MHTCET 17 Apr 2026 Shift 1 Paper

Concept:The effective inductance depends on how the coils are grouped: series combinations add directly, while parallel combinations follow the reciprocal rule.Explanation:Given: $L_{1}=2\ \mathrm{H}$, $L_{2}=3\ \mathrm{H}$, $L_{3}=6\ \mathrm{H}$.We check each circuit to find the equivalent inductance between points A and B.In circuit P, all three inductors are in series:$$L_{eq}=L_{1}+L_{2}+L_{3}=2+3+6=11\ \mathrm{H}$$This is not equal to $\frac{18}{11}\ \mathrm{H}$.In circuit Q, all three inductors are in parallel:$$\frac{1}{L_{eq}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}+\frac{1}{L_{3}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$$So, $L_{eq}=1\ \mathrm{H}$, which is not the required value.In circuit R, $L_{1}$ and $L_{2}$ are in series, and this combination is in parallel with $L_{3}$:$$L_{eq}=\frac{(L_{1}+L_{2})L_{3}}{(L_{1}+L_{2})+L_{3}}=\frac{5\times 6}{5+6}=\frac{30}{11}\ \mathrm{H}$$This does not match the required value either.In circuit S, $L_{2}$ and $L_{3}$ are in series, and this combination is in parallel with $L_{1}$:$$L_{eq}=\frac{L_{1}(L_{2}+L_{3})}{L_{1}+(L_{2}+L_{3})}=\frac{2\times 9}{2+9}=\frac{18}{11}\ \mathrm{H}$$This exactly matches the required effective inductance.Answer:The correct circuit is S, which corresponds to option B.