Help with Resistors!

missbossyy · 2008-07-22 10:49:45

Identify the number and location of each of the items below. Assume that the resistances and source voltage are known.

Use the gray highlighted letters. Identify voltage drops across resistor Ri as Vi.

- How many nodes are there in this circuit? Where? ________________________

- How many loops possible? Where? ________________________

- How many current variables are there? ________________________

- Is this system solvable? (i.e., can you solve ________________________

for all of the currents?) Why?

b.) Set up the loop and node equations using Kirkhoffs Laws (but do not solve them). Clearly identify the node or loop, including all directions, associated with each equation.

DRAW EACH CURRENT ON THE DIAGRAM ABOVE, CLEARLY SHOWING ITS DIRECTION!

Use the gray highlighted letters. Identify voltage drops across resistor Ri as Vi. (For example, the voltage drop across R1 is V1.)

c.) What is the algebraic method used for solving this system? (name?)

John E. Franklin · 2008-07-22 11:25:57

Did you forget to erase the line or wire or "short"-they-call-it, because
I think you don't want that line going right thru or else the other 4 resistors
will never get any current at all.
My power just went out, so I have very little time left on batteries.

John E. Franklin · 2008-07-22 11:32:59

R45combined = 1/(1/R4 + 1/R5)
R23combined = 1/(1/R2 + 1/R3)
R2345combined = 1/(1/R23 + 1/R45)
R12345alltogether = R1 + R2345combined, because it is in series there.
Hope that helps a bit.
Vtotal / R12345 = I1
Now the I1 gets shared among I2,3,4 and 5.
The bigger the resistor, the smaller the current, but I2 + I3 + I4 + I5 will = I1, okay?

John E. Franklin · 2008-07-22 11:35:36

So to clear my mind, I'll try numbers.
Say I1 is 10amps.
Say R2 is 2 ohms.
Say R3 is 4 ohms.
Say R4 is 8 ohms and say R8 is 16ohms.
Now we have to cut 10amps up into 4 parts, kinda reciprocal-like.
And I2 * R2 = I3 * R3 = I4 * R4 = I5 * R5, because the voltage is the same from node to node.

John E. Franklin · 2008-07-22 11:38:11

So 6 + 3 + 1.5 + 0.75 = what, for a guess, is it over 10?
Yes, too high.
Need algebra.

missbossyy · 2008-07-22 11:45:49

yes! I forgot to erase the lines.....there are no lines !

missbossyy · 2008-07-22 11:47:51

thanks! it sure does helped me....
also can you help me Set up the loop and node equations using Kirkhoffs Laws....i will really appericiate it!

John E. Franklin · 2008-07-22 11:52:36

I can't remember, are there 10 loops here, I can't remember the loop analysis.
And nodal analysis was a different method we learned after I think.
When you do the loops, do you make all the currents go clockwise in every loop,
can't recall?

John E. Franklin · 2008-07-22 12:04:30

http://en.wikipedia.org/wiki/Loop_analysis
Mesh analysis
Uh, I think there are 4 loops now, just the ones
you can see as spaces of air in the diagram.
Like if you spilt paint, the colored paint would fill one of the
loops, so there are 4 loops.

John E. Franklin · 2008-07-22 12:07:10

If you look just below each of the 1st four resistors, you have an
airy space which is the loop, so what should we call things? (I'm almost out of power; a black out)

John E. Franklin · 2008-07-22 12:14:02

Power just came back on at 8:10PM in Westmoreland, NH. (only out for 70 minutes)
Personally, I would use letters, like Ia and Ib and Ic and Id, but
if you want to use numbers, then when you solve all the equations,
the currents thru each resistor will also have numbers, and then
you get all confused because I3 will be I3 - I2, so do you want to
use numbers or letters or what?
When there are two loops in one little leg of circuit, then you subtract those
2 pretend currents and multiply it by the resistor to get the voltage, and then
add up all those voltage terms around the loop and the voltage should come
back to zero when you get all the way around, if I remember right,
since the voltage is like the height of a roller coaster, you have to
end at the same height when you get around, so set it equal to zero, right?

missbossyy · 2008-07-22 12:23:41

I figured 4 loops too!...going clockwise

missbossyy · 2008-07-22 12:24:36

answer is in letters

missbossyy · 2008-07-22 12:26:05

your explanation really helps can you solve one loop equation for me, so i would know i am on right track or not

John E. Franklin · 2008-07-22 12:41:47

Okay, let's use Ia for the big loop on the bottom.
then Ib for the top loop around R2 and R3.
Ic will be the middle one where you forgot to erase the line between R3 and R4.
Id will the loop with R4 and R5.
Now let me think...

