Q.
In the following circuit a $10\ m$ long potentiometer wire with resistance $1.2\ ohm/m,$ a resistance $R_{1}$ and an accumulator of emf $2\ V$ are connected in series. When the emf of thermocouple is $2.4\ mV$ then the deflection in galvanometer is zero. The current supplied by the accumulator will be
A) $4 \times 10^{- 4}A$
B) $8 \times 10^{- 4}A$
C) $4 \times 10^{- 3}A$
D) $8 \times 10^{- 3}A$
Explanation
(a) $E = xl = i\rho l \Rightarrow i = \frac{E}{\rho l} = \frac{2.4 \times 10^{- 3}}{1.2 \times 5} = 4 \times 10^{- 4}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.