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Ohm's law with mesh and nodal analysis
- Thread starter areebTAG
- Start date
Not quite, its first year uni electronic & electrical engineering. It's coursework for which none of the lectures have explained what to do.
If this is something for which you will receive a grade, we consider it homework.
You're not being asked to solve anything. Just write equations for the voltages across the resistors and the KCL equations for the nodes.
You have all of the voltages and currents already annotated on the drawing. What you need to do is apply Ohm's Law to each resistor in the circuit to come up with equations relating the voltages to the currents.
Remember that Ohm's Law relates the resistance to the voltage across that resistance and the current through that resistance.
Show your best attempt to get as far as you can and we will then have something more to work with to help you move in the right direction.
Since ohm's law : V=IR
R3=0.5/ I3
R6= 0.5/I6
R7=V2/I7
R10=V4/I10
R11=V5/I6
R12=V4/(I10+I7)
What do you think about this?
I have attached the full question for you to see. It says use MATLAB to solve the simultaneous equations which I can do however coming up with the equations is the hard part.
Let's focus on R3 for now.
Where does the 0.5 come from? That should also be "0.5 V" -- units matter!
The question says to write the equations in terms of the labeled currents and voltages. In terms of the node (wire) voltages, what is the voltage across R3 (consistent with I3 being the current through R3)?
Now let's consider R10. You have
R10 = V4/I10
Is V4 the voltage difference ACROSS R10?
This is a very common mistake. Remember that Ohm's Law uses the voltage difference ACROSS the resistor. You can't just pick the voltage on one side of the resistor (which is relative to some arbitrary node at some arbitrary location in the circuit that was arbitrarily picked as the reference (ground) node) and throw it at Ohm's Law.
I don't know how to find the voltage across it. I thought that R3 is connected in series with the other R7,R11,R6 but that does not seem to be the case. Is it V1 maybe?
You don't necessarily need units in the equation, only show units in the final answer.
What is required for two components to be in series?
The voltage across a resistor is the voltage difference between the two ends of the resistor. The voltage difference between Points A and B is the voltage at A minus the voltage at B. The voltage across a resistor that is consistent with the current through the resistor is the voltage at the end of the resistor that the current enters minus the voltage at the end of the resistor that the current exits.
This is how airliners filled with passengers run out of fuel in midflight and spacecraft get slammed into planets, I've seen someone get killed because they believed as you and just showed units in the final answer.
If potentially getting yourself or someone else killed isn't motivation enough, then I can almost guarantee that if you develop the habit of always properly tracking your units through your work from beginning to end, you will not only see a significant improvement in your grades, but you will also spend considerably less time doing your work overall.
So, I've done this again:
R3=(V2-V1)volts/I3 amperes
R6= (V5-2) v / I6 A
R7 = (V2-2) v/I7 A
R9= (V2-V1) v/I9 A
R10= (V4-V3)v / I10 A
R11 =(V1-V5) v/ I11 A
R12= (V4-2) v/ (I10-I7) A
What do you think about this? I added the units this time so planes don't crash.
Variables carry their own units. If I have
d = v*t
to give the distance traveled by an object traveling with a certain velocity for a certain amount of time, v might have units of mm/hr or it might have units of miles/s. Similarly, t might have units of minutes or units of years. The units are carried by the variable.
Consider if I write
d = v*3
and tell you have v is 20 m/s. How far did the object travel?
You have no idea, because "3" is not a time, it is just a number. If you evaluate v*3 you end up with 60 m/s, which is a velocity, not a distance. The units don't work out.
Notice that your results are very different if you learn that the "3" was really 3 years.
So your literal values must have the units -- they aren't proper quantities if they don't. Again, "3" is not a measure of time, but "3 years" is.
d = v * 3 years
is dimensionally sound.
Remember that both current and voltage are signed quantities (meaning that they have direction associated with them). The voltage used in Ohm's Law is the voltage drop across the resistor in the direction of the current flow, so it is the voltage at the end that the current enters minus the voltage at the end where the current exits. So this should be
R3 = (V1 - V2) / I3
Your volts units distribute so that you have (V5)*(1 V) - (2)*(1 V) giving you something that has units of volts-squared minus something that has units of volts.
What is the voltage on the wire connected to the right end of R7?
Where is I11 on that diagram?
How do you figure that the current flowing in R12 is different than the current flowing in R10?
The wire voltage is not given so, I thought it would be V2 minus the 2v voltage source.
I think I made a mistake. The current in R12 is the same as in R10.
I thought some of I7 will flow into that node on the right of R7 and into R12.
The current through R11 would be I6. There is no I11, my mistake.
The 2 V voltage source is establishing that the voltage on the bottom-right wire is 2 V higher than the voltage on the top-right (or middle-right, depending on your viewpoint) wire. It has NOTHING to do with the voltage across the resistor.
But they DO give you the voltage on the wire that is connected to the right end of R2 -- it is indicated by the 'ground' symbol and it means that we are declaring that the voltage on that wire is 0 V and that all other wire voltage will be the voltage difference between the other wire and the ground wire.
You equation didn't reflect SOME of I7 flowing up into R12, it would only be correct if ALL of I7 flowed upward into R12.
But if ANY of it flows up into R12, then the resistor has a different current flowing into in on one side than it has flowing out on the other, which can't happen.
By R2 I think you mean R7? Yeah I can see the ground symbol so voltage across R7 will be (0-V2) volts.
What is the voltage across R9. I legit can't figure that out. I think it might be V1-V2 volts but there is also V3 connected.
okay I understand
You are correct -- I meant R7.
But let me say this yet again. Ohm's Law requires that you use the voltage drop across the resistor IN THE DIRECTION of the current flowing through it. That means that you use the voltage at the node where the current you are using in Ohm's Law enters the resistor minus the voltage at the node where the current you are using Ohm's Law exits the resistor.
If you are using I7 at the current in Ohm's Law, then it enters the left side of the R7 and exits the right side of R7, so you need the voltage on the left side of R7 (which is V2) minus the voltage on the right side of R7 (which is 0 V), yielding
R7 = (V2 - 0V) / I7 = V2/R7
If you do not take care of the polarities of your quantities properly, you won't get a correct result unless you just happen to randomly assign all of the polarities correctly or (possibly) all of them incorrectly. If you have some correct and some not, it virtually guarantees a wrong answer will result.
You are correct -- and notice that this time you got the polarity correct which, coupled with the wrong polarity for R7 would have made getting the correct answers to the problem impossible.
As for V3, it is NOT connected to R9, it is connected to the other side of the current source that happens to be connected to R9.
What might help you visualize this better is to make each wire a different color.
So the voltage across R9 in the direction of I9 is the voltage on the red node minus the voltage on the blue node.
