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I asked a question here about calculating the resistance an unknown resistor using superposition.
As part of the answer I was given $$I_1=I_3\frac{R_2+R_i}{R_1+R_2+R_i}$$ and $$I_2=I_3\frac{R_1}{R_1+R_2+R_i}$$ as the third case of the superposition method involving a current source.
My confusion is that firstly, I understood that the current divider rule was $$\frac{I_0}{I_n}=\frac{R_{sum}}{R_n}$$ and the above answer is opposite.
Secondly I see \$R_1\$ in parallel with \$R_2+R_i\$ so I don't understand why $$I_1=I_3\frac{R_2+R_i}{R_1+R_2+R_i}$$

I have the feeling there's something that i have fundamentally misunderstood.

My understanding is $$I_n=\frac{V_n}{R_n}$$ and the voltage in a parallel circuit is $$V_n=V_{total}=R_{total}\cdot I_{total}$$ so that $$I_n=\frac{R_{total}\cdot I_{total}}{R_n}$$ which is just what is says here. What people are telling me on here seems to be that the resistance in the other branch divided by the total resistance is equal to the resistance I'm looking for.

I've looked for proofs, but I seem to just find the formula which I knew to begin with, which I'm being told here is wrong. What am I missing?

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For two resistors R1 and R2 in parallel the current divider rule is $$ I_1 = I \cdot \frac{R_2}{R1+R_2} $$ where I is the total current and I1 the current through R1.

For two resistors the current through one resistor is the total current times the current of the other resistor divided by the sum of the resistors.

The proof is trivial but here it is:

For two parallel resistors with the same voltage V across them the current through one of them is given by $$ I_1 = \frac{V}{R_1} = I \cdot \frac{R_1 R_2}{R_1 + R_2} \cdot \frac{1}{R_1}= I \cdot \frac{R_2}{R_1 + R_2} $$ since V can be expressed as the product of the total current I times the resistance of R1 and R2 in parallel.

In your first example the total current is I3. The current I1 flows through R1, the other resistor is R2+Ri, the sum is R1+R2+Ri which results in $$I_1=I_3\frac{R_2+R_i}{R_1+R_2+R_i}$$

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  • \$\begingroup\$ Ok, but why? See edited question above.. \$\endgroup\$ – Andy Grey Dec 12 '16 at 9:14
  • \$\begingroup\$ I've added the proof. I don't know if that's what you were after. \$\endgroup\$ – Mario Dec 12 '16 at 11:35
  • \$\begingroup\$ Cheers, I'd just realised this myself :) \$\endgroup\$ – Andy Grey Dec 12 '16 at 11:36

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