Question

In an electric circuit with two resistors of resistance R1 and R2 connected in parallel, the total resistance R is
1/R=1/R1+1/R2
Show that R1*dR/dR1+R2*dR/dR2=R

Answer 1

D represents the partial derivative

Answer 2

Any ideas?

Answer 3

lol i was about to ask that haha so what have you tried so far ?

Answer 4

looks like you just need to find the partials and plug in right ?

Answer 5

So find the partials and plug them in for what

Answer 6

plug them in the left hand side of what ever you want to prove

Answer 7

Is the partial of 1/R =-R since its R to the negative 1

Answer 8

\[\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\]

differentiate implicitly with.respect.to \(R_1\). i.e, assume \(R_2\) is a constant :
\[-\frac{1}{R^2} \frac{\partial R}{\partial R_1} = -\frac{1}{R_1^2}+0\]
\[\frac{\partial R}{\partial R_1} = \frac{R^2}{R_1^2}\]

Answer 9

similarly see if you can find the other partial \(\large \dfrac{\partial R}{\partial R_2}\)

Answer 10

Where did the -1/R^2 come from

Answer 11

recall the derivative of 1/x with respect to x

Answer 12

\[\dfrac{d}{dx}\left(\frac{1}{x}\right) =\dfrac{d}{dx}\left(x^{-1}\right) = -1x^{-1-1} = -\frac{1}{x^2}\]

Answer 13

Its -x^-2 but in this case we did the - (R1^-2) * R^2 why R^2

Answer 14

Is the partial derivative of R2= R^2/R2^2

Answer 15

yes thats right! work it out and see if you really get that

Answer 16

That's what I got. so the partial of R must be -1/R^2 or no?

Answer 17

parital of R with respect to what ?

Answer 18

How do you show what R=

Answer 19

plug the partials in left hand side of what ever you want to show

Answer 20

You want to show `R1*dR/dR1+R2*dR/dR2 = R` right ?

Answer 21

Yea not sure what to plug in

Answer 22

LHS = R1* `dR/dR1`+R2* `dR/dR2`

Answer 23

plugin the values of `dR/dR1` and `dR/dR2`

Answer 24

LHS = R1* `dR/dR1`+R2* `dR/dR2`

= R1 * `R^2/R1^2` + R2* `R/R2^2`
= R^2/R1 + R^2/R2
= R^2(1/R1 + 1/R2)
= R^2(1/R)
= R

Answer 25

there you have complete work
see if that makes sense more or less

Answer 26

Im still confused on why the partial of R1 is R^2/R1^2

Answer 27

are you fine wid this step :
http://gyazo.com/12fed2172b9ee8a71f90826b48088f97
?

Answer 28

No. The 0 is the partial of R2 and the -1/R1^2 is the R1 partial y do you have the -1/R^2 at the beginning

Answer 29

Ahh now I see what you're asking

Answer 30

let me ask you a side question

Answer 31

whats the derivative of \(\large \frac{1}{y}\) with respect to \(x\) ?

Answer 32

0

Answer 33

Right! another question :
whats the derivative of \(\dfrac{1}{y(x)}\) with respect to \(x\) ?

Answer 34

0

Answer 35

Notice, \(y\) is not longer a constant. It is a function of \(x\) now. So you can't say the derivative is 0

Answer 36

since \(y\) is not a constant, we have : \[\dfrac{d}{dx}\left(\dfrac{1}{y}\right) = -\dfrac{1}{y^2} \dfrac{dy}{dx}\]

Answer 37

Oh thanks for all your help

Answer 38

np :) let me knw if you have questions.. asking questions is a great way to learn fast !