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Thanks.

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- Thread starter jimmy42
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- #1

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Thanks.

- #2

tiny-tim

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apply Kirchhoff's rules …

the sum of the voltage drops going round the loop must be zero, so the ratio of the currents must be … ?

- #3

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So, is it just as simple as the wire with two resistors will have 4mA and the other 6mA?

- #4

tiny-tim

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if you're not guessing, then show us your Kirchhoff's equation to prove this!

- #5

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All I know is Ohms Law. so V=iR. Do I need that here?

I know the total resistance to be 12 k Ohm. The wires rejoin and then go onto other stuff.

- #6

tiny-tim

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All I know is Ohms Law. so V=iR.

oh i see … i assumed you'd know Kirchhoff's rules …

sorry, but you do need them here …

KCL (the current law, or junction law) says total current in = total current out at any junction;

KVL (the voltage law, or loop law) says total iR round any loop = 0 (or = the voltage of any battery in the loop, if there is one) …

in this case, if you mark the current with arrows, and go round the loop adding iR for each resistor (counting i as

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- #8

tiny-tim

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that's not possible (unless all the reisistors are on one branch) …

the resistors on one branch

draw it and you'll see

- #9

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-----R1-------R2-----

1------- -------1

----R3----R4---R5----

Those 1s will loop back to each other and attach to a battery, so how can it go in opposite directions?

- #10

tiny-tim

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So, lets say one wire with 10ma splits into two. On one of the wires there are two resistors each with 10k ohm and on the other wire three resistors each with 10k ohm. Does the current just split in two? Does the amount of resistance affect this?

… i assumed you meant that the current joined up again after the resistors, and carried on round the circuit to the other terminal of the battery

- #11

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The drawing did not turn out so well. What is the equation? Maybe I can work it out from that.

- #12

tiny-tim

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i

round the loop the other way …

-i

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