Question

The resistance of wire in a heater at room temperature is $65 \,\Omega$. When the heater is connected to a $220 \,V$ supply the current settles after a few seconds to $2.8 \,A$. 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
$R_{2}=R_{1}\left[1+\alpha\left(T_{2}-T_{1}\right)\right]$
$\therefore T_{2}-T_{1}=\frac{R_{2}-R_{1}}{R_{1}}\times\frac{1}{\alpha}$
$=\frac{78.6-65}{65}\times\frac{1}{1.7\times10^{-4}}$
$T_{2}-T_{1}=1231$
$T_{2}=1231+T_{1}=1231+27$
$=1258\,{}^{\circ}C$