# Homework with ciruit analysis help please

Discussion in 'Homework Help' started by dutch944, Oct 24, 2012.

Oct 24, 2012
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Apr 26, 2005
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3. ### dutch944 Thread Starter New Member

Oct 24, 2012
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I know I have to say how I tried to do it and failed for the first one I tried many times and I just couldnt get the answer ,the answer for the first question A) is = 300watt and I just dont know how they got 300 for the second question down not B,isA) =-25 B)260 C)16ohm

4. ### mlog Member

Feb 11, 2012
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For question 2.A, do you know how to find power dissipated by a resistor if you know the current through that resistor? Do you know how to use superposition of current sources? Do you know hot to combine resistors in parallel?

5. ### dutch944 Thread Starter New Member

Oct 24, 2012
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I know how to combine resistors in parallel but I dont really know how to do the superposition
EDITk the first question I know how to solve it now but the second one is just complicated because of the top current the 4A i dont know what to do with it

Last edited: Oct 25, 2012
6. ### mlog Member

Feb 11, 2012
276
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You can combine the current sources too since they're in parallel. Glad you solved it.

Apr 26, 2005
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8. ### dutch944 Thread Starter New Member

Oct 24, 2012
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Yeah thank you,but can you help me solve the other question

9. ### WBahn Moderator

Mar 31, 2012
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It would really help if you would post your attempt at a solution. Then we can determine what you are doing right and where you are going wrong.

10. ### dutch944 Thread Starter New Member

Oct 24, 2012
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well,I tried a lot but they are very bad attempts and i dont know what to do I Just need guidance on how to begin the question and solve it ,please help because Im studying for my retest on this for after 3 days.Thank you for your help

11. ### panic mode Senior Member

Oct 10, 2011
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it does not matter how bad your attempts are, just post what you tried. if you don't strain your brain, you will not learn. we need to know how you think this should be done. there are several methods that can be used to solve the problems. one of them was mentioned (superposition). there are others, such as node and mesh analysis, substitutions using Thevenin and Norton etc.

For example it would be a really good idea to mark each node, and each current.
Then you can write bunch of KVL/KCL equations.

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,401
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The problem states the node voltage method.

13. ### dutch944 Thread Starter New Member

Oct 24, 2012
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Well as I said before I do not know what to do with the upper 4A current ,and I have to use the node-voltage method ,so can you give me just the beginning and the Ill try to continue and solve it and I dont know what to to do when I get to the v3 node ,Please help as my exam is coming very soon

14. ### WBahn Moderator

Mar 31, 2012
18,087
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Since you are supposed to use the node voltage method, start by describing,in your own words, what the node voltage method is.

15. ### dutch944 Thread Starter New Member

Oct 24, 2012
13
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I will tell you what I tried to do but I know this is wrong
I did the node voltage method and got three equations:
1)v1/30 +v1/6 -30+(v1-v2)/25=0
2)(v2+v3)/50 +v2/3.75 -1+(v2-v2)/25=0
3)(v3+v2)/50 + v3/20=0
And I know this is wrong but I tried a lot of different ways and still couldnt get the right answer
But please people I dont mean to sound douchy or arrogant but please can you help me solve this and tell me how you did it my exam is tomorrow

16. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
That makes it all the more useful for you to show it, because it almost guarantees that we will be able to provide useful feedback that is specific to the problems YOU are having.

Thank you for finally giving us something to work with. Now we can start making progress.

So,it appears you are making the node with Point A your ground, which is a very reasonable choice.

If you've seen very many of my posts, you will know that I am a self-professed Units Nazi. If you ALWAYS track your units, you will catch a huge fraction of the errors you make and get valuable hints as to where the problems are. In this case, let's add the units and then examine the equation.

v1/30Ω + v1/6Ω - 30V + (v1-v2)/25Ω = 0

Notice that the first, second, and last terms are each a voltage divided by a resistance, which yields a resistance. But the third term is just a voltage. You can't add a voltage and a current, so you know this is wrong. No point going any further until get this resolved.

Now, before we can correct it, let's review what node voltage analysis is all about. At the end of the day, it is nothing more than a systematic way of applying KCL at each node of the circuit. So what is KCL? It merely states that the current into each node must equal the current out of that same node.

So let's examine your equation, term by term, and see how it relates to applying KCL at node v1:

v1/30Ω

This is the current flowing OUT of the node downward to Node A (which you are defining as 0V). Look's good.

v1/6Ω

This would be the current flowing OUT of the node toward the left IF the node on the left of the 6Ω resistor were 0V. But it's not, it's 30V. Note that it is 30V not just because there is a 30V source, it is because the 30V source makes the voltage on that node 30V higher than the voltage on Node A, which happens to be 0V. So you need (v1-30V)/6Ω.

