Resistors in a parallel circuit

[tehexchro]

I know this probably isn't the greatest place to ask, but can anyone explain resistors in a parallel circuit?

 

[Steve]

Isn't a resistor simpy anything that drains current? For example, if you hook a parallel battery up to a light bulb, the light bulb is a resistor....I think.

edit: damn it, it's been so long since physics...

 

[SpuD]

I'm trying to recall, but its been a while since electricity.

A resister simply cuts the amount of power down to match what the device needs, so it doesnt over do it. If its a parallel circuit, theres no other way around the resistor, it has to go through it.

I think....

 

[Phisionary]

I think you're thinking serial. As in in-line.

If you have resistors in parallel, the resistance in decreased. The current has more paths to travel through.

If you have a resistor in parallel to another component, the component will get more of the current if it is lower resistance, since the current will take the path of lowest resistance.

If you have a parallel circuit, the current draw of the cicuiot is higher, since all the components in the circuit draw power at once. if you have them in serial, the voltage to each component drops and you have similar current draw to a single component.

Well, I'm guess kind of guessing here. I _think_ what I've said is true..
Other than that... I'll get back to you in a couple years after I've taken some electronics classes.

Check teh intrawebs. There's lots of info and stuff ther :thumbs: :dork:

 

[tehexchro]

Isn't a resistor simpy anything that drains current? For example, if you hook a parallel battery up to a light bulb, the light bulb is a resistor....I think.

edit: damn it, it's been so long since physics...

Yes, any load would be a resistor, but I'm trying to figure out exactly how much a load in a parallel circuit affects it, as compared to a series circuit.

 

[ship]

Holy hell, I can't believe I remember this stuff.

I=V/R
V=I*R
R=V/I

A parralel circuit looks something like this:
http://www.tpub.com/neets/book1/chapter3/32NE0087.GIF

I'm not exactly sure what your question is, though.

 

[Revisedsoul]

so then based on that equation ship would the two feeds of .5 equal 1 or jsut .5?

 

[ship]

so then based on that equation ship would the two feeds of .5 equal 1 or jsut .5?

The total current of the circuit is 1 ampere.
The current moving across each resistor is .5 ampere, because it is a parralel circuit and the "1" ampere must be spread out among two branches. I'm pretty sure....
Uhhhh......

That's all I know! It's been too long since I did this stuff. Maybe someone else can elaborate.

 

[SpuD]

I think you're thinking serial. As in in-line.

If you have resistors in parallel, the resistance in decreased. The current has more paths to travel through.

If you have a resistor in parallel to another component, the component will get more of the current if it is lower resistance, since the current will take the path of lowest resistance.

If you have a parallel circuit, the current draw of the cicuiot is higher, since all the components in the circuit draw power at once. if you have them in serial, the voltage to each component drops and you have similar current draw to a single component.

Well, I'm guess kind of guessing here. I _think_ what I've said is true..
Other than that... I'll get back to you in a couple years after I've taken some electronics classes.

Check teh intrawebs. There's lots of info and stuff ther :thumbs: :dork:

Yeah you're right, I got parallel/serial mixed up

 

[Revisedsoul]

The total current of the circuit is 1 ampere.
The current moving across each resistor is .5 ampere, because it is a parralel circuit and the "1" ampere must be spread out among two branches. I'm pretty sure....
Uhhhh......

That's all I know! It's been too long since I did this stuff. Maybe someone else can elaborate.

yeah i got it, been a while myself.

 

[Parrot of doom]

Analogy time:

A troop of soldiers, marching in a line. They come to a river. The river has 2 very thin bridges (parallel). The soldiers split up and use both bridges to cross the river.

Now consider if there was only one of those very thin bridges, wouldn't it take longer for the soldiers to cross? Of course it would. This is in essence the difference between parallel and serial circuits.


If you want the maths, its easy:

If you have a serial circuit, 2x 5ohm resistors, the total circuit resistance is 10 ohms (5ohm + 5ohm = 2ohms).

If you have a parallel circuit, 2x 5 ohm resistors, the total circuit resistance is 2.5ohms:

__1__ = 2.5
1 + 1
5 .. 5

The dots between each 5 are there to format the message.


You can alter that formula to include any resistance value you wish, and you can have as many resistors as you like. The real tricky stuff comes when you start calculating impedance :/

 

[Shippi]

I really should know this. I really should, especially since I'm doing electronics as a GCSE. We revised parallel circuits last lesson.. we spent most the frikkin lesson on them, and you know what? I don't remember a damn thing

Heh, Parrot of doom, you just explained it better than my teacher ever has done, thanks

 

[HunterSeeker]

The total recistance of a resistors in a parrarel circuit is: (R1^-1 + R2^-1)^-1

Where R1 and R2 are the total resistanse of path one and two, add more if there are more paths.

 

[Tantalus]

The total recistance of a resistors in a parrarel circuit is: (R1^-1 + R2^-1)^-1

Where R1 and R2 are the total resistanse of path one and two, add more if there are more paths.

R1^-1 = 1/R , right?

 

[HunterSeeker]

R1^-1 = 1/R , right?

I think so, the formula was something like (1/R1 + 1R2), or something. The one I gave above is the one I know in my head!

 

[Parrot of doom]

*points to Parrot post*

 

[Tinneth]

Science wasnt my strongest subject :|

 

[nicrd]

The total recistance of a resistors in a parrarel circuit is: (R1^-1 + R2^-1)^-1

Where R1 and R2 are the total resistanse of path one and two, add more if there are more paths.

That formula is right, but a simpler one would be: (R1 + R2) / (R1 * R2) = total resistance.

 

[Marker]

Formula for total resistance in a parallel circuit is 1/Resistance(subscript) total = 1/Resistor (subscript) 1 + 1/Resistor (subscript) 2 + 1/Resistor (subscript) 3 + 1/Resistor (subscript) 4 ....... 1/Resistor (subscript) n

Resistance in a series circuit is the sum of all the resistors

 

[dream431ca]

Are you trying to add the resistance between 2 resistors in a parallel circuit?? Very easy:

The formula is 1/R1 + 1/R2 + 1/R3..... and so on.

Lets say R1 is 2Kohms and R2 is 10Kohms.

RT= 1/2k + 1/10k = 0.0005 + 0.0001 = 0.0006

Now invert the answer, because the previous answer is just the sum of 2 inversions.

1/0.0006 = 1.67kohms.

 

[sinkoman]

I know this probably isn't the greatest place to ask, but can anyone explain resistors in a parallel circuit?

Electricity goes the path of least resistance.

Now, you have two resistors hooked up in paralel, giving the electricity a choice between more or less resistance, guess which one it takes?

Seems simple, only you also still have some current flowing through the higher value resistor.

R = (R1+R2)/(R2*R2) Is the formula for total resistance between the two. That's for TWO resistors now. it's a diff formula for more than two.

EDIT: Damnt, beat me to it. Sorta.

 

[Adidajs]

This is funny, because we dealt with resistance in parallel (pretty much the only physics besides fluids in medical school). Had to calculate the resistance of some of the vessels of the body, eg. splanchnic circulation, because they're in parallel to the rest of the circulation. The GI(digestive tract) in effect stores blood and can mobilize it when there's a hemorrhage.

nice explanation Parrot.

 

[MiccyNarc]

Holy bump batman!

 

[Blakeb155]

"World's biggest bump"

 

[Parrot of doom]

Threadnomancy is gay, especially when done by a one post newbie.

 

[Shippi]

What a weird thread resurrection...

Thanks for reminding me of electronics GCSE btw.. just as I've purged the horrors of that lesson out my mind... *shudder*

Closed