Question: How do I prove the current division law for two resistors in parallel?

[RadiationX]

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]

I = V/R, correct? Please show us more work. Relax and focus.

 

[RadiationX]

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]

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]

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!

 

[hage567]

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]

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 $I_1$.