Question

A voltage V across a resistance R generates a current I=V/R.
A constant voltage of 8 volts is put across a resistance that is increasing at a rate of 0.4 ohms per second when the resistance is 5 ohms, at what rate is the current changing?

***round answer to three decimal places***

The current is changing at a rate of ________ amperes/second.

please explain? Thank you!!!

Answer 1

@iheartfood We have
\[I=\frac{V}{R}\]
Could you differentiate I with respect to T, treat V as constant and R as variable

Answer 2

-V/r^2 ?

Answer 3

Yes, but it's differentiated with respect to t, so you will will get
\[\frac{dI}{dT}=\frac{-V}{R^2}\times \frac{dR}{dT}\]

Answer 4

Do you follow?

Answer 5

yes:) ....so far haha :P

Answer 6

Just need to plug in the values on the left side and you get the rate of change of current

Answer 7

Could you try?

Answer 8

okay... :)

umm so is this close (ish? haha) ?

dI -8 dR
--- = ----- * ------
dt 5^2 dT

? :/

Answer 9

Yes, it's close
"increasing at a rate of 0.4 ohms per second when the resistance is 5 ohms"
When R=5 ohm
\[\frac{dR}{dT}=0.4 ohm/sec\]

Answer 10

ahh darn :P

ohh okay

so we get this?

dI/dT = -8
----- * 0.4
25

Answer 11

yes Sir. It does make sense with rate of resistance increase positive, rate of current will be negative.

Answer 12

ahh okay yay! :)

so then we get this?

-0.128

?

Answer 13

yes, correct

Answer 14

ahh!! woo! thank you so much!! :D

Answer 15

Welcome :)