Question

Please... does anyone know how to use kirchhoffs laws for circuits containing current and voltage sources?

Answer 1

yes

Answer 2

this should be asked in physics

Answer 3

@ehoax I have been asking for help with these on math and physics ALL DAY...

Answer 4

so what do you need to know? do you want me to help you from the beginning of the problem? do you not understand it at all?

Answer 5

I understand that the some of currents entering a node is =0, I know that the sum of voltages around a closed loop is =0. I just have not been able to solve these circuits on my own. I actually have a few pictures I can post, two of them (including the one above) are from the explanation in the textbook. They give you the solution but no explanation. Then I have a hand written example of the same Ideas from class where the Professor started to show us how to do them, but then he changed directions and showed us mesh analysis. But I really need an explanation as to how to approach these problems.... the whole nine yards. My professor does not like the way our text explains it so he uses his own method. I have bought two supplemental books (Schaums) to get help but I think I have just thoroughly confused myself. If you can give me an approach to take I would be forever grateful

Answer 6

this is another example from the book

Answer 7

ok lets start with this last example

Answer 8

another one from the book

Answer 9

and here is what my professor showed us... all of this on the hand written one he gave us

Answer 10

anywhere you want to start is good for me

Answer 11

ok ill do .27 example first, first step is to draw "loops" inside any direction doesn't matter

sometimes it is a good idea to do one around the outside too i will post pictures

Answer 12

Just to start.... you are using KVL???... what tells you to take that approach?

Answer 13

I am only asking cause I want to understand the thought process you are having when you solve these

Answer 14

KVL?

Answer 15

yes

Answer 16

I thought you draw the loops when you are using KIrchhoff's Voltage Law... why did you decide to start with that route?

Answer 17

i am sorry i didnt understand what you meant at firt by kvl

Answer 18

No problem... so is it that when you see only current sources you use KVL?

Answer 19

not sure what you mean by that

Answer 20

Nevermind I am asking too many questions so I will just try to follow what you do

Answer 21

do you see what I did here?

Answer 22

which problem are you doing?

Answer 23

2.8 fig 2.27

Answer 24

ok I am sorry I was looking at a different one.. give me two seconds

Answer 25

Ok I see what you wrote... do you have t assign polarities to the resistors? And if so how do you assign them?

Answer 26

and what is that on the top of the page? is it \[\pm \rightarrow \frac{ 1 }{ t }\]

Answer 27

at the top of the page is another way to write the voltage

Answer 28

the reason why the voltage source is negative is because the loop passes through it from positive to negative, if it was from negative to positive we would put a positive sign on it

Answer 29

I understand the - on the source. Do I have to assign polarities to the resistors before doing this?

Answer 30

the reason why the (8ohm)i1 is positive is because the current and loop are going in opposite directions if they were in the same direction then you would use a negative sign

Answer 31

Ok so when you say the current and the loop you mean the loop you drew and the assignment of \[i_{1}\] that was given?

Answer 32

the only reason is to show which direction the current is travelling

Answer 33

yes

Answer 34

ok

Answer 35

current in a resistor will flow from + to -

Answer 36

ok what next?

Answer 37

loop 2 try and see if you can do it on your own

Answer 38

come up with an equation

Answer 39

I believe I can do it in the same fashion you did...\[-3\Omega i _{2}+6\Omega i _{3}=0\]

Answer 40

If that is correct I have a question...

Answer 41

good job! it is correct

Answer 42

That would be easy enough on this one.... because it is two equations and two unknowns so it is a simple linear system to solve....but....

Answer 43

My look at 4.jpg from above..... my professor said that when you consider a loop that has two currents traveling on it that you have to consider them both

Answer 44

could you do 1.jpg?

Answer 45

your professor should have drawn the currents also

Answer 46

this is the exact example off the chalk board.

Answer 47

Can you show me how to approach the first picture i posted? It only has current sources in it.

Answer 48

so i4 = 9A

Answer 49

now you would do the loops, and then find equations

Answer 50

could you show me how to do the loops on this one? I actually did the current parts..... that I was comfortable with... but I don't know how to show the current sources as a voltage. Because v=ir and I have no clue what the resistance for those are.

Answer 51

ok you can only do a loop on the second box and on the last one as well as around the whole system box

Answer 52

no!

Answer 53

sorry ill post it right now

Answer 54

ok

Answer 55

try 1 loop inside the last two boxes which gives

-i0(2ohm)+i2(8ohm)=0

Answer 56

so you are saying draw a loop that goes down the \[i _{0}\]] element and up the \[v_{0} \] element?

Answer 57

yes

Answer 58

Ok I have run into a problem.... I will load a picture...

Answer 59

what does the diamond shape mean?

Answer 60

The circle is an electrical symbol for an independent source (voltage or current) in this case it is current. The diamond shape means it is a dependent source (voltage or current) in this case current

Answer 61

do you see how that did't work?

Answer 62

I am sure I did something wrong but I don't know what?

Answer 63

your i2 should be positive beacuse it is going in the opposite direction of the loop

Answer 64

I am confused about that.... the way he said it was that if the current you were looking at went into the "assumed" positive polarity it was a positive current, if it went into the assumed negative polarity then it was a negative current. However when he explained the loop analysis he said when we arbitrarily chose a direction for the "loop" we were assuming current to flow in that direction for that loop. That is why I chose i2 to be negative because the loop is going into the negative polarity. Is that not what the "loop" is? An assumed current direction?

Answer 65

yeah if you do that then your current equations will also change

Answer 66

See this is what I am saying.... I am so confused .... what do you mean by the last statement? Yeah he is correct? And how would that change the current equations?

Answer 67

i was not taught with the polarity of the resistance, and it has always worked for me, try doing it without that see if it helps sorry that i dont understand the way your professor did it

Answer 68

Did you ever have to determine if Power was absorbed in an element or if it was expelled?

Answer 69

no, my class was 25% taught and 75% self taught

Answer 70

ok well thanks