At room temperature- 27-C- the resistance of a heating element is 100 - What is the temperaturein -C of the element if the resistance is found to be 117 - Given that the temperature coefficient of the material of the resistor is 1-70- 10-4C-1-

We know that R(T)=R(T_0)(1+αΔ T)=R(T_0)[1+αΔ (T T_0)] Here, 117=100[1+1.70× 10^ 4(T 27)] or #160; T 27=(1.17 1)(10^41.70)=1000 or T=1000+2 ...

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Standard X

Physics

Factors Affecting Resistance & How They Affect

At room tempe...

Question

At room temperature, $27^{\circ} C$ , the resistance of a heating element is $100 Ω$ . What is the temperature(in $^{\circ} C$ ) of the element if the resistance is found to be $117 Ω$ ? Given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{- 4} C^{- 1}$ .

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Solution

We know that $R (T) = R (T_{0}) (1 + α Δ T) = R (T_{0}) [1 + α Δ (T - T_{0})]$
Here, $117 = 100 [1 + 1.70 \times 10^{- 4} (T - 27)]$
or $T - 27 = (1.17 - 1) (\frac{10^{4}}{1.70}) = 1000$
or $T = 1000 + 27 = 1027^{o} C$

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At room temperature, 27∘C, the resistance of a heating element is 100Ω. What is the temperature(in ∘C) of the element if the resistance is found to be 117Ω? Given that the temperature coefficient of the material of the resistor is 1.70×10−4C−1.

We know that R(T)=R(T0)(1+αΔT)=R(T0)[1+αΔ(T−T0)]Here, 117=100[1+1.70×10−4(T−27)]or T−27=(1.17−1)(1041.70)=1000or T=1000+27=1027oC

At room temperature, $27^{\circ} C$ , the resistance of a heating element is $100 Ω$ . What is the temperature(in $^{\circ} C$ ) of the element if the resistance is found to be $117 Ω$ ? Given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{- 4} C^{- 1}$ .

We know that $R (T) = R (T_{0}) (1 + α Δ T) = R (T_{0}) [1 + α Δ (T - T_{0})]$
Here, $117 = 100 [1 + 1.70 \times 10^{- 4} (T - 27)]$
or $T - 27 = (1.17 - 1) (\frac{10^{4}}{1.70}) = 1000$
or $T = 1000 + 27 = 1027^{o} C$