# Problem 4.2-2 and 4.6-5

Discussion in 'Homework Help' started by fire_lizard, Jun 19, 2013.

1. ### fire_lizard Thread Starter New Member

Apr 4, 2013
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Hi everyone from newbie (in electronics and this forum)

I am trying to solve problem 4.2-2 (Node Voltage Method) from Introduction to Electric Circuits, 9th Edition

and I wrote following equation:

1. V1/5 + 1 + (V1 - V2)/20 = 0
2. -2 + (V2 - V1)/20 + (V2 - V3)/10 = 0
3. -1 + (V3 - V2)/10 + V3/15 = 0

and get these results:

V1 = - 0.1818182 ; V2 = 19.090909 ; V3 = 17.454545

but the textbook and SPICE tell me that results are:

V1 = 2; V2 = 30; V3 = 24

Could you tell what I am doing wrong?

And the second problem is 4.6-5 (Mesh Currents Method) from the same textbook.

I am familiar with Mesh Currents Method but could anyone guide me how to write equation in this case?

2. ### Brownout Well-Known Member

Jan 10, 2012
2,373
1,003
I solved your equations in problem #1 for V2 and got V2=30. Therefore, your equations are correct, but you have a mistake solving for the unknowns.

Don't be surprised if your thread gets moved to the homework forum.

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3. ### fire_lizard Thread Starter New Member

Apr 4, 2013
9
0
Yes, I should have used Scilab before

Now I see that I wrote right equations for 4.2-2.

Any clue about 4.6-5? How should I start?

4. ### Brownout Well-Known Member

Jan 10, 2012
2,373
1,003
4.6-5 is similar, but this time you're doing voltages in loops rather than currents in nodes. So, for example, in the left most mesh, and identifying the current as I1, you would write:

12 - 8 - 5I1 = 0.
Write all mesh equations and solve as before.

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5. ### fire_lizard Thread Starter New Member

Apr 4, 2013
9
0
Thank you for helping me. Just one more question about 4.6-5

Shouldn't we take into account mesh current I2?... I mean I thought we should write it as:

12 - 8 - 5*I1 + 5*I2 = 0

6. ### Brownout Well-Known Member

Jan 10, 2012
2,373
1,003
You are correct.

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7. ### WBahn Moderator

Mar 31, 2012
23,400
7,109
I would recommend that you develop the habit of using and tracking units religiously through your work. Most quantities in engineering have dimensions and those dimensions are an inherent part of the quantity. By tracking them, you enable an extremely powerful error detection capability in your work since most mistakes you make will disrupt the units and if the units are wrong, then you KNOW the answer is wrong and there is no point spending any more time banging on an equation that has no change of working out when you can instead find and fix the problem right near the point it arises.

So your node equations for the first problem should be:

1. V1/5Ω + 1A + (V1 - V2)/20Ω = 0
2. -2A + (V2 - V1)/20Ω + (V2 - V3)/10Ω = 0
3. -1Ω + (V3 - V2)/10Ω + V3/15Ω = 0

Carry those units throughout your work, frequently asking if the units still work out. As soon as they don't, STOP. You've made an error. Track it down and fix it.

Also, maybe I'm missing it, but I don't see where you indicate which mesh current is which (i.e., which is I1, I2, etc.). Don't make people pull out their crystal balls and guess -- especially the grader!

The first thing you should do is check if either set of answers is actually correct. Since their answers are simpler, let's check them first. For KCL at V1, you have the following currents flowing OUT of the node:

1A + 2V/5Ω = 1A + 0.4A = 1.4A

And the following current flowing INTO the node:

(30V-2V)/20Ω = 28V/20Ω = 14V/10Ω = 1.4A

So they balance. This does NOT prove that they are correct, merely that they are consistent with the equation for the first node.

Now check yours:

OUT of the node:

1A

INTO the node:

(0V - -0.182V/5Ω + (19.09V- -0.182V)/20Ω = 0.182V/5Ω + 19.27V/20Ω = 1.00V

So your results are also consistent with the first node equation.

So now do the same for one of the other nodes.

The key point is to learn to check your own work -- and one of the wonderful things about many, if not most, engineering problems is that you can usually check the validity of an answer from the answer itself.

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