# Current divider law help

1. Sep 29, 2010

### arkturus

1. The problem statement, all variables and given/known data
I'm just trying to calculate Ix using the current divider law.

http://i273.photobucket.com/albums/jj224/illway17/circuitS.jpg

2. Relevant equations
Ix = I_total * (R_total / R_x)

3. The attempt at a solution

I've had an issue with the voltage & current divider law for a while. They seem to require some common sense which I'm lacking.

Anyway, I know that I can find the total current by condensing the circuit down to the voltage source and a single resistors. I_total = 10 A.

The tricky part for me is figuring out what R_total and R_x equal. I've always assumed that R_total is the resistance of the entire circuit while R_x is the resistance of the item in question (in this case the 80 ohm resistor).

I'm not getting the correct answer based off of my work, so any help would be appreciated.

2. Sep 29, 2010

### Zryn

3. Sep 29, 2010

### vk6kro

You have 3 resistors in parallel, so you can reduce them to a single resistor.

Then you have two resistors in series, so you add them together to get the total resistance and from this you could work out the total current.

OK. You have already done this, to get 10 amps.

Now, you can work out the voltage across the 3 resistors in parallel.

It is Vrparallel = (rparallel / rtotal) * 100 volts.
(Or, you could just subtract the voltage across the 2 ohm resistor from 100 volts (10 amps * 2 ohms = 20 volts).

If you know the voltage across each of the resistors, you can work out how much current flows in each of them.

And that's it.

4. Sep 29, 2010

### arkturus

I like your method, but I'm trying to do it through the current divider law.

The correct answer is Ix = ((16||20) / (16||20 + 80)) * 10A

I understand where the 10A comes from, that's just the total current. However, I don't understand why they are using 16||20 as Rtotal and 16||20 + 80 as Rx.

5. Sep 29, 2010

### Zryn

There are several ways of doing this. Your way is the current divider when you have any number of parallel branches (Ix = I_total * R_total / R_x), their way is a shorthand method when you have only two parallel branches (Ix = I_total * R_parallel / R_total).

Try it on a variety of different combination's and see how it works out, you should get the same answers.

6. Sep 29, 2010

### d-rock

Alright, let's start from the beginning.

Req = [(80 || 20) || 16] + 2

80 || 20 = 80(20)/(80 + 20) = 16
((80 || 20) || 16) = 16(16)/(16 + 16) = 8
and lastly, + 2, so:

Req = 10 ohms

Then Ieq = 100 V / 10 ohms = 10 A. You had this correct.

Now, the current divider, let's do this for all the resistors. Assume that Ia, Ib, and Ic, are the 2, 16, and 20 ohms respectively.

Ia = 10 A * 10 ohms / 2 ohms = 50 A
Ib = 10 A * 10 ohms / 16 ohms = 6.25 A
Ic = 10 A * 10 ohms / 20 ohms = 5 A
Ix = 10 A * 10 ohms / 80 ohms = 1.25 A

Now, let's test the voltages:

Va = 50 A * 2 ohms = 100 V
Vb = 6.25 A * 16 ohms = 100 V
Vc = 5 A * 20 ohms = 100 V
Vx = 1.25 A * 80 ohms = 100 V.

And so these are correct. There are two rules you have to remember about resistors, as far as series and parallel:

1. Resistors in series have the same current
2. Resistors in parallel have the same voltage drop

Which is what is used to be able to do these laws. And as all 4 of them are in parallel, then all 4 have a 100 V drop across them.

7. Sep 29, 2010

### vk6kro

This is the Wikipedia entry:

[PLAIN]http://dl.dropbox.com/u/4222062/current%20divider.PNG [Broken]

So, in this case,

Current in 80 ohm resistor = Total resistance of other parallel resistors (16 ohms // 20 ohms) / ( 80 ohms + ( 16 ohms // 20 ohms ) times 10 amps.

So current in 80 ohms = 8.89 / (80 + 8.89) *10 = 1 amp

Last edited by a moderator: May 5, 2017
8. Sep 29, 2010

### arkturus

Thanks a lot everyone