Question

The resistance of wire in a heater at room temperature is $65\Omega$. When the heater is connected to a $220V$ supply the current settles after a few seconds to $2.8A$. What is the steady temperature of the wire. (Temperature coefficient of resistance $\alpha =1.70\times 10^{-4}{}^{\circ }{C}^{-1}$)

• A. $955{}^{\circ }C$
• B. $1055{}^{\circ }C$
• C. $1155{}^{\circ }C$
• D. $1258{}^{\circ }C$

Answer 1

Answer: (D)

Here, $T_1=27{}^{\circ }C$, $R_1=65\Omega$
$R_2=\frac{\text{Supply voltage}}{\text{Steady current}}=\frac{220}{2.8}$
$=78.6\Omega$
Now, using the relation
$R_2=R_1\left[1+\alpha \left(T_2-T_1\right)\right]$
$\therefore T_2-T_1=\frac{R_2-R_1}{R_1}\times \frac{1}{\alpha }$
$=\frac{78.6-65}{65}\times \frac{1}{1.7\times 10^{-4}}$
$T_2-T_1=1231$
$T_2=1231+T_1=1231+27$
$=1258{}^{\circ }C$