Question

\(\color{blue}{\text{Originally Posted by}}\) @Abhisar
Two resistances \(\sf R_1\) and \(\sf R_2\) are made of different materials. the temperature coefficient of the material of \(\sf R_1~is~\alpha\) and of \(\sf R_2~is~- \beta\). The resistance of the series combination of \(\sf R_1~and~R_2\) will not change with temperature, then ratio of resistance of two wire at 0°C will be ?
\(\color{blue}{\text{End of Quote}}\)

Answer 1

@nincompoop

Answer 2

@abb0t @kropot72

Answer 3

@Somy

Answer 4

and where is the circuit

Answer 5

There is no circuit

Answer 6

@iambatman can u help ?

Answer 7

I would but I'm about to go buy myself a new carpet xD.

Answer 8

aaaahhh come on it's ur left hand's play XD

Answer 9

answer is \(\sf \frac{\beta}{\alpha}\) btw

Answer 10

We got the answers here, but not able to understand
http://openstudy.com/updates/53e2690be4b0045ba5512d19

Answer 11

Is it the coefficient you don't understand?

Answer 12

yes...

Answer 13

This line
If things don't change with temperature, then \(\alpha - \beta = 0 \Rightarrow \alpha = \beta\)?

Answer 14

I am not even able to understand how he solved the equation ?

Answer 15

You have to ignore any changes due to thermal expansion, the resistance of each section changes with temperature, so the total resistance of the materials doesn't change with temperature so that's how he got it basically. (If that makes sense)

Answer 16

Anyways PK did a good job :P, I'll get back to you later if you're still having trouble.

Answer 17

Just tell me one thing...if resistance does not change then what is its effect on coefficient ?

Answer 18

Just tell me how he solved the equation ?

Answer 19

This one

\[R_1 + R_2 = r_1 + r_2 + \Delta T \left(r_1 \alpha - r_2 \beta\right)\]\[\Rightarrow r_1 \alpha = r_2\beta\]And blah blah.
\(\color{blue}{\text{End of Quote}}\)

Answer 20

There's a dependance of the resistance for the series combination of resistance at To, do you know the formula?

Answer 21

no !
What is it ?

Answer 22

I think you should give this a read before you attack these sort of problems haha :P http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html

Answer 23

It's pretty neat stuff

Answer 24

aaahhh..i know this...:D
just not able to how he deduced the last line \(\sf r_1 \alpha =r_2\beta\)

Answer 25

*able to understand

Answer 26

It simplifies to that because delta T is the same for each material

Answer 27

Idk if I'm confusing you more or if it's helping XD

Answer 28

Hmmm...i think i'll have to give some more time on it.

Answer 29

Maybe I'll give you an example which is some what similar but a bit simpler and it will help you understand.

Answer 30

yeah..that will be great !

Answer 31

Alright, well here is a problem: Two wires have the same cross-sectional area and are joined end to end to form a single wire. One is tungsten, which has a temperature coefficient of resistivity of \[\alpha \] 0.0045 (C degrees)^-1. The other is carbon, for which \[\alpha \]= -0.0005 (C degrees)^-1. The total resistance of the composite wire is the sum of the resistances of the pieces. The total resistance of the composite does not change with temperature. What is the ratio of the lengths of the tungsten and carbon sections? Ignore any changes in length due to thermal expansion.

Yeah I suck at LaTeX, but I'm too lazy to make it fancy.

Answer 32

It's ok...i can understand it without latex too :)

Answer 33

How will u solve it ?

Answer 34

Well ok it's like the same process, so the total resistance of the composite wire doesn't change...
\[(R_t)_0+(R_c)_0 = R_t+R_c\]

so left side is at room temp, and right side at temp T

Answer 35

The temperature dependence of the resistance of the wire of resistance \[R_0\] at temperature \[T_0\] is given by \[R=R_0=[1+\alpha(T-T_0)]\]

\[(R_t)_0+(R_c)_0=(R_t)_0(1+ \alpha_t \Delta T)+(R_c)_0(1+\alpha_c \Delta T)\]

and for this delta T is same for each wire, amg the similarities with the other question XD

Answer 36

This is cutting into my carpet buying time but I will finish!

Answer 37

so \(\sf \alpha _t=\alpha _c\) ?

Answer 38

\[(R_t)_0 \alpha _t=-(R_c) \alpha _c\]

oh btw the subscript t = tungsten and subscript c = carbon

So with this expression you can find the ratio of the resistances. And once that's known, you can easily find the ratio of the lengths of the sections using the equation \[L = \frac{ RA }{ p }\]

\[\frac{ (R_t)_0 }{ (R_c)_0 }=-\frac{ \alpha_c }{ \alpha_t }=-\frac{ -0.0005 C degrees ^{-1} }{ 0.0045 C degrees ^{-1} }\]

Answer 39

\[= \frac{ 1 }{ 9 }\]

Answer 40

\[\frac{ L_t }{ L_c }=\frac{ (R_0A/p)_t }{ (R_0 A/p)_c }=\frac{ (R_t)_0 }{ (R_c)_0 }\ \left( \frac{ p_c }{ p_t } \right)\] ok i lied this was harder than your question lmao.

Answer 41

;(

Answer 42

Did that help at all ? :S

Answer 43

Yes it did !

Answer 44

Alright, cool haha, I think it just requires practice and eventually you'll understand, just do a few problems related to these kinds of things.

Answer 45

Yes..i tdo think so !

Answer 46

Thanx for ur tremendous Help !

Answer 47

Alright AB, I gtg now, take care and good luck!! ^_^

Answer 48

No problemo

Answer 49

Now i think u can go buy ur Carpet xD

Answer 50

Yeah haha xD

Answer 51

Byeee ! :)
tcu2

Answer 52

ahaaan !
I think i really got it.......... :D
Thank you @iambatman ...wish i could gift uh a carpet XD