At room temperature \({(27.0 {\degree}C)}\) the resistance of a heating element is \(\text{100 $\Omega $.}\) What is the temperature of the element if the resistance is found to be \(\text{17 $\Omega $,}\) given that the temperature coefficient of the material of the resistor is \({1.70 {\times} 10^{-4} {\degree}C}.\)
Answer
Let, Room temperature as \(T1\)
Resistance of heating element at room temperature as \(R1\)
Resistance of heating element at final temperature \((T2 )\) as \(R2\)
Here, we are given \(T1\) as \(27^{\circ} {C}\). Resistance \(R1\) as \(100 \;\text{ohms}\) ad Resistance \(R2\) as \(17 \;\text{ohms}.\)
The \(\alpha\) is \(1.70 \times 10^{-4}{^\circ} {C}\)
\(T 1=27^{\circ} C\)
\(R1=100\; \text{ohm}\)
\(R2=17\; \text{ohm}\)
\(\alpha=1.70 \times 10^{-4 \ \circ} C\)
We have the relation-
\(R2=R1 \times[1+\alpha \times(T 2-T 1)]\)
\(17=10\left [1+(1.7 \times 10^{-4})(T 2-27)\right]\)
On solving,
\(T 2=1027^{\circ} {C}\)
Resistance of heating element at room temperature as \(R1\)
Resistance of heating element at final temperature \((T2 )\) as \(R2\)
Here, we are given \(T1\) as \(27^{\circ} {C}\). Resistance \(R1\) as \(100 \;\text{ohms}\) ad Resistance \(R2\) as \(17 \;\text{ohms}.\)
The \(\alpha\) is \(1.70 \times 10^{-4}{^\circ} {C}\)
\(T 1=27^{\circ} C\)
\(R1=100\; \text{ohm}\)
\(R2=17\; \text{ohm}\)
\(\alpha=1.70 \times 10^{-4 \ \circ} C\)
We have the relation-
\(R2=R1 \times[1+\alpha \times(T 2-T 1)]\)
\(17=10\left [1+(1.7 \times 10^{-4})(T 2-27)\right]\)
On solving,
\(T 2=1027^{\circ} {C}\)
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