Question

$\longrightarrow \huge {\tt Question :-}$
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 ofnichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1.​

Answer 1

$\bigstar\huge\mathtt{\red{Solution:}}$

As per the given question ,

Supply voltage = 230 V

Current drawn by the nichrome initially  = 3.2 A

Lets find the resistance of the nichrome by using the given data ,

we know that ,

$\rightarrow\mathtt{R = \dfrac{V}{I}}$

$\rightarrow\mathtt{R = \dfrac{230}{3.2}}$

$\rightarrow\mathtt{\blue{R =71.87\: \Omega }}$

___________________

At steady state

Supply voltage = 230 V

current drawn by the nichrome = 2.8 A

Lets find the resistance of the nichrome by using the given data ,

we know that ,

$\rightarrow\mathtt{R _1= \dfrac{V}{I_1}}$

$\rightarrow\mathtt{R _1= \dfrac{230}{2.8}}$

$\rightarrow\mathtt{\orange{R _1= 82.14 \: \Omega}}$

____________________

Temperature coefficient of resistance = 1.7 x 10^-4 C-1

Initially the temperature of the nichrome = 27  °C

Initial resistance of nichrome = 71.87 Ω

resistance of the nichrome at steady state = 82.14  Ω

As per the formula ,

$\rightarrow \mathtt{\Delta R=\alpha R 1(\Delta T)}$

$\rightarrow \mathtt{ R_2 - R_1 =\alpha R 1(T_2 - T_1)}$

$\rightarrow \mathtt{ 82.14 - 71.87 =1.7 \times 10^{-4}\times 71.87(T_2 - 27)}$

$\rightarrow \mathtt{ 10.27 =122.17\times 10^{-4}(T_2 - 27)}$

$\rightarrow \mathtt{ \dfrac{10.27}{ 122.17\times 10^{-4}}=(T_2 - 27)}$

$\rightarrow \mathtt{ \dfrac{102700}{ 122.17}=T_2 - 27}$

$\rightarrow \mathtt{ 840.63=(T_2 - 27)}$

$\rightarrow \mathtt{T_2= 840.63+27}$

$\rightarrow \mathtt{\pink{T_2= 847.63 \:°C }}$

The steady temperature of the heating element is 847.63  °C

Answer 2

Answer:

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