The temperature coefficient of resistance at | vedclass

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

Question

(B) The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$,where $R_0$ is the resistance at $0\,^{\circ}	ext{C}$.Give

Vedclass · Accepted Answer

$190\,^{\circ}	ext{C}$

Vedclass · Answer

(B) The resistance of a conductor at temperature $T$ is given by $R_T = R_0(1 + \alpha T)$,where $R_0$ is the resistance at $0\,^{\circ}	ext{C}$.Given $\alpha = 0.00125\,^{\circ}	ext{C}^{-1}$.At $T_1 = 25\,^{\circ}	ext{C}$,$R_1 = 1\,\Omega$. So,$1 = R_0(1 + 0.00125 	imes 25) = R_0(1 + 0.03125) = R_0(1.03125)$.Thus,$R_0 = \frac{1}{1.03125} \approx 0.9697\,\Omega$.We want to find $T_2$ such that $R_2 = 1.2\,\Omega$.Using $R_2 = R_0(1 + \alpha T_2)$,we have $1.2 = R_0(1 + 0.00125 T_2)$.Substituting $R_0$: $1.2 = \frac{1}{1.03125} (1 + 0.00125 T_2)$.$1.2 	imes 1.03125 = 1 + 0.00125 T_2$.$1.2375 = 1 + 0.00125 T_2$.$0.2375 = 0.00125 T_2$.$T_2 = \frac{0.2375}{0.00125} = 190\,^{\circ}	ext{C}$.

The temperature coefficient of resistance at $0\,^{\circ}\text{C}$ is $\alpha = 0.00125\,^{\circ}\text{C}^{-1}$. At a temperature of $25\,^{\circ}\text{C}$,the resistance of a conductor is $1\,\Omega$. Find the temperature at which the resistance becomes $1.2\,\Omega$.

Similar Questions

Write the equation representing the relation between resistivity and temperature.

Which of the following has a negative temperature coefficient?

In order to quadruple the resistance of a uniform wire,a part of it is uniformly stretched so that the final length of the wire becomes $1.5$ times the original length. The fractional length of the stretched part is

$A$ cylindrical metallic wire is stretched to increase its length. If the resistance of the wire is increased by $4 \%$,then the percentage increase in its length is (in $\%$)

The resistance of the bulb filament is $100 \ \Omega$ at a temperature of $100^{\circ} C$. If its temperature coefficient of resistance is $0.005 \ ^{\circ} C^{-1}$,at what temperature will its resistance become $200 \ \Omega$ (in $^{\circ} C$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module