Answer
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Hint: As resistances increases the temperature of metallic conductors also increases therefore the resistance is directly proportional to the temperature. If voltage and current increase then the heat temperature will also increase.
Complete answer:
Now from the question
V=230Volt ${I_1} = 3.2$ A , ${I_2} = 2.8$A and
$\alpha = 1.70 \times {10^{ - 4}}{C^{ - 1}}$ , ${t_1} = 27^\circ C$
As we know resistance is stated as the ratio of voltage to current so,
$\begin{gathered}
{R_1} = \dfrac{V}{{{I_1}}} = \dfrac{{230}}{{3.2}}\Omega = 71.875\Omega \\
{R_2} = \dfrac{V}{{{I_2}}} = \dfrac{{230}}{{2.8}}\Omega = 82.143\Omega \\
\end{gathered} $
Now the temperature coefficient of resistor is:
$\alpha = \dfrac{{{R_2} - {R_1}}}{{{R_1}\left( {{t_2} - {t_1}} \right)}}$
$1.70 \times {10^{ - 4}} = \dfrac{{82.143 - 71.875}}{{71.875\left( {{t_2} - 27} \right)}}$
\[71.875\left( {{t_2} - 27} \right) = \dfrac{{82.143 - 71.875}}{{1.70 \times {{10}^{ - 4}}}}\]
${t_2} - 27 = \dfrac{{10.268}}{{1.70 \times 71.875 \times {{10}^{ - 4}}}}$
${t_2} - 27 = 840.35$
${t_2} = 840.35 + 27$
${t_2} = 867.35^\circ C$ is the steady temperature.
Additional Information:
Nichrome is used as a heating element because it has high resistivity and resistance to oxidation at high temperatures. When it is used as a heating element it is usually wound into coils. When it is heated in red- hot temperature, it develops an outer layer of chromium oxide, which is thermodynamically stable in air. It is not used in fuse wire because it may lead to damage to home appliances. Generally, nichrome is composed of 80% of nickel and 20% of chromium, moreover it makes a great heating element because this material yields a pretty high resistance.
Note:
Nichrome is a non-magnetic alloy of nickel and chromium, it is used as the most common resistance wire for heating purposes. It is good in making heaters but not light lamps because if you increase the voltage to get a brighter colour, then the nichrome will burn or get melted.
Complete answer:
Now from the question
V=230Volt ${I_1} = 3.2$ A , ${I_2} = 2.8$A and
$\alpha = 1.70 \times {10^{ - 4}}{C^{ - 1}}$ , ${t_1} = 27^\circ C$
As we know resistance is stated as the ratio of voltage to current so,
$\begin{gathered}
{R_1} = \dfrac{V}{{{I_1}}} = \dfrac{{230}}{{3.2}}\Omega = 71.875\Omega \\
{R_2} = \dfrac{V}{{{I_2}}} = \dfrac{{230}}{{2.8}}\Omega = 82.143\Omega \\
\end{gathered} $
Now the temperature coefficient of resistor is:
$\alpha = \dfrac{{{R_2} - {R_1}}}{{{R_1}\left( {{t_2} - {t_1}} \right)}}$
$1.70 \times {10^{ - 4}} = \dfrac{{82.143 - 71.875}}{{71.875\left( {{t_2} - 27} \right)}}$
\[71.875\left( {{t_2} - 27} \right) = \dfrac{{82.143 - 71.875}}{{1.70 \times {{10}^{ - 4}}}}\]
${t_2} - 27 = \dfrac{{10.268}}{{1.70 \times 71.875 \times {{10}^{ - 4}}}}$
${t_2} - 27 = 840.35$
${t_2} = 840.35 + 27$
${t_2} = 867.35^\circ C$ is the steady temperature.
Additional Information:
Nichrome is used as a heating element because it has high resistivity and resistance to oxidation at high temperatures. When it is used as a heating element it is usually wound into coils. When it is heated in red- hot temperature, it develops an outer layer of chromium oxide, which is thermodynamically stable in air. It is not used in fuse wire because it may lead to damage to home appliances. Generally, nichrome is composed of 80% of nickel and 20% of chromium, moreover it makes a great heating element because this material yields a pretty high resistance.
Note:
Nichrome is a non-magnetic alloy of nickel and chromium, it is used as the most common resistance wire for heating purposes. It is good in making heaters but not light lamps because if you increase the voltage to get a brighter colour, then the nichrome will burn or get melted.
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A heating element using nichrome connected to a 230V supply draws an initial current of 3.2 A which settles after a few seconds to steady a value of 2.8 A. what is the steady temperature of the heating element if the room temperature is $27^\circ C$ ?temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times {10^{ - 4}}{C^{ - 1}}$ .
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Current Electricity Class 12 Physics Chapter 3 | NCERT EXERCISE 3.6 | Vishal Kumar Sir
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