Question

Sol. [Key Concept: Each switch gives $tan\varphi = (X_{L} - X_{C})/R$; solve the two for $3L_{1} - L_{2}$.]

For $S_{1}$ is closed and $S_{2}$ open

$${V = V_{0}sin(300t) }{\frac{X_{L_{1}} - X_{C}}{R} = tan\left( 30^{\circ} \right) = \frac{1}{\sqrt{3}}}$$

$$\begin{array}{r} \sqrt{3}\left( X_{L_{1}} - X_{C} \right) = R\#(1) \end{array}$$

For $S_{1}$ is open and $S_{2}$ is closed.

$$\frac{\left| X_{L_{2}} - X_{C} \right|}{R} = tan60^{\circ} = \sqrt{3}$$

$$\begin{array}{r} \frac{\left| X_{L_{2}} - X_{C} \right|}{\sqrt{3}} = R\#(2) \end{array}$$

from Eq. (1) and (2)

$${\sqrt{3}\left( X_{L_{1}} - X_{C} \right) = \frac{\left( X_{L_{2}} - X_{C} \right)}{\sqrt{3}} }{3\left( \omega{\text{ }L}_{1} - \frac{1}{\omega C} \right) = \left( \omega L_{2} - \frac{1}{\omega C} \right) }{\left| 3{\text{ }L}_{1} - L_{2} \right| = \frac{3}{\omega^{2}C} - \frac{1}{\omega^{2}C} = \frac{2}{9 \times 10^{4} \times 100 \times 10^{- 6}} }{\left. \ \mid 3{\text{ }L}_{1} - L_{2} \right\rbrack = \frac{2}{9} }$$