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

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Discussion Overview

The discussion revolves around calculating the current Ix in a circuit using the current divider law. Participants explore various methods to apply this law, including the use of equivalent resistances in parallel and series configurations. The conversation includes attempts to clarify the relationships between total current, total resistance, and individual resistances within the circuit.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in applying the current divider law and understanding the definitions of R_total and R_x.
  • Another participant suggests reducing the three resistors in parallel to a single equivalent resistor to simplify calculations.
  • A different approach is proposed involving the calculation of voltages across resistors to determine current flow.
  • One participant provides a detailed calculation of equivalent resistance and current using the current divider law, showing specific values for each branch.
  • There is a mention of different methods for calculating current in parallel branches, highlighting that both the current divider law and a shorthand method can yield the same results.
  • A participant shares a reference image from Wikipedia to illustrate the current divider concept, providing a specific calculation for the current in the 80 ohm resistor.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to apply the current divider law, with multiple approaches and calculations presented. Some participants agree on the total current being 10 A, but there are differing interpretations of how to apply the law and calculate individual currents.

Contextual Notes

There are unresolved assumptions regarding the definitions of equivalent resistances and the application of the current divider law in different configurations. Some calculations rely on specific interpretations of circuit arrangements that may not be universally agreed upon.

arkturus
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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.
 
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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
 

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