if two or more impedances are in parallel, the current that enters the combination will be split between them in inverse proportion to their impedances (according to Ohm's law). It also follows that if the impedances have the same value the current is split equally.
In the circuit, you can use current division carefully. For example, on the right, if you combine the 9k and 3k into 12k, and you know I3, then you can calculate I4 and I5 via current division. In other words, you have to have a known current flowing into 2 or more parallel resistors. NOT series parallel combo's, just parallel resistors.
For more than 2 parallel resistors, say R1.R2,R3, and you want the current into R1, you have to calculate the parallel equivalent to R2 and R3, ie, you reduce it to two resistors in parallel. Then apply the formula.
bottom line, current division (and also voltage division) gets you nothing you can't get with simple ohms law and the rules of series and parallel resistors. Some people find them confusion, and skip them entirely. They gain you no more info, as I said, than ohms law and the rules of series and parallel resistors.
What are the conditions that must be must in order for you to use current division?
Use current division only for two paths. If you have more than two paths, you cannot use current division with one exception, if the other paths can be reduced to a single equivalent resistance, then you may use current division.
Some circuits I have seen are neither parallel nor series so does current division still work for those?
No; the circuit described sounds like a Δ configuration; you should use Δ - Y conversion to make that circuit become a more analyzable (Please see reference)
Is current division STRICTLY for resistors in parallel only?
See first answer.
LASTLY, what do you do when you want the current going into a resistor that is in parallel with 10 other resistors.....what is the current division formula for that?
You use the same current division formula but you use equivalent resistances for the formula:
To find I2 and I3, you need to compute an equivalent resistance R_i3
R_i3 = 3k + R_i4
R_i4 = 1/(1/4k + 1/(9k + 3k) = (36k + 12k)/(4k + 9k + 3k) = 48k/(16k) = 3k
R_i3 = 6k
R_i2 = 1/(1/6k + 1/6k) = 3k
Now use the voltage divider rule:
Va = (12)3k/(9k + 3k) = 3V
This means that the current entering node 2 is:
I1 = 9V/9k = 0.001 A
I2 flows through the 6k and the rest of 1 mA flows through the R_i3 equivalent which is, also, 6k. You do not need to formula for this the current divides in half:
I2 = 0.0005 A
I3 = 0.0005 A
I4 = (0.0005 A)(12)/(4 + 12) = 0.000375 A
I5 = 0.0005 A - 0.000375 A = 0.000125 A
Vb = 4k(I4) = 4k(0.000375 A) = 1.5 V
Vc = 3k( 0.000125 A) = 0.375 V
the principle you are using is: for each node Iin=Iout
To find the current entering node 3 you must first analyze node 2
at node 2 you can say that I2=(I1*Req)/6 where Req is the equivalent reistance of everything except the V1 resistor because I1 already went through that one. I3=I1-I2
Now that you know I2 and I3 you can find I4 and I5 by saying that I4=(I3(Req)/4, but now Req is the equivalent resistance of everything to the right of the V3 resistor. I5=I3-I4
V divide rule is much easier to use when the circuit is only paralell resistors because you can use Isource and the same Req for each path (In=IsourceReq/Rn). This problem is tougher because you need to calculate a new Req and use a different Isource each time you analyze a node.
Current divides in a parallel circuit becaus there are two or more paths to take.
In a series circuit all the current goes through all the components as there is nowhere else to go and it has to get back to earth.
Calculating the current in both circuits together then you have to work out the parallel resistances to make them one resistance. then use that one resistance to calculate the series part of the circuit.
I know this concept is suppose to be reallly simple but I have some particular questions that I can't find answers to on the web....
What are the conditions that must be must in order for you to use current division?
Some circuits I have seen are neither parallel nor series so does current division still work for those?
For example, look at the current entering node 3 in below picture....2 resistors are in series and those 2 are in parallel with another resistor....so in their case do you add the series one up to get only 2 resistors so you can do current division ???????
also, is current division STRICTLY for resistors in parallel only?
LASTLY, what do you do when you want the current going into a resistor that is in parallel with 10 other resistors.....what is the current division formula for that? My general formula I was taught was:
I=Itotal *(Ropposite/(Ropposite+Ri))
http://i.imgur.com/jfht4.png
