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Practicing Mesh Analysis I came across this problem and te answer is really alluding me and I believe is just one polarity that I'm not understanding, the problem is the following...

enter image description here

Vs​=10V, R1=100Ω, R2=50Ω, R3=25Ω, Is=2A.

Calculate the current I1 in amps that goes through R3​ from left to right. Enter >only the numerical answer for I1​ in the text box, omit the units. Note: this >problem can be done different ways, but try using Mesh Analysis.

Now, I'm setting up my equation using the passive sign convention as following.

Vs voltage rise R2 voltage drop R3 voltage rise (since current is flowing clockwise entering the "negative" terminal of R3)

so it would look something like this:

-Vs+R2(I1-I2)-R3(I1)=0

-10+50(I1-(-2))-25(I1)=0

Now the answer to the question is 1.2 which I could achieve if the voltage across R3 was in fact " 25(I1) " instead of negative as in my equation, the thing is that I don't understand what I'm doing wrong and I don't Understand why it is positive if it is a voltage rise?

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  • \$\begingroup\$ You have two meshes so you need two equations. \$\endgroup\$ – Oldfart Aug 25 '18 at 10:50
  • \$\begingroup\$ I am going to delete my earlier comment! I2 is as you said. I have not yet checked the signs in your equation. \$\endgroup\$ – Oldfart Aug 25 '18 at 11:10
  • \$\begingroup\$ Ok, got it. I f you use +R2(I1.. then you must also use +R3I1 as I1 goes in the same direction through R2 as it goes through R3. \$\endgroup\$ – Oldfart Aug 25 '18 at 11:13
  • \$\begingroup\$ @Oldfart that is what I thought but it is really confusing in how to identify that because according to the passive sign convention, R2 has the positive terminal at the top (that is, resistors that are not in the left most branch are consider voltage drop unless specified otherwise like in the case of R3) which makes me think that current goes in the opposite direction of R3 when I'm doing the analysis, so how can I identify that if nothing is specified? \$\endgroup\$ – Marco Castro Aug 25 '18 at 11:18
  • \$\begingroup\$ Sorry I don know " passive sign convention". Just know that once you have drawn a current arrow e.g. In in mesh loop all the In * R on that loop must have the same sign. \$\endgroup\$ – Oldfart Aug 25 '18 at 11:26
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When you assume the current direction in the loop you automatically set the voltage polarity across the resistors (current flow from + to - in the resistor). Hence, you assumed the clockwise flow. Therefore this forces you to stick to this assumed direction and the voltage across the resistors. And you should forget about the VR3 polarities shown on the diagram.

Case one:

enter image description here

And the equation (notice that in this case only one equation is needed)

$$(I_1+ I_S) R_2 + I_1R_3 - V_S = 0$$

And the solution is

\$I_1 = -1.2A\$

which means the \$I_1\$ current is flowing in opposite direction than we have assumed.

Case two

enter image description here

$$V_S + I_1 R_3 + (I_1 + I_S)R_2 = 0$$

Additional we see that \$I_S = -2A\$

So, the solution is \$I_1 = 1.2A\$

EDIT

For each individual mesh, you can pick the loop current direction arbitrarily.

Look at this example

Loop one and two have the same loop current direction (clockwise).
enter image description here

So for loop one we have

I start at point B $$I_13\Omega + 2V + (I_1 - I_2)10\Omega + I_14\Omega - 10V = 0 $$

(notice that I1 is first here (I1 - I2)*10 )

And the second loop (start at point A)

$$ I_28\Omega - 15V + (I2 - I1)10\Omega - 2V = 0$$

In this case loop I2 "is first" (I2 - I1)*10

And the solution is:

\$I_1 = 1.52427A\$ , \$I_2 = 1.79126A\$

And now in this example, I pick the loop current direction this way:

enter image description here

As you can see I1 is clockwise but I2 is counterclockwise.

And the equations look like this:

Loop one

$$I_13\Omega + 2V + (I_1 + I_2)10\Omega + I_14\Omega - 10V = 0 $$

Do you see the defense?

Loop two:

$$2V + (I_2 + I_1)10\Omega + 15V + I_28\Omega = 0$$

And the result is:

\$I_1 = 1.52427A \$

\$I_2 = -1.79126A\$

And this minus sign in the final result tell us the I2 current is, in fact, flowing in the opposite direction then I assume.

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  • \$\begingroup\$ It makes a lot of sense, I was not aware that you're supposed to treat all the resistors with the same polarity. Now there is only one thing confusing me, when you define the direction youre going to use for the mesh analysis, aren't you supposed to treat all current loops the same? (have the same direction, clockwise in this case), why is the second one reversed? is it because the direction of the current source? If there was another mesh between mesh 1 and mesh 2, should I treat the current as clockwise in that case? \$\endgroup\$ – Marco Castro Aug 25 '18 at 21:36
  • \$\begingroup\$ @MarcoCastro I update my answer. \$\endgroup\$ – G36 Aug 26 '18 at 6:52

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