# Thread: Given I_3 * R_3 = I_1 * R_1, I_x * R_x = I_2 * R_2, solve for R_x

1. ## Given I_3 * R_3 = I_1 * R_1, I_x * R_x = I_2 * R_2, solve for R_x

I am trying to find out the math used to divide the 2 equations and come up with the solution below and isolate Rx. Any help is appreciated.

. . . . .$I_3\, \cdot\, R_3\, =\, I_1\, \cdot\, R_1$

. . . . .$I_x\, \cdot\, R_x\, =\, I_2\, \cdot\, R_2$

The equations are divided and rearranged, giving the below. Not sure how the match was done?

. . . . .$R_x\, =\, \dfrac{R_2\, \cdot\, I_2\, \cdot\, I_3\, \cdot\, R_3}{R_1\, \cdot\, I_1\, \cdot\, I_x}$

Thanks

2. Originally Posted by ccbike
I am trying to find out the math used to divide the 2 equations and come up with the solution below and isolate Rx. Any help is appreciated.

. . . . .$I_3\, \cdot\, R_3\, =\, I_1\, \cdot\, R_1$

. . . . .$I_x\, \cdot\, R_x\, =\, I_2\, \cdot\, R_2$

The equations are divided and rearranged, giving the below. Not sure how the match was done?

. . . . .$R_x\, =\, \dfrac{R_2\, \cdot\, I_2\, \cdot\, I_3\, \cdot\, R_3}{R_1\, \cdot\, I_1\, \cdot\, I_x}$

Thanks
Actually, the first equation tells you that the result you show could be simplified by cancelling $I_3 \cdot R_3$ and $I_1 \cdot R_1$.

If you do that, you will find that the result you want is just what you get if you solve the second equation (alone) for Rx.

If for some reason you wanted the overly complicated final form, you would multiply what you get by $\dfrac{I_3 \cdot R_3}{I_1 \cdot R_1} = 1$.

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