How to Use Current Divider Law to Calculate Ix in a Circuit

  • Thread starter Thread starter arkturus
  • Start date Start date
  • Tags Tags
    Current Law
AI Thread Summary
The discussion centers on calculating the current Ix using the current divider law in a circuit with multiple resistors. The total current is established as 10 A, and the challenge lies in determining the correct values for R_total and R_x. Participants clarify that R_total refers to the equivalent resistance of the parallel resistors, while R_x is the resistance of the specific resistor in question, which is 80 ohms. The correct application of the current divider formula is emphasized, with examples showing how to compute the current through each resistor based on their configurations. Ultimately, the calculations confirm that the current in the 80-ohm resistor is approximately 1 A.
arkturus
Messages
27
Reaction score
0

Homework Statement


I'm just trying to calculate Ix using the current divider law.

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


Homework Equations


Ix = I_total * (R_total / R_x)


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.
 
Physics news on Phys.org
What answer did you get and what does your reference say the correct answer is?
 
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.
 
vk6kro said:
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.

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.
 
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.
 
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.
 
This is the Wikipedia entry:

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

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:
Thanks a lot everyone
 

Similar threads

Replies
3
Views
2K
Replies
4
Views
2K
Replies
22
Views
2K
Replies
24
Views
3K
Replies
17
Views
12K
Replies
2
Views
2K
Replies
4
Views
2K
Back
Top