Question

The resistance of wire in a heater at room temperature is $6.5 \times 10^{- 6}m^{2}V^{- 1}s^{- 1}$. 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$2.5 \times 10^{6}m^{2}V^{- 1}S^{- 1}$

• A. 955 $v_{d} \propto E$
• B. 1055$v_{d} \propto E^{2}$
• C. 1155 $v_{d} \propto \sqrt{E}$
• D. 1258 $v_{d} \propto \frac{1}{E}$

Answer 1

Answer: (D)

1258 $v_{d} \propto \frac{1}{E}$

Answer 2

: 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

$R_{2} = R_{1}\lbrack 1 + \alpha(T_{2} - T_{1})\rbrack$

$\therefore \mathrm { T } _ { 2 } - \mathrm { T } _ { 1 } = \frac { \mathrm { R } _ { 2 } - \mathrm { R } _ { 1 } } { \mathrm { R } _ { 1 } } \times \frac { 1 } { \alpha } = \frac { 78.6 - 65 } { 65 } \times \frac { 1 } { 1.7 \times 10 ^ { - 4 } }$ $T_{2} - T_{1} = 1231$

$T_{2} = 1231 + T_{1} = 1231 + 27 = 1258^{o}C$