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Home » A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heatingelement if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1.
CBSE | Class 12 | Current Electricity | NCERT | Physics
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1.

Answer –

According to the question;

Supply voltage,  V = 230 V

initial current drawn is given by I 1 = 3.2 A

Let the initial resistance be given by R 1, which can be determined by the following relation –

    \[{{R}_{1}}=\frac{V}{I}=\frac{230}{3.2}\]

    \[{{R}_{1}}=71.87\Omega \]

Current at steady state is , I 2 = 2.8 A

Let the resistance at steady state is  R 2

R 2 can be easily determined by the following equation –

    \[{{R}_{2}}=\frac{230}{2.8}=82.14\Omega \]

According to the statement, the temperature coefficient of nichrome is 1.70 x 10 – 4 ° C – 1

Initial temperature of nichrome , T 1 = 27.0 ° C

Let the steady state temperature reached by nichrome be T 2

This temperature T 2 can be obtained by making use of the following formula –

    \[\alpha =\frac{{{R}_{1}}-R}{R({{T}_{2}}-{{T}_{1}})}\]

Upon re arranging we get –

    \[({{T}_{2}}-{{T}_{1}})=\frac{{{R}_{1}}-R}{\alpha R}\]

Upon substituting values, we get –

    \[({{T}_{2}}-27)=\frac{82.14-71.87}{71.87\times 1.7\times {{10}^{-4}}}=840.5\]

    \[{{T}_{1}}=867.5{}^{\circ }C\]

Hence, the steady temperature of the heating element is 867.5 ° C.

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