PDA

View Full Version : Proof of current division


RadiationX
May16-07, 11:44 PM
My professor gave us the task of proving the current division law for two resistors in parallel. I know that the voltage across parallel resistors is the same, and that the sum of the branch currents adds to the sum of the total current.

Here is the question:

Prove that I_1= \frac{I_sR_2}{R_1 + R_2} ; I_s = the source current.

Now, this is what I know: I_s= I_1 +I_2

That is about as far as I can get. I know that I need to manipulate I_s= I_1 +I_2 somehow to derive I_1= \frac{I_sR_2}{R_1 + R_2}.

This is the first time I have ever been assigned a proof and I don't have any experience solving them, so my tool set is kind of lacking.

As a side note I would like to be able to prove 'simple' theorems like this. I think it would give me a level of insight that I don't currently have.

berkeman
May17-07, 12:18 AM
I = V/R, correct? Please show us more work. Relax and focus.

RadiationX
May17-07, 12:37 AM
yes I = V/R so...

I_s= \frac{V_s}{R_1} +\frac{V_s}{R_2}

It appears that the thing to do here is get rid of the fractions. I have done this on paper, but I am not seeing a critical relationship when I work through it.

berkeman
May17-07, 01:42 AM
yes I = V/R so...

I_s= \frac{V_s}{R_1} +\frac{V_s}{R_2}

It appears that the thing to do here is get rid of the fractions. I have done this on paper, but I am not seeing a critical relationship when I work through it.

Good. So write the equation for each leg....

RadiationX
May17-07, 06:44 AM
The equation for each leg? Well, each leg is just back to the current again. Right? \frac{V_S}{R_1} + \frac{V_S}{R_2} = I_1 + I_2.

I'm running in circles!:cry:

hage567
May17-07, 07:53 AM
Try to work out an expression for Vs in terms of the total resistance of the circuit and Is (so find the equivalent resistance of two resistors in parallel).

Fredrik
May17-07, 09:15 AM
You can eliminate I_2 from the equation I_s=I_1+I_2, if you use the V=RI law to express I_2 as a function of I_1.