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 $\alpha =1.70\times {10}^{-4}$ ${}^o{C}^{-1}$)

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

Answer 1

Answer: (D)

$1258$ ${}^{o}C$

Here, $T_1=27$ ${}^{o}C$, $R_1=65\Omega$

$R_2=\frac{Supplyvoltage}{Steadycurrent}=\frac{220}{2.8}=78.6\Omega$

Now, using the relation
$R_2=R_1[1+\alpha (T_2-T_1\left)\right]$

$\therefore T_2-T_1=\frac{R_2-R_1}{R_1}\times \frac{1}{\alpha }=\frac{78.6-65}{65}\times \frac{1}{1.7\times {10}^{-4}}$

$T_2-T_1=1231$
$T_2=1231+T_1=1231+27={1258}^o$C