Three coils of inductance $\mathrm{L}_1=2 \mathrm{H}, \mathrm{L}_2=3 \mathrm{H}$ and $\mathrm{L}_3=6…
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
S
P
R
Q
Solution
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 $\mathrm{R}$,
$\begin{aligned}
& \mathrm{L}_1+\mathrm{L}_2=5 \mathrm{H} \\
& \frac{1}{\mathrm{~L}_1+\mathrm{L}_2}+\frac{\mathrm{L}}{\mathrm{L}_3}=\frac{1}{5}+\frac{1}{6}=\frac{6+5}{30}=\frac{11}{30}
\end{aligned}$
$\therefore \quad \mathrm{L}_{\mathrm{eff}}=\frac{30}{11}=2.72 \mathrm{H}$
$\therefore \quad$ The correct combination would be Q.
$\begin{array}{ll}
& \frac{1}{\mathrm{~L}_1}+\frac{1}{\mathrm{~L}_2}+\frac{1}{\mathrm{~L}_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6} \\
\therefore \quad & \mathrm{L}_{\mathrm{eff}}=1 \mathrm{H}
\end{array}$
