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

In summary, the homework statement is trying to find the total current through a circuit by condensing it down to the voltage and resistance source. They are having trouble figuring out what the correct answer is based off of common sense and need help from someone with more knowledge. The attempted solution is to use the current divider law when there are multiple parallel branches, and the shorthand method when there are only two parallel branches. They found that by knowing the voltage across each resistor, they could work out the current flowing through each resistor.
  • #1
arkturus
27
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.
 
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  • #2
What answer did you get and what does your reference say the correct answer is?
 
  • #3
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
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.
 
  • #5
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
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
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
 
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  • #8
Thanks a lot everyone
 

1. What is the current divider law?

The current divider law is a principle in electrical engineering that explains how current is divided between parallel branches in a circuit. It states that the current flowing through each branch is inversely proportional to the resistance of that branch.

2. How is the current divider law derived?

The current divider law is derived from Ohm's law and Kirchhoff's current law. It can also be derived using the concept of equivalent resistance.

3. What are the applications of the current divider law?

The current divider law is commonly used in the design and analysis of electrical circuits. It is also used in the calculation of power dissipation in parallel circuits and in the design of voltage regulators.

4. How do I apply the current divider law in a circuit?

To apply the current divider law, you need to identify the parallel branches in the circuit and determine the resistance of each branch. Then, use the formula Ii = (IT * RT) / Ri, where Ii is the current in the ith branch, IT is the total current, RT is the total resistance, and Ri is the resistance of the ith branch.

5. Are there any limitations to the current divider law?

The current divider law assumes that the branches in the circuit have equal voltage sources and that there are no other factors affecting the current flow. It also assumes that the resistance of the branches remains constant. Therefore, it may not be accurate in circuits with non-ideal components or varying conditions.

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