Q 12-03-215JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
If you are provided a set of resistances, $2\ \Omega$, $4\ \Omega$, $6\ \Omega$ and $8\ \Omega$. Connect these resistances to obtain an equivalent resistance of $\frac{46}{3}\ \Omega$.
Answer: (C) $2\ \Omega$ and $4\ \Omega$ are in parallel with $6\ \Omega$ and $8\ \Omega$ in series
Each option means: the first two resistors in parallel, and this combination in series with the other two.
$2\ \Omega \parallel 4\ \Omega = \dfrac{8}{6} = \dfrac{4}{3}\ \Omega$; adding $6 + 8 = 14\ \Omega$ in series:
$$R = \frac{4}{3} + 14 = \frac{46}{3}\ \Omega$$
(The others give $13.5\ \Omega$, $12.4\ \Omega$ and $\frac{66}{7}\ \Omega$.)
Solution by Sreeraj P, M.Sc Physics