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Question: An electric toaster uses nichrome for its heating element. When a negligibly small current passes th...

An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27∘C{{27}^{\circ }}C) is found to be 75.3Ω75.3\Omega . When the toaster is connected to a 230V supply, the current settles, after a few seconds to a steady value of 2.68A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70×10−4∘C−11.70\times {{10}^{-4}}{}^{\circ }{{C}^{-1}}.

Explanation

Solution

Provided that the electric toaster uses nichrome wire. The resistance at a particular value is given . The temperature coefficient of resistance is also given. Hence by using Ohm’s law and formula of resistance variation with temperature we will get the solution.

Complete step-by-step solution: -
First of all we can define ohm’s law. Ohm’s can be stated as, at constant temperature the voltage is directly proportional to the current flowing through the conductor. And the resistance is here the proportionality constant.
According to Ohm’s law,
V=IR ………….(1)
Where V is the potential difference
I is the current flow through the circuit
R is the resistance
The resistance variation at some particular value is given by,
Rt=R0[1+α(T−T0)]{{R}_{t}}={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right] …………(2)
Where Rt{{R}_{_{t}}}is the resistance at a given temperature
R0{{R}_{0}}room temperature resistance
α\alpha temperature coefficient of resistance
T&T0T\And {{T}_{0}}are the given and room temperatures respectively
Applying Ohm’s law,
R=VIR=\dfrac{V}{I}
R=2302.68  \begin{aligned} & R=\dfrac{230}{2.68} \\\ & \\\ \end{aligned}
∴R=85.82Ω\therefore R=85.82\Omega
Then using equation (2) we get,
Rt=R0[1+α(T−T0)]{{R}_{t}}={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]
85.82=75.3[1+1.7×10−4(T−27)]85.82=75.3\left[ 1+1.7\times {{10}^{-4}}\left( T-27 \right) \right]
85.8275.3=[1+1.7×10−4(T−27)]\dfrac{85.82}{75.3}=\left[ 1+1.7\times {{10}^{-4}}\left( T-27 \right) \right]
1.13=[1+1.7×10−4(T−27)]1.13=\left[ 1+1.7\times {{10}^{-4}}\left( T-27 \right) \right]
0.13=[1.7×10−4(T−27)]0.13=\left[ 1.7\times {{10}^{-4}}\left( T-27 \right) \right]
0.131.7×10−4=(T−27)\dfrac{0.13}{1.7\times {{10}^{-4}}}=\left( T-27 \right)
∴T=848.81∘C\therefore T=848.81{}^{\circ }C
Hence the steady temperature of nichrome is 848.81∘C848.81{}^{\circ }C.

Note: As the temperature of a material increases resistance of the material also increases. That is defined as the positive value of temperature coefficient of resistance that a material can have.