The following circuit diagram shows the set up for measurement of emf generated in a thermocouple connected between X and Y. The cell E of emf 2 V has negligible internal resistance. The potentiometer wire of length 1 m has a resistance of 10 Ω. The balance point S is found to be 400 mm from point P. Calculate the emf generated by the thermocouple.

Solution

Resistance of potentiometer wire PQ, r = 10 Ω
Length of the potentiometer wire, l = 1m

Current through the wire PQ is
$I = \frac{2 V}{(990 + 10) Ω} = \frac{2}{1000} = 2 \times 10^{- 3} A$

Potential drop across the wire PQ = V_r= I_rr
V_r= 2 x 10^–3 x 10
= 0.02 V

Potential gradient (k) along the wire PQ is $= \frac{V_{r}}{L}$
$k = \frac{0.02 V}{1 m} = \frac{0.02 V}{1000 mm}$

Potential drop across the wire PS = Potential drop across XY
$= \frac{0.02 V}{1000 mm} \times 400 mm = 0.008 V .$
Thus, emf generated by thermocouple = 0.008 V.

For Daily Free Study Material Join wiredfaculty

Current Electricity

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction