Rate of change: Resistors and rate of change in R1

[wind]

Three resistors, R1, R2 and R3, are connected in parallel in an electrical circuit. The total resistance, R, in ohms is given by:

. . .1/R = 1/R1 + 1/R2 + 1/ R3

R, R2, and R3 are increasing at rates of 0.1, 0.5 and 0.7 ohms/s, respectively. Determine the rate at which R1 is changing when R = 50 ohms, R2 = 70 ohms, and R3 = 60 ohms.

d/dt 1/R = d/dt 1/R1 + d/dt 1/R2 + d/dt 1/ R3

(0.1) 1/50 = d/dt 1/R1 + (0.5) 1/70 + (0.7) 1/ 60
0.002 = d/dt 1/R1 + 0.007 + 0.011
-0.016 = d/dt 1/R1

What do I do now? Thank you!
______________________________________________
Edited by stapel -- Reason for edit: spelling, punctuation, capitalization, etc

 

[soroban]

Re: Rate of change- Resistors

Hello, wind!

Sorry, I don't follow your notation . . .

Three resistors, \(\displaystyle R_1,\:R_2,\:R_3\) are connected in parallel in an electrical circuit.

The total resistance \(\displaystyle R\) in ohms is given by: \(\displaystyle \L\:\frac{1}{R}\;=\;\frac{1}{R_1}\,+\,\frac{1}{R_2}\,+\,\frac{1}{R_3}\)

\(\displaystyle R,\:R_2,\:R_3\) are increasing at rates of \(\displaystyle 0.1,\:0.5,\:0.7\) ohms/s, respectively.

Determine the rate as which \(\displaystyle R_1\) is changing when: \(\displaystyle R\,=\,50,\:R_2\,=\,70,\:R_3\,=\,60\)

We have: \(\displaystyle \R)^{-1} \;=\;\left(R_1\right)^{-1}\,+\,\left(R_2\right)^{-1}\,+\,\left(R_3\right)^{-1}\)

Differentiate with respect to time:

. . \(\displaystyle \,-(R)^{-2}\left(\frac{dR}{dt}\right) \;=\;-\left(R_1\right)^{-2}\left(\frac{dR_1}{dt}\right) \,-\,\left(R_2\right)^{-2}\left(\frac{dR_2}{dt}\right)\,-\,\left(R_3\right)^{-2}\left(\frac{dR_3}{dt}\right)\)

. . . . . \(\displaystyle \frac{1}{R^2}\left(\frac{dR}{dt}\right)\;=\;\frac{1}{R_1^2}\left(\frac{dR_1}{dt}\right)\,+\,\frac{1}{R_2^2}\left(\frac{dR_2}{dt}\right)\,+\,\frac{1}{R_3^2}\left(\frac{dR_3}{dt}\right)\;\) [1]


We are given: \(\displaystyle \,\frac{dR}{dt}\,=\,0.1,\:\frac{dR_2}{at}\,=\,0.5,\:\frac{dR_3}{dt}\,=\,0.7\)

When \(\displaystyle R\,=\,50,\:r_2\,=\,70,\:R_3\,=\,60\), we have: \(\displaystyle \:\frac{1}{50}\:=\:\frac{1}{R_1}\,+\,\frac{1}{70}\,+\,\frac{1}{60}\;\;\Rightarrow\;\;R_1\,=\,52.5\)

Then [1] becomes: \(\displaystyle \:\frac{1}{50^2}(0.1)\;=\;\frac{1}{52.5^2}\left(\frac{dR_1}{dt}\right)\,+\,\frac{1}{70^2}(0.5)\,+\,\frac{1}{60^2}(0.7)\)

We have: \(\displaystyle \:\frac{1}{52.5^2}\left(\frac{dR_1}{dt}\right) \;= \;\frac{0.1}{50^2}\,-\,\frac{0.5}{70^2}\,-\,\frac{0.7}{60^2}\)

. . . . . . . . . . . . \(\displaystyle \frac{dR_1}{dt}\;=\;52.5^2\left(\frac{0.1}{50^2}\,-\,\frac{0.5}{70^2}\,-\,\frac{0.7}{60^2}\right) \;=\;-0.7069375\)

Therefore, \(\displaystyle R_1\) is decreasing at the rate of about 0.7 ohms per second.

 

[skeeter]

d/dt[1/R= 1/R₁ + 1/R₂ + 1/R₃]

(-1/R²)(dR/dt) = (-1/R₁²)(dR₁/dt) + (-1/R₂²)(dR₂/dt) + (-1/R₃²)(dR₃/dt)

now substitute in your given values and find (dR₁/dt).

 

[wind]

Thanks, so soroban's method is right except for the negative

 

[skeeter]

I didn't simplify anything in my post ... all the negative signs will cancel, won't they?

 

[wind]

Then, is'nt your post and soroban's line "[1]" the same? but the answer in the back of the book, 0.128, is different form soroban's. So what was wrong? Sorry I don't understand

 

[stapel]

Why don't you try the exercise, following along from what you've been given by the tutors in this thread and have been taught in the various other threads. See if you can find any errors.

(The book might have a typo, or the tutor might have made a mistake. Either way, you need to learn how to do these.)

Please reply clearly showing all of your work. Thank you.

Eliz.

 

[skeeter]

recheck the given values you posted for the resistors ...

R = 50 ohms, R2 = 70 ohms, and R3 = 60 ohms.

using those values, R1 works out to be a negative value ... which cannot be true.

 

[wind]

using those values, R1 works out to be a negative value ... which cannot be true.

thoes values are correct, the answer is negative

 

[skeeter]

I'm not talking about R1's rate of change.

R1 cannot have a resistance value less than 0.
in other words ...

1/50 = 1/R1 + 1/70 + 1/60

1/50 - 1/70 - 1/60 = 1/R1

-23/2100 = 1/R1

-2100/23 = R1

no such thing as negative resistance.

 

[wind]

oh, but thoes values are right:

R=50
R2=70
R3=60