electrics help - work much confused :D

[barwel1992]

ok so got some work to do for college but im a little lost can some on confirm this for me

ok so scenario: have a 12v parallel circuit no resistors and grounded to the chassis with 2 100w bulbs wired in parallel.

so my understand is both bulbs get 12v, i need to work out the amps for the diagram so 100w/12v= 8.33amp's for each wire going to the bulb. The wire that comes from the battery's + then later on splits off to the two bulbs +'s would be double that of each wire going to the bulbs because its supplying both bulbs ?

and the resistance for each 100w bulb is 12v/8.33amp = 1.44ohms ?

i truly hate electrics ......

thanks for any help if a diagram would help i can make one up

[beabert]

Im not sure i understand what your question is lol, then again i am tired.

[tigersaw]

Each lamp consuming 8.33 Amps.
Total consumption is twice that as its a parallel circuit of two lamps.
Resistance of lamps.. tricky, as they are non linear devices. But in the circumstances you describe you can assume they behave similar to a resistor of 1.44 ohms

[timwilky]

I am not going to do it for you but instead tell you what you need to do.

You need to work out the resistance of the circuit, that is the two resisters in parallel

Then the current for the total circuit can be established from good old V=IR

Do it publish it and I will tell you if you are right.

[21QUEST]

What tigersaw said ie you are correct.

But timwilky has got the right idea ;)

[21QUEST]

Re timwilky's post, a little pointer

V/R x V=W .

Do that for each...

[timwilky]

OK guys I will do it for him, my way

He has already worked out the resistance of a lamp as 1.44 ohms

Total resistance of parallel resistors bulbs = 1/((1/1.44) +(1/1.44)) =0.72 ohms

12/0.72 = 16.66..A

[barwel1992]

ok so i think this is right ps sorry bout the drawing/image paint for you :D

http://i357.photobucket.com/albums/oo16/barwell1992/wiringdiagram.jpg

EDIT beaten to it by Tim

[timwilky]

That is the most weird diagram I have ever seen, definitely not how to wire a parallel set of bulbs back to earth.

Your current would remain constant over the circuit, you do not have different sections at different current.

Your total resistance on a parallel circuit is the reciprocal of the sum of the recipricals of the parallel resistances. So 0.72 ohms.

[21QUEST]

Hi mate,

Mine would have been slightly different/simpler diagram :) .

Total resistance is wrong though(refer to timwilky's last post). You've calcilated it as as you would for resistors in series.

[tigersaw]

you've used a single pole double throw switch and both lamps have twin elements, like in a car / bike , is this delierate?

[barwel1992]

that how we were told to do it lol

the bulbs are dual filament ( lets just say there are both 100w filaments so went switch is thrown it just changes filament (both filaments do not come on at once)) hence the 2 + and one - on the bulb

what we were told to do to find ohms for one bulb was to

100w / 12v = 8.33amp (?)

then

12v / 8.33amp = 1.44 ohms resistance for that bulb(?)

im sort of lost now :/

[barwel1992]

you've used a single pole double throw switch and both lamps have twin elements, like in a car / bike , is this delierate?

yes its deliberate, so the system can never be off only low or high beam and only one element is on in each bulb at a time

PS this is the first time doing this on my own, and have only had one lesson on it

[21QUEST]

that how we were told to do it lol

the bulbs are dual filament ( lets just say there are both 100w filaments so went switch is thrown it just changes filament (both filaments do not come on at once)) hence the 2 + and one - on the bulb

what we were told to do to find ohms for one bulb was to

100w / 12v = 8.33amp (?)

then

12v / 8.33amp = 1.44 ohms resistance for that bulb(?)

im sort of lost now :/

I see...

All you've done is correct. Different ways to skin a cat 'n all that :) . It's only the total resistance that you've calculated wrong.

What you just need to remember in the future is ,

Resistors in parallel... 1/Rt=1/R1+1/R2

Resistors in series... Rt=R1+R2

In the old days when we didn't have calculators lol, you needed to know how to add fractions(in the case of parallel resitors)but I'm guessing, all that is needed these days, is a calculator lol

[barwel1992]

I see...

All you've done is correct. Different ways to skin a cat 'n all that :) . It's only the total resistance that you've calculated wrong.

What you just need to remember in the future is ,

Resistors in parallel... 1/Rt=1/R1+1/R2

Resistors in series... Rt=R1+R2

In the old days when we didn't have calculators lol, you needed to know how to add fractions(in the case of parallel resitors)but I'm guessing, all that is needed these days, is a calculator lol

so the 16.66 amp coming from the battery on the diagram is correct ?

ok if i try what you have there 1/RT=1/r1+1/R2

1/rt = 1/1.44 + 1/1.44 = 2/2.88 then another source says turn it on its head so 2.88/2 then calculate 2.88/2 = 1.44ohms ::confused:

[21QUEST]

so the 16.66 amp coming from the battery on the diagram is correct ?

ok if i try what you have there 1/RT=1/r1+1/R2

1/rt = 1/1.44 + 1/1.44 = 2/2.88 then another source says turn it on its head so 2.88/2 then calculate 2.88/2 = 1.44ohms ::confused:
Yep, the total current is correct.

The formular/rule I gave ie 1/Rt=R1+R2 is actually the same as what timwilky said.

1/1.44 + 1/1.44 is not 2/1.44...that's where the bit of being able to add fractions comes in. Though, to be honest, you only need to use a calculator. You have one of those so called 'scientific calculators' right ?

How about you try doing it this way. Find the answer for 1/1.44 (same for each bulb). Now add both together.

The sum is equal to 1/Rt

BUT you aren't finished yet...since you are actually want Rt and not 1/Rt. To now find Rt you need to find the reciprocal of I/Rt and then also need to do the same for the answer you got from 1/1.44 + 1/1.44 .

To this, just flip 1/Rt and then divide 1 by the result(see red bit) you got to earlier ....just keeping it uncomplicated.

Sorry, I'm not always the best at explaining such stuff.... :)

[barwel1992]

got it :D

1/1.44 + 1/1.44 = 2/1.44 = 1.44(divide by)2 = 0.72ohms resistance

[21QUEST]

In that example, yup.

How about this to double make sure :) .

Whats 1/1.44 + 1/2.88 .

[barwel1992]

In that example, yup.

How about this to double make sure :) .

Whats 1/1.44 + 1/2.88 .

1/1.44 + 1/2.88 = 3/1.44 = 1.44/3 = 0.48ohms ?

:)

[21QUEST]

Should be 0.96 .


Best is to learn how to add fractions...so you actually know the methology.

Alternative, just do each dividson seperately(1 divided by whatever figure) using calculator, add all the results together and then flip over.... dividing 1 by the result got from adding them together.

[barwel1992]

ok then so where i went wrong was the 1/2.88 that should be left alone and the 1/1.44 be changed to 2/2.88 ?

if so i did it wrong and changed the 1/2.88 to 2/2.88 (idk why i did that)

so 1/1.44 + 1/2.88 change to 2/2.88 + 1/2.88 = 3/2.88 = 2.88/3 = 0.96