# Missing some basics here

Discussion in 'Homework Help' started by flexmanlet55, Jun 29, 2013.

1. ### flexmanlet55 Thread Starter New Member

Jun 29, 2013
3
0
I seem to have forgotten some simple stuff here and its bugging me.

How do I figure out by looking at currents/voltage drops etc, which direction my current is going or rather how can I tell when a current source or voltage source is absorbing or generating power..?

i am sure this has to do with passive sign convention etc, but I am confused I guess. I am taking Electric Network Analysis for summer school and I just finished up the 2nd half of physics.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You are better advised to actually analyze the circuit for yourself and determine the direction of the currents and relative potential differences.

If you are adopting conventional current for analysis then for a DC circuit as you have ...

1. Current flowing out of the positive (into the negative) terminal of a voltage source indicates the source is delivering power to the circuit.
2. Current flowing into the positive (out of the negative) terminal of a voltage source indicates the source is absorbing power from the circuit.
3. A positive potential difference between the source terminal driving current into the circuit and the other return terminal indicates the source is delivering power to the circuit.
4. A negative potential difference between the source terminal driving current into the circuit and the other return terminal indicates the source is absorbing power from the circuit.
Points 3 & 4 are probably sufficient to make that determination anyway regardless of whether one is considering a voltage or a current source.

If the polarities indicated on the circuit annotations are contrary to the source convention by virtue of a negative sign, then one must take this into account when considering which of conditions 1-4 above apply.

For instance, source I2 in your attachment is indicated with a negative value (-9A) which implies it is driving current in the opposite direction to that shown by the source convention annotation. So source I2 is actually driving a current of 9A upwards rather than downwards. Noting that the upper I2 source terminal is at a higher potential [152 V] than its bottom terminal would lead one to conclude it is actually delivering power to the circuit - point 3 above.

Relying on a simulation to resolve this for you is probably not advisable unless you are absolutely sure how the simulator presents the indicated values as you have them in the attached schematic. Some simulators will actually allow you to probe the source or circuit element for power dissipation. This is the case in the one I use. I my case circuit elements absorbing or dissipating power (such as resistors) are shown to have a positive power value when probed for that parameter. Again (in my case) a source delivering power to the circuit would indicate a negative value when probed for power.

Last edited: Jun 30, 2013
3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
If you just finished Physics II, then use what you have learned.

How can you tell if a charge gains or loses energy as it goes from one voltage to another voltage?

4. ### flexmanlet55 Thread Starter New Member

Jun 29, 2013
3
0
All jokes aside, my physics teacher was terrible and I was taking too many other hard classes at the same time and I didn't retain as much as I should have.

5. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
And so....what? Does that mean that you should forever not be expected to know the foundational knowledge that is part of Physics II? Would you accept an excuse like that from your doctor? "Hey, doc, how could you possibly miss the cancerous tumor on my back for years to the point that now it is untreatable and I have just six months to live?" "Well, I had a really horrible oncology professor and I didn't retain as much as I should have. Sorry."

We have all had poor teachers here and there. We have all had other issues at one time or another -- a too-heavy course load, health issues, family problems, working full-time-plus, you name it -- that have resulted in us not taking away from a course everything that we should have. But once you receive credit for a course, it because YOUR responsibility to have a working knowledge of the material in that course. So if you know that you don't, you need to do something to correct it. You can retake the course, you can pull out the text (or get another one if you've compounded the problem by selling yours), or you can change direction and choose a path for which that foundational knowledge is not a prerequisite.

What I recommend, right now, is pulling out that text (or looking at online resources) and gaining the understanding necessary to answer the question I asked. It should be contained in the first chapter or two of your Physics II material. What you are looking for is an understanding of what voltage is -- how it is defined and what it tells us.

6. ### tbinder3 Member

Jun 30, 2013
30
1
You dont want to say Charge, that's a whole nother ball game lol
In order to figure out the voltages at certain nodes must watch the component voltage drops. You can find voltage drops by taking the current through or into the resistor multiplied by the resistor value and that will give you voltage across resistor. Also advise to check out Kirchoff's Current Law, and Voltade Divider.

Anyways... Ripped, and babbling

But start from the power source and do a Voltage Walkabout (google if you don't know) As you find values on your schematic it will make things alot easier.

EDIT: I would also start using polarities, it will help with the voltage walkabout and which way current is going, cause (depenging on current source) you have to remember as long as there is positive current, there is negative current flowing the opposite way. Also I would brush your history up on conventional current and electron flow.

Last edited: Jul 2, 2013
7. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
No, it's not.

What is the definition of voltage?

It is the work done by an electric field per unit charge in moving a charge from point A to point B.

What is the definition of the electric field?

It is a vector field that yields the force per unit charge at every point in space.

Since work is force through a distance, the voltage between two points is the line integral of the electric field between those two points.

Notice that nothing in any of this refers to the polarity of the charge or what sign Ben Franklin arbitrarily gave to what turned out to be electrons. The definitions make everything work out fine as long as you treat charge as the signed quantity that it is.

8. ### tbinder3 Member

Jun 30, 2013
30
1
Bro I seen you in other threads and we are dis-agreeing a lot. So, I don't want to argue, I am putting my opinion out there same as you are.

I know exactly what Charge is, but in the thread creator's case, no need for all that.

Charge is expressed in Coulombs Q=CV and also Coulombs law

Where Q = Charge
C = Capacitor rating

@Flex Also, When Im working my voltage drops I use Conventional method

Last edited: Jul 3, 2013
9. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
It is very relevant to the OP's case. The OP is trying to understand how to determine if something is absorbing or generating power based on the directions and polarities of the voltages and currents involved. That is directly related and answerable by understanding what voltage is and how it relates to the work done on charges moving between two points. If the OP understands that, then they won't have to rely on just memorizing facts by rote.

Why are you bringing capacitance into this?

10. ### flexmanlet55 Thread Starter New Member

Jun 29, 2013
3
0
Wbahn I am not making excuses, I am telling you the reason why I don't (yet) have the right understanding for something. That being said, try to remember I am here asking for help instead of just saying "screw it ill figure it out sometime"

I have noticed that I learn different then most of my class mates. I seem to do better when I can see a problem worked backwards, (or from the start) in full glorious and bloody detail. Once I understand it my way, then I have no problem with application.

Normally I have to find a person in class that can communicate well with me, and we do homework etc together. When I get stuck, we talk it out, and I learn.

Also sadly sometimes I don't connect two things together that are glaringly obvious, I seem to be bad (or scared) at making leaps on my own with some of this stuff. Everybody learns differently.

11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Where did I say screw it and go figure it out for yourself? I asked a very specific question to help focus your efforts on the principles you need to learn. I am waiting for you to make SOME kind of effort to ATTEMPT to answer that. Then I will be more than happy to point out where you are going right, where you are going wrong, and where to look to take the next step. If we need to take a step further back, then fine, I'm more than willing to do that and go as slow as needed. But you have made zero effort to make any attempt at all.

12. ### â¬Hunter New Member

Jul 6, 2013
33
3
An interesting exercise is to make a cube out of resistors and figure out the total resistance at various points. So make up 4 squares with 4 resistors in each and solder them together. Calculate what you think and then verify with a multimeter.

13. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
What do either of these, particularly the first one, have to do with anything that is going to be of any help to the OP?

14. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Socrates liked to teach by asking questions. It is a good technique, give it a chance.

tbinder3 likes this.