The thermo emf of a hypothetical thermocouple | vedclass

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(D) The thermo emf is given by $E = a	heta + b	heta^2$. To find the neutral temperature $(T_n)$,we differentiate $E$ with respect to $	heta$ and se

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None of these

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(D) The thermo emf is given by $E = a	heta + b	heta^2$. To find the neutral temperature $(T_n)$,we differentiate $E$ with respect to $	heta$ and set it to zero,as the thermo emf is maximum at the neutral temperature. $\frac{dE}{d	heta} = a + 2b	heta$. Setting $\frac{dE}{d	heta} = 0$,we get $a + 2b	heta = 0$. Therefore,$T_n = -\frac{a}{2b}$. Given that $a/b = 700^{\circ}C$,we substitute this into the expression: $T_n = -\frac{1}{2} 	imes (700^{\circ}C) = -350^{\circ}C$. Since the calculated value $-350^{\circ}C$ is not among the options,the correct choice is $D$.

The thermo emf of a hypothetical thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700^{\circ}C$. If the cold junction is kept at $0^{\circ}C$,then the neutral temperature is:

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