- 30V

This is not a current, but a voltage. So it doesn't belong here since KCL deals with summing up currents. You probably meant to write the expression I just provided, but got sloppy. If so, carrying and checking your units and you will catch things like that.

(v1-v2)/25Ω

This is the current flowing OUT of the node toward the right and is correct.

But what about the current flowing OUT of the node upward through the current source? You've neglected that entirely. Since the 4A is actually flowing INTO the node, it is the same as -4A flowing OUT of it.

Thus your first equation should be:

v1/30Ω + (v1-30V)/6Ω + (v1-v2)/25Ω - 4A = 0

Again, let's track the units:

(v2+v3)/50Ω + v2/3.75Ω -1A + (v2-v2)/25Ω = 0

All four terms have units of current, so the units are not catching any errors at this point. But it is important to remember that while if the units don't work out, there MUST be a mistake, that does NOT mean that if the units DO work out that there ISN'T a mistake -- it simply means that whatever mistakes were made just didn't happen to affect the units.

So let's again examine it term by term:

(v2+v3)/50Ω

Remember that Ohm's Law relates the voltage DIFFERENCE across a resistor, not the SUM, to the current flowing through it. So you need either (v2-v3) or (v3-v2). Which one you need is determine by the fact that you want the current flowing OUT of node v2 and when you use (Va-Vb)/R, you are finding the current flowing from Node a to Node b.

v2/3.75Ω

This is a current flowing from v2 through a 3.75Ω resistor IF the other side of that resistor is at 0V. You have no idea what voltage the other side of this resistor is at.

-1A

This is correct.

An important point to note here is that when you have a current source in a branch, the presence of resistors in that branch doesn't matter. The current sources FORCES the specified current in the node and the only thing that the resistors do is determine the voltage across the current source required to make this happen -- but you don't care about the voltage across the current source (at least not yet).

(v2-v2)/25Ω

Does this make sense? (v2-v2) is identically zero. But more to the point, this is NOT the voltage ACROSS the 50Ω resistor.

Once again, you have neglected the current flowing upward through the 4A current source.

Try to take these into account and build up the correct equation for this node.

Similar observations as above. You need voltage differences across resistors and you need to take the currents in ALL branches leaving a node into account.

Come up with new equations and post them and let's make sure they are correct before you proceed (but, since there might be a delay in getting feedback from any forum, feel free to proceed anyway to see if you can get the answer).

All I can say to that is, had you shown your work the first time (or after any of the times that you were asked to do so), we could have gotten moving a lot sooner. Not doing so was a choice YOU made....

dutch944 likes this.
17. ### dutch944 Thread Starter New Member

Oct 24, 2012
13
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So for the second equation it would be:
for the first one(v2-v3)/50Ω

for this one it would be v2-3.75(because I did v=i.r so that there would be a voltage there?or should it be just +v2 and I ignore the resistance)/3.75Ω

-1A

(v2-v1)/25Ω
I meant to say here v2-v1 sorry

so the second equation should be (v2-v3)/50Ω +v2-1A +(v2-v1)/25Ω=0 I think
and third one is I think (v3-v2)/50 + v3/20=0
and I thank you very much for your help Couldnt do it without your help ,Im sorry for not writting my attemps from the beginning and I realized how much more it would help if I write my attempts rather than you just saying the answer

18. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
v2-3.75Ω (don't leave off the units!) has to be wrong because you can't subtract ohms from volts.

"doing" v=i.r is more than just grabbing some v and some r and throwing them into a formula. That formula only works if the v is the voltage difference across THAT resistor, the i is the current flowing in THAT resistor, and the r is the resistance of THAT resistor.

v2 is a voltage and you can't add it to a bunch of currents! Check your units! Always! Always! ALWAYS!

The node v2 has four branches coming off of it. Go through branch by branch and express the current going away from the node along that branch:

Upward:
Downward:
Rightward:
Leftward:

If the current is flowing toward the node, just make it negative.

Now add them together and set them equal to zero.

DON'T LEAVE OFF THE UNITS!!!!

You keep ignoring the current sources!

Do the same as above - list each branch current explicityly and then add them together.

dutch944 likes this.
19. ### dutch944 Thread Starter New Member

Oct 24, 2012
13
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Hey,I thank you very much for this but is it possible for you to help me with the other two equations because I dont have a lot of time and I have to study the other materials for the test if not I could try to do that but I would really appreciate that if you help me

20. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I really don't think that is going to help you. I walked through the first equation term by term and described all of the issues. There are no new issues in the second and third equations and I've pointed out the specific problems in your attempts so far. So just giving you the other two equations will provide you with absolutely nothing that providing you with the first didn't. At some point you have to start paddling for yourself if you are ever to learn how to swim.

Post your attempts and I will be happy to continue pointing out where you are going astray.