If the cold junction is held at 0^{circ} C, t | vedclass

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

Question

(D) Given the thermo emf equation: $V = 10 	imes 10^{-6} t - \frac{1}{40} 	imes 10^{-6} t^2$. To find the neutral temperature $(t_n)$, we differenti

Vedclass · Accepted Answer

$200^{\circ} C ; 1 	ext{ mV}$

Vedclass · Answer

(D) Given the thermo emf equation: $V = 10 	imes 10^{-6} t - \frac{1}{40} 	imes 10^{-6} t^2$. To find the neutral temperature $(t_n)$, we differentiate $V$ with respect to $t$ and set it to zero: $\frac{dV}{dt} = 10 	imes 10^{-6} - \frac{2}{40} 	imes 10^{-6} t = 0$. $10 	imes 10^{-6} = \frac{1}{20} 	imes 10^{-6} t_n$. $t_n = 200^{\circ} C$. To find the maximum thermo emf $(V_{\max})$, we substitute $t_n = 200^{\circ} C$ into the original equation: $V_{\max} = 10 	imes 10^{-6} (200) - \frac{1}{40} 	imes 10^{-6} (200)^2$. $V_{\max} = 2 	imes 10^{-3} - \frac{40000}{40} 	imes 10^{-6} = 2 	imes 10^{-3} - 1 	imes 10^{-3} = 1 	imes 10^{-3} 	ext{ V} = 1 	ext{ mV}$.

If the cold junction is held at $0^{\circ} C$, the thermo emf $V$ of a thermocouple varies as $V = 10 \times 10^{-6} t - \frac{1}{40} \times 10^{-6} t^2$, where $t$ is the temperature of the hot junction in $^{\circ} C$. The neutral temperature and the maximum value of thermo emf are respectively:

Similar Questions

For a thermocouple,the inversion temperature is $600^{\circ} C$ and the neutral temperature is $320^{\circ} C$. Find the temperature of the cold junction (in $^{\circ} C$)?

Two wires with resistances $R$ and $2R$ are connected in parallel. The ratio of heat generated in $2R$ and $R$ is:

The Thomson coefficient of a conductor is $10 \mu V/K$. The two ends of the conductor are kept at $50^{\circ} C$ and $60^{\circ} C$, respectively. The amount of heat absorbed by the conductor when a charge of $10 C$ flows through it is:

$A$ cell of emf $E$ has an internal resistance $r$ and is connected to a rheostat. When the resistance $R$ of the rheostat is changed,which is the correct graph of the potential difference $V$ across it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module