Question

The temperature of hot junction of a thermo-couple changes from ${80}^{\circ }C$ to ${100}^{\circ }C$.The percentage change in thermo-electric power is

• A. 80%
• B. 10%
• C. 20%
• D. 25%

Answer 1

Answer: (A)

80%

We know the thermoelectric power (for thermocouple) is given by

$P=\frac{dE}{dT}=\frac{\pi }{T}$

when, the temperature of hot junction of a thermocouple changes from ${80}^{\circ }C$ to ${100}^{\circ }C$

Then $p_1=\frac{\pi }{80}$ and $P_2=\frac{\pi }{100}$

The change in thermoelectric power is $\delta =P_1-P_2$

Now, the percentage change in thermoelectric power is

$\frac{\delta P\times 100}{P}=\frac{P_1-P_2}{P_1}$

$\frac{\delta P\times 100}{P}=20$