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

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(D) Given the thermo emf equation: $V = 10 	imes 10^{-6} t - \frac{1}{40} 	imes 10^{-6} t^2$.The neutral temperature $(t_n)$ occurs when the thermo

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

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(D) Given the thermo emf equation: $V = 10 	imes 10^{-6} t - \frac{1}{40} 	imes 10^{-6} t^2$.The neutral temperature $(t_n)$ occurs when the thermo emf is maximum,which implies $\frac{dV}{dt} = 0$.Differentiating $V$ with respect to $t$: $\frac{dV}{dt} = 10 	imes 10^{-6} - \frac{2}{40} 	imes 10^{-6} t = 10 	imes 10^{-6} - \frac{1}{20} 	imes 10^{-6} t$.Setting $\frac{dV}{dt} = 0$: $10 	imes 10^{-6} = \frac{1}{20} 	imes 10^{-6} t_n \implies t_n = 200^{\circ} C$.Now,substitute $t_n = 200^{\circ} C$ into the equation for $V$ to find $V_{\max}$:$V_{\max} = 10 	imes 10^{-6} (200) - \frac{1}{40} 	imes 10^{-6} (200)^2$.$V_{\max} = 2000 	imes 10^{-6} - \frac{40000}{40} 	imes 10^{-6} = 2 	imes 10^{-3} - 1 	imes 10^{-3} = 1 	imes 10^{-3} 	ext{ V} = 1 	ext{ mV}$.

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:

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