Q 12-03-210JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
A $16\ \Omega$ wire is bent to form a square loop. A $9$ V supply having an internal resistance of $1\ \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is ______ $\times10^{-1}$ V.
Numerical value type. Enter your answer.
Answer: 45
Each side is $4\ \Omega$. Across the chosen side, $4\ \Omega$ is in parallel with the other three sides ($12\ \Omega$): $R = 3\ \Omega$.
$I = \dfrac{9}{3 + 1} = 2.25$ A, so the voltage across the side is $6.75$ V.
The current in the $12\ \Omega$ path is $\dfrac{6.75}{12} = 0.5625$ A, giving $2.25$ V per side. Each diagonal spans two sides of this path:
$$V_{diagonal} = 2\times2.25 = 4.5\ \text{V} = 45\times10^{-1}\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics