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January 27, 2015

January 27, 2015

Posted by **archi** on Monday, November 7, 2011 at 6:07pm.

1/R=1/R1+1/R2

If the resistance of R1 and R2 are increasing at 330 and 660 ohms per second respectively, then the rate of increase of R when R1=5000 and R2=5000 is how many ohms per second?

- calculus - please help! -
**Steve**, Monday, November 7, 2011 at 8:06pmJust plug and chug

1/R = 1/R1 + 1/R2

1/R = 1/5000 + 1/5000 = 2/5000

R = 2500

-1/R^2 R' = -1/R1^2 R1' - 1/R2^2 R2'

-1/625E4 R' = -1/25E6 * 330 - 1/25E6 * 660

1/625E4 R' = 990/25E6

R' = 990*625E4/25E6 = 990*625/2500

R' = 247.5 ω/s

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