John E. Franklin · 2008-07-22 12:49:25

Big "a" loop on bottom:
0 = -Vsource + R1Ia + R5(Ia-Id)

Top "b" loop:
0 = R2Ib + R3(Ib-Ic)

Middle "c" loop:
0 = R4(Ic-Id) + R3(Ic-Ib) (see how b and d pretend currents go against the clockwise Ic in the middle?)

missbossyy · 2008-07-22 12:51:19

hmm smart!

John E. Franklin · 2008-07-22 12:52:28

Thanks!! It's been many years since 1986 & 1987.

Bottom little "d" loop above the biggest loop:
0 = R5(Id-Ia) + R4(Id-Ic)

missbossyy · 2008-07-22 12:56:19

God Bless you for taking out time and sharing your knowledge:-)

John E. Franklin · 2008-07-22 13:04:56

Thanks a lot.
And if you have to do the node equations, well that will
be a lot of work too, do we really have to do that?

missbossyy · 2008-07-22 13:58:20

yes i do :-(

John E. Franklin · 2008-07-22 15:14:37

I will attempt to write the node equations.
See new drawing.

missbossyy · 2008-07-22 15:22:51

thanks!

John E. Franklin · 2008-07-22 15:23:34

V1 = V0 + Vs (I read in wikipedia that for voltage sources, you just add the voltage to get to other side)
Now all the current entering V2 is zero.
there are 5 wires attached like a * or something.
If you pretend all 5 currents go into V2, then they have to
add up to zero. Actually, the current goes into one wire and
out the other 4 wires, but if you turn those other 4 wire currents
around and point them into the dot (v2 dot), then they all add to zero.
That's how it works anyway. Instead of the in equals the out, since you
don't know sometimes, then you pretend all the 5 currents are going
in, and nothing out, so you set it equal to zero. Like say 2 people
walk in a room and say 2 other people leave the room at another door.
2 in = 2 out or really 2 in = -2 out I guess. Who knows. But anyway.
2 - 2 = 0, and somehow it all works, if you get my drift.

John E. Franklin · 2008-07-22 15:29:50

So current = Voltage over resistor. When I say "over", I mean divided-by.
You subtract 2 numbers to get the distance between them, so
you do the same between V1 and V2 to get the voltage across R1.
(V1 - V2)/R1 I happen to know is a positive number because I know the
current is gonna stay the same all the time, no weird stuff here, no capacitors
or inductors, that's when calculus comes in!!
V1 is the highest voltage in the circuit.
V2 is a little lower.
V0 is the lowest voltage in the circuit.
I just know this from lots of schooling.

Math Is Fun Forum

#1 2008-07-22 10:49:45

Help with Resistors!

#2 2008-07-22 11:25:57

Re: Help with Resistors!

#3 2008-07-22 11:32:59

Re: Help with Resistors!

#4 2008-07-22 11:35:36

Re: Help with Resistors!

#5 2008-07-22 11:38:11

Re: Help with Resistors!

#6 2008-07-22 11:45:49

Re: Help with Resistors!

#7 2008-07-22 11:47:51

Re: Help with Resistors!

#8 2008-07-22 11:52:36

Re: Help with Resistors!

#9 2008-07-22 12:04:30

Re: Help with Resistors!

#10 2008-07-22 12:07:10

Re: Help with Resistors!

#11 2008-07-22 12:14:02

Re: Help with Resistors!

#12 2008-07-22 12:23:41

Re: Help with Resistors!

#13 2008-07-22 12:24:36

Re: Help with Resistors!

#14 2008-07-22 12:26:05

Re: Help with Resistors!

#15 2008-07-22 12:41:47

Re: Help with Resistors!

#16 2008-07-22 12:49:25

Re: Help with Resistors!

#17 2008-07-22 12:51:19

Re: Help with Resistors!

#18 2008-07-22 12:52:28

Re: Help with Resistors!

#19 2008-07-22 12:56:19

Re: Help with Resistors!

#20 2008-07-22 13:04:56

Re: Help with Resistors!

#21 2008-07-22 13:58:20

Re: Help with Resistors!

#22 2008-07-22 15:14:37

Re: Help with Resistors!

#23 2008-07-22 15:22:51

Re: Help with Resistors!

#24 2008-07-22 15:23:34

Re: Help with Resistors!

#25 2008-07-22 15:29:50

Re: Help with Resistors!

Board footer