Q 12-03-209JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
The given potentiometer has its wire of resistance $10\ \Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across $2\ \Omega$ resistor is:
Answer: (C) $\frac{40}{9}$ V
The slider divides the wire into two $5\ \Omega$ halves. The $2\ \Omega$ resistor is in parallel with the half near $A$:
$$R_p = \frac{5\times2}{5 + 2} = \frac{10}{7}\ \Omega,\qquad R_{total} = 5 + \frac{10}{7} = \frac{45}{7}\ \Omega$$
$I = \dfrac{20}{45/7} = \dfrac{28}{9}$ A, so the drop across the parallel part (and the $2\ \Omega$ resistor) is
$$V = \frac{28}{9}\times\frac{10}{7} = \frac{40}{9}\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics