# Basic Current Divider Question

Discussion in 'Homework Help' started by Hitman6267, May 6, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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Never mind the not connected branches a and b in the middle. They are part of another exercise which isn't related to my question.

Lets say I want to calculate the current in the branch that contains AB. (I know its Isrc/2 but I want to calculate it using the formula.)

Iab = Req Isrc / Rab

Req is the equivalent resistance of the two branches in parallel.
Req= (R1+R2 ) /2

My question is when is Req the equivalent of Rab and Rcd only ? When the branch ab (the ones I told you to forget) is connected for example? And why is that ?

2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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In your diagram, you've used capital letters (A,B,C,D) to designate certain nodes, and lower case letters (a,b) to designate certain nodes.

Then you've given an algebraic expression where I believe you've used those designators in a confused manner.

Iab = Req Isrc / Rab

i think that should have been:

IAB = Req Isrc / RAB

because RAB is a different resistance than Rab; the branch AB is a different branch than branch ab.

Now, as to your question. This is a Wheatstone bridge; see:

http://en.wikipedia.org/wiki/Wheatstone_bridge

When the voltage at node a is the same as the voltage at node b the bridge is said to be balanced, and if a resistor Rab is connected to a balanced bridge, no current will flow in that resistor, because the voltage at each end of the resistor is the same.

In that circumstance, Req will be the equivalent of RAB and RCD only. If the bridge is unbalanced, then that will not be true.

It's also true that if the the bridge is initially balanced, a short can be placed across a-b and the currents in all the resistors will remain the same. That's why you only need to consider the top two resistors in the total left hand branch and right hand branch to get the current division. You could short a-b and the the bottom of RAB and the bottom of RCD would be connected together; then the bottom resistors have no effect on the current division.

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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Why would is short-circuiting a b not affect the currents ? I have a basic understanding of the Wheatstone bridge and I don't see how the analogy helps.

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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355
If the bridge is balanced, the voltage at a is the same as the voltage at b. If you connect a resistor between two nodes that are at the same voltage, then the voltage across the resistor is zero, and no current flows. A short is just the limiting case of a resistor where the resistance approaches zero; the current through a short connected between two nodes which are at the same voltage is zero.

If no current flows through a short which is connected from a to b, then the other currents in the network will be unchanged.

If the bridge is not balanced, then connecting a short between a and b will change other currents in the network.

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
May be this will clear up my question. I'm talking about a general case. I just realized that the circuit I posted is the circuit used to explain the wheatstone bridge and that's why your answers were always related to the bridge.

In this case:

If we want to apply current divider between the 5ohm resistor and 5j impedance. Why do we only consider those impedances and not the whole circuit ? What condition does that satisfy that the other circuit doesn't ?

P.S: I realize that logically you will only consider the 5ohm resistor and 5j impedance but I'm just trying to figure out what makes it logical.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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355
You can understand it if you go through the derivation of the current divider rule yourself; it goes like this:

Replace the two 5 ohm impedances with Z1 and Z2 so you get a general solution.

First lets calculate the voltage across the parallel combination of Z1 and Z2.

Their equivalent impedance is $\frac{Z1*Z2}{Z1+Z2}$

Since Z1 and Z2 are in parallel, the voltage across each of them is the same as the voltage across the parallel combination; let's call it Vp.
By ohm's law, the voltage across the combination is $Vp=Ig*\frac{Z1*Z2}{Z1+Z2}$

Since the voltage across Z1 is Vp, by ohm's law the current through Z1 is:

$I_1=\frac{Vp}{Z1} = Ig*\frac{Z1*Z2}{Z1+Z2}*\frac{1}{Z1}=Ig*\frac{Z2}{Z1+Z2}$

Similarly:

$I_2=\frac{Vp}{Z2} = Ig*\frac{Z1*Z2}{Z1+Z2}*\frac{1}{Z2}=Ig*\frac{Z1}{Z1+Z2}$

These two results are just the current divider rule.

Other impedances in the circuit determine the current Ig, but only Z1 and Z2 determine the way the current splits through parallel combination of Z1 and Z2.

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