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

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.

$$\alpha_{\mathrm{1}} \:{and}\:\alpha_{\mathrm{2}} \:{are}\:{the}\:{temperature} \\ $$$${coefficients}\:{of}\:{the}\:{two}\:{resistors} \\ $$$${R}_{\mathrm{1}} \:{and}\:{R}_{\mathrm{2}} \:{at}\:{any}\:{temperature}\:{T}_{\mathrm{0}} \\ $$$$\mathrm{0}°{C}.{Find}\:{the}\:{equivalent}\:{resistance} \\ $$$${if}\:{both}\:{R}_{\mathrm{1}} \:{and}\:{R}_{\mathrm{2}\:} \:{are}\:{connected}\:{in} \\ $$$${series}\:{combination}.{Assume}\:{that} \\ $$$$\alpha_{\mathrm{1}} \:{and}\:\alpha_{\mathrm{2}} \:{remain}\:{same}\:{with}\:{change} \\ $$$${in}\:{temperature}. \\ $$

Commented by Necxx last updated on 26/Sep/18

I′ve tried this problem severally  but got stucked at some point.please  help.Thanks in advance.

$${I}'{ve}\:{tried}\:{this}\:{problem}\:{severally} \\ $$$${but}\:{got}\:{stucked}\:{at}\:{some}\:{point}.{please} \\ $$$${help}.{Thanks}\:{in}\:{advance}. \\ $$

Commented by Necxx last updated on 26/Sep/18

please help

$${please}\:{help} \\ $$

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

$${R}=\rho\frac{{L}}{{A}}=\rho\frac{{L}^{\mathrm{2}} }{{AL}}=\rho\frac{{L}^{\mathrm{2}} }{{V}}=\rho{d}\frac{{L}^{\mathrm{2}} }{{m}} \\ $$$${m}={Vd} \\ $$$${L}={L}_{\mathrm{0}} \left(\mathrm{1}+\alpha\bigtriangleup{T}\right) \\ $$$${R}_{\mathrm{1}} =\frac{\rho_{\mathrm{1}} {d}_{\mathrm{1}} }{{m}_{\mathrm{1}} }×\left\{{L}_{\mathrm{01}} \left(\mathrm{1}+\alpha_{\mathrm{1}} \bigtriangleup{T}\right)\right\}^{\mathrm{2}} \\ $$$${R}_{\mathrm{2}} =\frac{\rho_{\mathrm{2}} {d}_{\mathrm{2}} }{{m}_{\mathrm{2}} }×\left\{{L}_{\mathrm{02}} \left(\mathrm{1}+\alpha_{\mathrm{2}} \bigtriangleup{T}\right)\right\}^{\mathrm{2}} \\ $$$${R}_{{e}_{} {q}} ={R}_{\mathrm{1}} +{R}_{\mathrm{2}} \\ $$$$\:\:\:=\frac{\rho_{\mathrm{1}} {d}_{\mathrm{1}} }{{m}_{\mathrm{1}} }×\left\{{L}_{\mathrm{01}} \left(\mathrm{1}+\alpha_{\mathrm{1}} \bigtriangleup{T}\right)\right\}^{\mathrm{2}} +\frac{\rho_{\mathrm{2}} {d}_{\mathrm{2}} }{{m}_{\mathrm{2}} }×\left\{{L}_{\mathrm{02}} \left(\mathrm{1}+\alpha_{\mathrm{2}} \bigtriangleup{T}\right)\right\}^{\mathrm{2}} \\ $$

Commented by Necxx last updated on 27/Sep/18

wow.... I love this...Thanks

$${wow}....\:{I}\:{love}\:{this}...{Thanks} \\ $$

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