The thermo emf of a thermocouple is given by | vedclass

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

Question

(D) The thermo emf is given by $E = aT + bT^2$. At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf

Vedclass · Accepted Answer

$443$

Vedclass · Answer

(D) The thermo emf is given by $E = aT + bT^2$. At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf becomes zero. Setting $E = 0$: $aT_i + bT_i^2 = 0$ $T_i(a + bT_i) = 0$ Since $T_i 
eq 0$,we have $T_i = -\frac{a}{b}$. Given $\frac{a}{b} = -200^{\circ}C$,therefore $T_i = -(-200^{\circ}C) = 200^{\circ}C$. This $T_i$ is the temperature relative to the cold junction at $0^{\circ}C$. If the cold junction is at $T_c = 30^{\circ}C$,the inversion temperature relative to the cold junction is $T_i = 200^{\circ}C$. However,the standard relation for inversion temperature is $T_i - T_c = T_n - T_i$,where $T_n = -\frac{a}{2b} = 100^{\circ}C$. Thus,$T_i = 2T_n - T_c = 2(100) - 30 = 170^{\circ}C$. Converting to Kelvin: $T = 170 + 273 = 443 \ K$.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(P)$ $M^1 L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(Q)$ $M^1 L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(R)$ $M^1 L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(S)$ None of these

The thermo emf of a thermocouple is given by $E = aT + bT^2$,where $\frac{a}{b} = -200^{\circ}C$. If the cold junction is kept at $30^{\circ}C$,then the inversion temperature is ($\varepsilon$ in volt,$T$ in centigrade). (in $K$)

Similar Questions

Match the following two columns:
Column-$I$ Column-$II$
$(A)$ Electrical resistance $(P)$ $M^1 L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential $(Q)$ $M^1 L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance $(R)$ $M^1 L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance $(S)$ None of these

When a current $I$ is passed through a wire of constant resistance,it produces a potential difference $V$ across its ends. The graph drawn between $\log I$ and $\log V$ will be:

In the following circuit,the $18\,\Omega$ resistor develops $2\,J/s$ due to the current flowing through it. The power developed across the $10\,\Omega$ resistor is .............. $W$.

In the given circuit,the reading of the voltmeter connected across the $400\,\Omega$ resistor is $30\,V$. If it is connected across the $300\,\Omega$ resistor,what will be its reading in $V$?

Two voltmeters $V_1$ and $V_2$ are connected in series in a $D.C.$ circuit. The reading of $V_1$ is $80 \, V$ and its sensitivity is $200 \, \Omega/V$. If the resistance of $V_2$ is $32 \, k\Omega$,what is the total voltage of the circuit in $V$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module