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Problem 39

If two resistors with resistances $ R_1 $ and $ R_2 $ are connected in parallel, as in the figure, then the total resistance $ R, $ measured in ohms $ (\Omega), $ is given by

$ \frac {1}{R} = \frac {1}{R_1} + \frac {1}{R_2} $

If $ R_1 $ and $ R_2 $ are increasing at rates of $ 0.3 \Omega/s $ and $ 0.2 \Omega/s, $ respectively, how fast is $ R $ changing when $ R_1 = 80 \Omega $ and $ R_2 = 100 \Omega? $

Answer

0.132$\Omega / \mathrm{s}$

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## Discussion

## Video Transcript

Okay. Recall the fact that we have our run R two d r one over d, t and D are two over DT in this problem. Therefore, we know we can write out an equation. One over R is one of 11 over 80 plus one over r two. This simplifies to nine over 400 if we hop eight common denominators simplified which gives us our is 400 over nine because of one over ours. Not over 400 men are over. One of just our is 400 over nine. Therefore, we know we can multiply 400 over nine squared comes one over 80 squared times 0.3 plus one over 100 squared times 0.2 and then we know we can simplify to get 0.132 on the November. This is in terms of the units for resistance