and two should u use the polarity assigned using the first loop regardless of the direction of the current I2?
Yes I forgot to labeled the polarities for 1ohm resistor. Maybe I'll start again.
First lets assume conventional current flow, and a mesh current may be assigned to each mesh in an arbitrary direction. I assume that each mesh current flows clockwise except third mesh when I assume counterclockwise direction.
Additional I assume that if the mash current enters + side of a resistor I put a minus sign.

mesh 1
I start from A to B , current enter the resistor so i give the minus sign.

- 6Ω*I1 - 1Ω*(I1- I2) + 10V = 0

For mesh 2 I start from B

-2Ω*I2 - 3Ω*(I2 + I3) - 1Ω*(I2 - I1) = 0

And for mesh 3 and start from C to E in mesh direction

- 3Ω*(I2 + I3) + 20V - 10Ω*I3 = 0

But we can assume that if mesh current enter the "+" terminal of a component we give the + sign

So we can write

6Ω*I1 + 1Ω*(I1- I2) - 10V = 0
(1)

2Ω*I2 + 3Ω*(I2 + I3) + 1Ω*(I2 - I1) = 0 (2)

3Ω*(I2 + I3) - 20V + 10Ω*I3 = 0 (3)

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• devilstormz