A platinum wire has a resistance of 10 - at 0-C and 20 - at 273-C- The value of temperature coefficient of platinum is given as 1xoC-1- Find x

We know that R=R_0(1+αΔ T) Thus , the nbsp;temperature coefficient , α=R R_0/R_0(T T_0) or α=20 10/10(273 0)=1/273 ^oC^ 1 ...

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Standard XII

Physics

Dependence of Resistance on Temperature

A platinum wi...

Question

A platinum wire has a resistance of $10 Ω$ at $0^{\circ} C$ and $20 Ω$ at $273^{\circ} C$ . The value of temperature coefficient of platinum is given as ${\frac{1}{x}}^{o} C^{- 1}$ . Find $x$

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Solution

We know that $R = R_{0} (1 + α Δ T)$
Thus , the temperature coefficient , $α = \frac{R - R_{0}}{R_{0} (T - T_{0})}$
or $α = \frac{20 - 10}{10 (273 - 0)} = {\frac{1}{273}}^{o} C^{- 1}$

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A platinum wire has a resistance of 10Ω at 0∘C and 20Ω at 273∘C. The value of temperature coefficient of platinum is given as 1xoC−1. Find x

We know that R=R0(1+αΔT)Thus , the temperature coefficient , α=R−R0R0(T−T0)or α=20−1010(273−0)=1273oC−1

A platinum wire has a resistance of $10 Ω$ at $0^{\circ} C$ and $20 Ω$ at $273^{\circ} C$ . The value of temperature coefficient of platinum is given as ${\frac{1}{x}}^{o} C^{- 1}$ . Find $x$

We know that $R = R_{0} (1 + α Δ T)$
Thus , the temperature coefficient , $α = \frac{R - R_{0}}{R_{0} (T - T_{0})}$
or $α = \frac{20 - 10}{10 (273 - 0)} = {\frac{1}{273}}^{o} C^{- 1}$