α_1 and α_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 combination.Assume that α_1 and α_2 remain same with change in temperature.


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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
	$R=ρ(L/A)=ρ(L^2 /(AL))=ρ(L^2 /V)=ρd(L^2 /m) m=Vd L=L_0 (1+α△T) R_1 =((ρ_1 d_1 )/m_1 )×{L_(01) (1+α_1 △T)}^2 R_2 =((ρ_2 d_2 )/m_2 )×{L_(02) (1+α_2 △T)}^2 R_(e_ q) =R_1 +R_2 =((ρ_1 d_1 )/m_1 )×{L_(01) (1+α_1 △T)}^2 +((ρ_2 d_2 )/m_2 )×{L_(02) (1+α_2 △T)}^2$
	$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{A L} = ρ \frac{L^{2}}{V} = ρ d \frac{L^{2}}{m}$ $m = V d$ $L = L_{0} (1 + α △ T)$ $R_{1} = \frac{ρ_{1} d_{1}}{m_{1}} \times {L_{01} (1 + α_{1} △ T)}^{2}$ $R_{2} = \frac{ρ_{2} d_{2}}{m_{2}} \times {L_{02} (1 + α_{2} △ T)}^{2}$ $R_{e_{} q} = R_{1} + R_{2}$ $= \frac{ρ_{1} d_{1}}{m_{1}} \times {L_{01} (1 + α_{1} △ T)}^{2} + \frac{ρ_{2} d_{2}}{m_{2}} \times {L_{02} (1 + α_{2} △ T)}^{2}$
	Commented by Necxx last updated on 27/Sep/18

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