An alternating voltage given by V = 120 (2000t) is applied across a...

Given $V = 120 \sin(2000t)$, the angular frequency of the source is $\omega = 2000\text{ rad/s}$.

The resonant angular frequency is given by $\omega_0 = \dfrac{1}{\sqrt{LC}}$.

Substituting the given values $L = 0.1\text{ H}$ and $C = 10\ \mu\text{F} = 10^{-5}\text{ F}$:

$\omega_0 = \dfrac{1}{\sqrt{0.1 \times 10^{-5}}} = \dfrac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}$

The resonant frequency $\nu_0$ is:

$\nu_0 = \dfrac{\omega_0}{2\pi} = \dfrac{1000}{2\pi} = \dfrac{500}{\pi}\text{ Hz}$

To find the phase relationship at the source frequency $\omega = 2000\text{ rad/s}$, we calculate the inductive and capacitive reactances:

$X_L = \omega L = 2000 \times 0.1 = 200\ \Omega$

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2000 \times 10^{-5}} = 50\ \Omega$

Since $X_L > X_C$, the circuit is predominantly inductive at this frequency.

In an inductive circuit, the current lags the applied voltage.

Answer: $\nu_0 = \dfrac{500}{\pi}\text{ Hz}$; current lags the voltage