The resistance of a platinum resistance therm | vedclass

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

Question

(D) The resistance $R$ at temperature $	heta$ is given by the formula: $R = R_0(1 + \alpha 	heta)$.Given:$R_0 = 5\,\Omega$ (at $0\,^oC$)$R_{100} = 5

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The resistance $R$ at temperature $	heta$ is given by the formula: $R = R_0(1 + \alpha 	heta)$.Given:$R_0 = 5\,\Omega$ (at $0\,^oC$)$R_{100} = 5.75\,\Omega$ (at $100\,^oC$)$R_{	heta} = 5.15\,\Omega$ (at unknown temperature $	heta$)First,calculate the temperature coefficient of resistance $\alpha$:$R_{100} = R_0(1 + \alpha 	imes 100)$$5.75 = 5(1 + 100\alpha)$$1.15 = 1 + 100\alpha$$0.15 = 100\alpha$$\alpha = 0.0015\,\,^oC^{-1} = 1.5 	imes 10^{-3}\,\,^oC^{-1}$.Now,find the unknown temperature $	heta$ for $R_{	heta} = 5.15\,\Omega$:$R_{	heta} = R_0(1 + \alpha 	heta)$$5.15 = 5(1 + 0.0015 	imes 	heta)$$1.03 = 1 + 0.0015 	imes 	heta$$0.03 = 0.0015 	imes 	heta$$	heta = \frac{0.03}{0.0015} = 20\,^oC$.

The resistance of a platinum resistance thermometer at $0\,^oC$ is $5\,\Omega$ and at $100\,^oC$ is $5.75\,\Omega$. When the resistance is $5.15\,\Omega$,the unknown temperature is ............ $^oC$.

Similar Questions

$A$ wire of a certain material is stretched slowly by $10\%$. Its new resistance and specific resistance become respectively:

What is electrical resistance $(R)$? On which factors does the resistance value depend?

The resistance of a wire at $0^{\circ} C$ is $20 \Omega$. If the temperature coefficient of resistance is $5 \times 10^{-3} {}^{\circ} C^{-1}$,the temperature at which the resistance will be double that at $0^{\circ} C$ is: (in $^{\circ} C$)

$A$ conducting wire has length $L_1$ and diameter $d_1$. After stretching,the same wire's length becomes $L_2$ and diameter $d_2$. The ratio of resistance before and after stretching is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module