A wire of cross-section area A- length L1- resistivity - 1 and temperature coefficient of resistivity - 1 is connected to a second wire of length L2- resistivity - 2 - temperature coefficient of resistivity - 2 and the same area A- so that wire carries same current- Total resistance R is independent of temperature for small temperature change if Thermal expansion effect is negligible

The correct option is C ρ _1L_1α _1+ρ _2L_2α _2=0 R_1=P_10 L_1Aα_1 R_2=P_20 L_2Aα_2 R_eq=R_1+R_2 R_eq(1+ αΔ T) =R_1 (1+ α_1 Δ T)+ R_2 ...

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Standard XII

Physics

Resistance and Resistors

A wire of cro...

Question

A wire of cross-section area A, length $L_{1}$ , resistivity $ρ_{1}$ and temperature coefficient of resistivity $α_{1}$ is connected to a second wire of length $L_{2}$ , resistivity $ρ_{2}$ , temperature coefficient of resistivity $α_{2}$ and the same area A, so that wire carries same current. Total resistance R is independent of temperature for small temperature change if (Thermal expansion effect is negligible)

α_{1} α_{2}

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ρ_{1} L_{1} α_{1} + ρ_{2} L_{2} α_{2} = 0

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L_{1} α_{1} + L_{2} α_{2} = 0

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Solution

The correct option is C $ρ_{1} L_{1} α_{1} + ρ_{2} L_{2} α_{2} = 0$
$R_{1} = \frac{P_{10} L_{1}}{A} α_{1}$
$R_{2} = \frac{P_{20} L_{2}}{A} α_{2}$
$R_{e q} = R_{1} + R_{2}$
$R_{e q} (1 + α Δ T) = R_{1} (1 + α_{1} Δ T) + R_{2} (1 + α_{2} Δ T) = R_{1} + R_{2}$
$R_{1} α_{1} Δ T = R_{2} α_{2} Δ T = v$
$R_{1} α_{1} = R_{2} α_{2}$
$\frac{R_{1} L_{1}}{A} α_{1} = - \frac{α_{2} ρ_{2} L_{2}}{A}$
$α_{1} L_{1} ρ_{1} + α_{2} ρ_{2} L_{2} = 0$

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The correct option is C ρ1L1α1+ρ2L2α2=0R1=P10L1Aα1R2=P20L2Aα2Req=R1+R2Req(1+αΔT)=R1(1+α1ΔT)+R2(1+α2ΔT)=R1+R2R1α1ΔT=R2α2ΔT=vR1α1=R2α2R1L1Aα1=−α2ρ2L2Aα1L1ρ1+α2ρ2 L2=0