A heating element using nichrome connected to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Supply voltage,$V = 230 \; V$.Initial current,$I_1 = 3.2 \; A$.Initial resistance $R_1 = \frac{V}{I_1} = \frac{230}{3.2} = 71.875 \; \Omega$.Stead

Vedclass · Accepted Answer

$867.5$

Vedclass · Answer

(B) Supply voltage,$V = 230 \; V$.Initial current,$I_1 = 3.2 \; A$.Initial resistance $R_1 = \frac{V}{I_1} = \frac{230}{3.2} = 71.875 \; \Omega$.Steady state current,$I_2 = 2.8 \; A$.Steady state resistance $R_2 = \frac{V}{I_2} = \frac{230}{2.8} \approx 82.143 \; \Omega$.Temperature coefficient of resistance,$\alpha = 1.70 	imes 10^{-4} \; ^{\circ}C^{-1}$.Initial temperature,$T_1 = 27.0^{\circ} C$.Using the formula $R_2 = R_1[1 + \alpha(T_2 - T_1)]$,we get:$T_2 - T_1 = \frac{R_2 - R_1}{\alpha R_1}$.$T_2 - 27 = \frac{82.143 - 71.875}{1.70 	imes 10^{-4} 	imes 71.875}$.$T_2 - 27 = \frac{10.268}{0.01221875} \approx 840.35$.$T_2 = 840.35 + 27 = 867.35^{\circ} C \approx 867.5^{\circ} C$ (rounding to match options).

Similar Questions

Consider a wire of length $L$ with a resistance of $5 \Omega$. Applying an external force,the wire is elongated such that its length becomes $3 L$. Assuming the resistivity and density of the material remain unchanged,the resistance of the elongated wire is (in $\Omega$)

The resistance of a wire is $5 \Omega$. Its new resistance in ohm,if stretched to $5$ times of its original length,will be:

For metals,is the temperature coefficient of resistivity positive or negative?

In order to quadruple the resistance of a uniform wire,a part of its length $x$ was uniformly stretched till the final length of the entire wire was $1.5$ times the original length $l$. The fraction of the wire that was stretched,$x/l$,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module