At room temperature (27^{circ} C),the resista | vedclass

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

Question

(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\c

Vedclass · Accepted Answer

$1027$

Vedclass · Answer

(B) The resistance of a conductor at a temperature $T$ is given by the formula: $R = R_0(1 + \alpha \Delta T)$.Here,$R_0 = 50 \Omega$ at $T_0 = 27^{\circ} C$,$R = 62 \Omega$,and $\alpha = 2.4 	imes 10^{-4} { }^{\circ} C^{-1}$.Substituting the values into the equation:$62 = 50(1 + 2.4 	imes 10^{-4} \Delta T)$$1.24 = 1 + 2.4 	imes 10^{-4} \Delta T$$0.24 = 2.4 	imes 10^{-4} \Delta T$$\Delta T = \frac{0.24}{2.4 	imes 10^{-4}} = 1000^{\circ} C$.Since $\Delta T = T - T_0$,we have $T = T_0 + \Delta T = 27^{\circ} C + 1000^{\circ} C = 1027^{\circ} C$.

At room temperature $(27^{\circ} C)$,the resistance of a heating element is $50 \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4} { }^{\circ} C^{-1}$. The temperature of the element,when its resistance is $62 \Omega$,is $\qquad$ ${ }^{\circ} C$.

Similar Questions

Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be

The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled,then

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be..... $\%$.

For which of the following does the resistance decrease upon increasing the temperature?

The alloys constantan and manganin are used to make standard resistance due to they have

Vedclass Test Series

Exam Paper Generator

Online Exam Module