Q 12-03-239JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
Four resistances $40\ \Omega$, $60\ \Omega$, $90\ \Omega$ and $110\ \Omega$ make the arms of a quadrilateral $ABCD$ ($AB = 40\ \Omega$, $BC = 60\ \Omega$, $AD = 90\ \Omega$, $DC = 110\ \Omega$). Across $AC$ is a battery of emf $40\ \text{V}$ and internal resistance negligible. The potential difference across $BD$ in V is ______.
Numerical value type. Enter your answer.
Answer: 2
Branch $ABC$ ($100\ \Omega$) and branch $ADC$ ($200\ \Omega$) are each directly across $40$ V.
$V_A - V_B = 40\times\dfrac{40}{100} = 16$ V and $V_A - V_D = 40\times\dfrac{90}{200} = 18$ V.
$V_B - V_D = 18 - 16 = 2$ V.
Solution by Sreeraj P, M.Sc Physics