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

ccbike

New member
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.

. . . . .\(\displaystyle I_3\, \cdot\, R_3\, =\, I_1\, \cdot\, R_1\)

. . . . .\(\displaystyle I_x\, \cdot\, R_x\, =\, I_2\, \cdot\, R_2\)

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

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

Thanks
 

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Dr.Peterson

Active member
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.

. . . . .\(\displaystyle I_3\, \cdot\, R_3\, =\, I_1\, \cdot\, R_1\)

. . . . .\(\displaystyle I_x\, \cdot\, R_x\, =\, I_2\, \cdot\, R_2\)

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

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

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

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 R[SUB]x[/SUB].

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