stupid parallel circuits

[undesiredshoe]

Im having trouble figuring out the total resistance for a parallel circuit. The webpages i visit make it really difficult to understand. I know there are people who should know how to figure it out and give an explanation that i can understand.

http://i54.photobucket.com/albums/g90/undesiredshoe/untitled.jpg

what would the total resistance be for that?? I got alot of answers, 1, 6/11, 6, .54545454 and some other ones...im not sure what im doing wrong.

[EchoOfSilence]

alright, this is quite simple.

for parallel circuits:

it's 1/[(1/resistor1)+(1/resistor2)+(1/resistor3)]

which is 1/(1/1 + 1/3 + 1/2) = 6/11 or 0.4545

tell me if you get stuck


EDIT: i might as well go into more detail...
In parallel circuits, the voltage is the same for all resistors that are in parallel, but the current gets divided.

so i1 being current through resistor1, i2 through r2, and so on...
i1= v/R1
i2=v/R2
i3=v/R3

since the current gets divided, the total current must equal the current divisions, aka i_total= i1+i2+i3

now, substitute the previous equations and get: i_total= v/R1 + v/R2 + v/R3
factor v out and get: i_total= v * (1/R1 + 1/R2 + 1/R3)
note that it's very similar to i= v/r (or i= v * 1/r)

have you made the connection yet?

the 1/r in i=v/r is equal to (1/r1 + 1/r2 + 1/r3)....... (aka 1/R_equivalent = (1/R1 + 1/R2 + 1/R3)

but that's 1/R_equivalent and you want just R_equivalent, right?
well, just put it all under one (take the reciprocal), like:

R_equivalent= 1/[(1/R1 + 1/R2 + 1/R3)]


fuck, I hope that was a good enough explanation... lol

[kandyflip445]

There are 3 ways to find out.

I like Rt=V/It

So first we have to find the I for each branch.

I=V/R

So for branch 1 we get 14.2 amps

For branch 2 we get 4.73 amps

For branch 3 we get 7.1 amps

Add those all together for the total amperage It=26.03A

So then we take V/It and we get .54552 Ohms

[NemeGuero]

Ewww.. why do current loops?
In english...
the equivalent resistance is:

1/ R eq = 1/ r + 1/r.. blah blah

[kandyflip445]

You have to remember that the total resistance in a parallel circuit is always less than any branch.

Also you can use Rt= R1xR2/R1+R2

For circuits with more branches do the initial 2 and then plug that number in the R1 place.

So for yours we take 1x3=3 3/1+3 or 3/4=.75 then you plug that into R1 so

.75x2=1.5 1.5/.75+2 or 1.5/2.75=.5454

[kandyflip445]

I was just gonna help educate on different ways to solve it....sorry.... :/

[projekt_s13]

Im lost.........??? lol

[kandyflip445]

Where are you lost at?

[ALTRNTV]

Oohh! Teacher! Pick me, pick me! I know! I know!

1+1=2

:)

[undesiredshoe]

Ok, I was reading a page and it said you had to take the reciprocal of the final answer, 6/11, which would be 11/6 and i got 1.833.

http://buphy.bu.edu/~duffy/PY106/9d.GIF
i got this picture from...http://physics.bu.edu/py106/notes/Circuits.html

im super confused

[EchoOfSilence]

6/11 is the answer after taking the reciprocal.

read my explanation again

the shorthand way to do the calculation with the reciprocal-taking already built in is : 1/[(1/resistor1)+(1/resistor2)+(1/resistor3)]

[undesiredshoe]

hah yea i got it...thanks for the help

[fliprayzin240sx]

Wow, i used to know this...this is what happens when you havent had to fuckin troubleshoot a piece of equipment in 3 yrs.

[FRpilot]

what is all this electrical engineering [email protected]

[BOROSUN]

i cant see the pic...

but for parrallel ,easiest for me is finding the (i=e/r) current/amp first and then add it all up. then r=e/i, unless its a series-parrallel. ;p

[P4rD0nM3]

I thought you guys were talkign about NASCAR or something...

[NemeGuero]

Just go left, you can't miss it.

[P4rD0nM3]

Hahaha! Omg, no left turn here, please!