Q. Figure shows a simple potentiometer circuit for measuring a small e.m.f. produced by a thermocouple. The meter wire $PQ$ has a resistance $5\ \Omega$ and the driver cell has an e.m.f. of $2\ V.$ If a balance point is obtained $0.600m$ along $PQ$ when measuring an e.m.f. of $6.00\ mV,$ what is the value of resistance $R$
A) $995\ \Omega$
B) $1995\ \Omega$
C) $2995\ \Omega$
D) None of these
Explanation
(a) The voltage per unit light of the metre wire $PQ$ is $\left( \frac{6.00\ mV}{0.600\ m} \right)\ i.e.\ 10\ mV/m.$ Hence potential difference across the metre wire is $10mV/m \times 1m = 10\ mV.$ The current drawn from the driver cell is $i = \frac{10\ mV}{5\ \Omega} = 2\ mA$
The resistance $R = \frac{(2V - 10mV)}{2\ mA} = \frac{1990\ mV}{2\ mA} = 995\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.