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# Physics question 2) watch

1. a similar question, but I need some help on this
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2. Have you uploaded the correct image?
3. The current splits in the ratio of resistances.
5. (Original post by SpikeSpiegel)
In the first junction, the current will split in the ratio 1:7.
Continue working in this way for each junction.
6. Hi I still don't understand . can anyone help pls?
7. (Original post by SpikeSpiegel)
Hi I still don't understand . can anyone help pls?
You can either use the rule that the current splits it the ratio of the resistances and apply this to every junction and the sum them if you have multiple currents going through a wire (as you would in the middle one).

Or you can apply the same analysis as in the last question: Label all currents into a junction I1, I2, I3 etc and write down equations that link them using Kirchoff's current law. You can then write down voltage equations for each loop and simplify with V=IR and finally solve simultaneous equations for the currents.

The first way is easier, but the second provides practice at a method which always (afaik) works.
8. (Original post by SpikeSpiegel)
Hi I still don't understand . can anyone help pls?
Start at the first junction, the resistors there are 2 and 14, so you need to split the current 1:7. Since you have 8A, you can write in 1A going through the 2R resistor and 7 going through the 14A one.

Do this for all junctions, and if you get two currents going through the same wire, add them.
9. ok so, total current in R1 and R2 is 8A?

so i have do the same with R4 and R5?
10. (Original post by SpikeSpiegel)
ok so, total current in R1 and R2 is 8A?

so i have do the same with R4 and R5?
Yes, that should work.
11. I've annoted the currents on. At each junction, you can write an equation for the currents, e.g.:

(sorry for not putting arrows on for direction. Hopefully you can work those out; if you want something to go in the opposite direction just put a -ve sign in front)

Start by doing this at every junction. Sum of currents in = sum of currents out.
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12. At any junction, the current going in is equal to the current going out.

You can also do this at the other junctions eg.

Also in parallel circuits, the potential difference is the same for each branch.
So eg.
13. so what is it that i have to work out exactly? Current values for r1,2,3,4,5?? thats all?
14. (Original post by SpikeSpiegel)
so what is it that i have to work out exactly? Current values for r1,2,3,4,5?? thats all?
Yes
15. so if r1= 1A and r2 = 7A

R3 = -6A

right?
16. Kirchoff's first law is what you want:

17. ok i finally have an answer... im 99.9% sure i haven't done this right.. but can someone please check this

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18. (Original post by SpikeSpiegel)
ok i finally have an answer... im 99.9% sure i haven't done this right.. but can someone please check this

I don't think it says that the current labelled I4 is 8Amps. It says the two currents labelled I1 are 8A each.
19. I'm not quite sure why, but the 'splitting in ratio of resistances' doesn't seem to work for this case.

Probably the quickest way is to form simultaneous equations: write down equations which link I1,2,3 etc together and then replace I1 with 8. Next, write down two voltage equations (you need two because you've got two points where current splits). The voltge equations should basically read like this:

Those are for example- you need to make sure your equations are in terms of resistance and current. Then, fiddle around with simultaneous equations until you have two equations with only three currents in (I used I2,3 and I4). You should be able to eliminate all but one current if you use the fact that I1+I2=8. Once you have one current, you can solve for all of them using the current equations you formed earlier.

I know I phrases that quite badly, so if you need any help on this just ask (and just for a hint, the currents are all integers, and none of them are 7 or 8 or higher).
20. i still dont get it

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