Three coils of inductance \(\mathrm{L}_1=2 \mathrm{H}, \mathrm{L}_2=3 \mathrm{H}\) and \(\mathrm{L}_3=6 \mathrm{H}\) are connected such that they are separated from each other. To obtain the effective inductance of 1 henry, out of the following combinations as shown in figure, the correct one is

The combinations \(\mathrm{P}\) and \(\mathrm{S}\) are series combinations. Hence, the effective inductance cannot be \(1 \mathrm{H}\).
This leaves combinations \(\mathrm{R}\) and \(\mathrm{Q}\).
For combination \(\mathrm{R}\),
\(
\begin{aligned}
& \mathrm{L}_1+\mathrm{L}_2=5 \mathrm{H} \\
& \frac{1}{\mathrm{~L}_1+\mathrm{L}_2}+\frac{\mathrm{L}}{\mathrm{L}_3}=\frac{1}{5}+\frac{1}{6}=\frac{6+5}{30}=\frac{11}{30}
\end{aligned}
\)
\(
\therefore \quad \mathrm{L}_{\mathrm{eff}}=\frac{30}{11}=2.72 \mathrm{H}
\)
\(\therefore\) The correct combination would be Q.
\(
\begin{array}{ll}
& \frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2}+\frac{1}{\mathrm{~L}_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6} \\
\therefore & \mathrm{L}_{\mathrm{eff}}=1 \mathrm{H}
\end{array}
\)