1-and-2-are-the-temperature-coefficients-of-the-two-resistors-R-1-and-R-2-at-any-temperature-T-0-0-C-Find-the-equivalent-resistance-if-both-R-1-and-R-2-are-connected-in-series-combinatio

Question Number 44298 by Necxx last updated on 26/Sep/18

α_{1} a n d α_{2} a r e t h e t e m p e r a t u r e

c o e f f i c i e n t s o f t h e t w o r e s i s t o r s

R_{1} a n d R_{2} a t a n y t e m p e r a t u r e T_{0}

0 ° C . F i n d t h e e q u i v a l e n t r e s i s t a n c e

i f b o t h R_{1} a n d R_{2} a r e c o n n e c t e d i n

s e r i e s c o m b i n a t i o n . A s s u m e t h a t

α_{1} a n d α_{2} r e m a i n s a m e w i t h c h a n g e

i n t e m p e r a t u r e .

Commented by Necxx last updated on 26/Sep/18

I^{'} v e t r i e d t h i s p r o b l e m s e v e r a l l y

b u t g o t s t u c k e d a t s o m e p o i n t . p l e a s e

h e l p . T h a n k s i n a d v a n c e .

Commented by Necxx last updated on 26/Sep/18

p l e a s e h e l p

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

R = ρ \frac{L}{A} = ρ \frac{L^{2}}{A L} = ρ \frac{L^{2}}{V} = ρ d \frac{L^{2}}{m}

m = V d

L = L_{0} (1 + α △ T)

R_{1} = \frac{ρ_{1} d_{1}}{m_{1}} \times {L_{01} (1 + α_{1} △ T)}^{2}

R_{2} = \frac{ρ_{2} d_{2}}{m_{2}} \times {L_{02} (1 + α_{2} △ T)}^{2}

R_{e_{} q} = R_{1} + R_{2}

= \frac{ρ_{1} d_{1}}{m_{1}} \times {L_{01} (1 + α_{1} △ T)}^{2} + \frac{ρ_{2} d_{2}}{m_{2}} \times {L_{02} (1 + α_{2} △ T)}^{2}

Commented by Necxx last updated on 27/Sep/18

w o w \dots . I l o v e t h i s \dots T h a n k s

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