When a 60 mH inductor and resistor are connected in series with an...

For the LR circuit, the phase angle $\phi_1$ is given by $\tan \phi_1 = \dfrac{X_L}{R} = \dfrac{\omega L}{R}$.

Given \phi_1 = 60^{}^{\circ}, we have \tan 60^{}^{\circ} = \dfrac{\omega L}{R} \Rightarrow \sqrt{3} = \dfrac{\omega L}{R} \Rightarrow R = \dfrac{\omega L}{\sqrt{3}}.

For the CR circuit, the phase angle $\phi_2$ is given by $\tan \phi_2 = \dfrac{X_C}{R} = \dfrac{1}{\omega C R}$.

Given \phi_2 = 30^{}^{\circ}, we have \tan 30^{}^{\circ} = \dfrac{1}{\omega C R} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{1}{\omega C R} \Rightarrow R = \dfrac{\sqrt{3}}{\omega C}.

Equating the two expressions for $R$:

$\dfrac{\omega L}{\sqrt{3}} = \dfrac{\sqrt{3}}{\omega C}$

$\omega^2 = \dfrac{3}{LC}$

Substituting the given values $L = 60 \times 10^{-3}$ H and $C = 0.5 \times 10^{-6}$ F:

$\omega^2 = \dfrac{3}{60 \times 10^{-3} \times 0.5 \times 10^{-6}} = \dfrac{3}{30 \times 10^{-9}} = \dfrac{3}{3 \times 10^{-8}} = 10^8$

$\omega = 10^4$ rad/s

The frequency of the AC supply is:

$f = \dfrac{\omega}{2\pi} = \dfrac{10^4}{2\pi}$ Hz

Answer: $\dfrac{10^4}{2\pi}$