Question

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 ?

Answer 1

@ash2326

Answer 2

@Abmon98

Answer 3

@ParthKohli

Answer 4

\[R_1 = r_1 \left(1 + \alpha\Delta T \right)\]\[R_2 = r_2\left(1 - \beta \Delta T\right)\]\[R_1 + R_2 = r_1 + r_2 + \left(\alpha- \beta\right)\Delta T\]If things don't change with temperature, then \(\alpha - \beta = 0 \Rightarrow \alpha = \beta\)?

I'm not really sure though... lol

Answer 5

Whoops... just a little mistake there.

Answer 6

\[R_1 + R_2 = r_1 + r_2 + \Delta T \left(r_1 \alpha - r_2 \beta\right)\]\[\Rightarrow r_1 \alpha = r_2\beta\]And blah blah.

Answer 7

i didn't got ur last equation

Answer 8

Hmm, it is given that change in temperature does not affect the sum of resistances. So the term containing \(\Delta T\) has to be zero.

Answer 9

I've gotta hit the bed now. Good night.

Answer 10

LOL !
u solved the problem

Answer 11

\(\frac{r_1}{r_2}=\frac{\beta}{\alpha}\)

Answer 12

Thank you !
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Answer 13

Let \(\sf R_1~and~R_2~be~the~temperature~at~t°C\\R^0_1~and~R^0_2~be~the~temperature~at~0°C\\ \because resistance~of~the~series~combination~of~R1~and~R2~will~not~change~with~ temperature,\\ \Rightarrow R_1+R_2=R^0_1+R^0_2\\ =>R^0_1(1+\alpha \Delta T)+R^0_2(1-\beta \Delta T)=R^0_1+R^0_2\\ \Rightarrow \cancel{R^0_1+R^0_2}+R^0_1\alpha\ \Delta T-R^0_2\beta \Delta T=\cancel{R^0_1+R^0_2}\\ \Rightarrow R^0_1\alpha \cancel{\Delta T}=R^0_2\beta \cancel{\Delta T}\\~\\~\\~\\ \huge \therefore ~~~~~~~~~~~~~~~~~~~~~\boxed{\huge \frac{R^0_1}{R^0_2}=\frac{\beta}{\alpha}}\)