Answer
(a) $291.8\Omega$
(b) greater than
Work Step by Step
(a) We can find the required resistance as
$R=Zcos\phi$
We plug in the known values to obtain:
$R=(337\Omega) cos(30^{\circ})$
$\implies R=291.8\Omega$
(b) We know that
$tan 30^{\circ}=\frac{X_L-X_C}{R}$
$\implies X_L-X_C=R\frac{1}{\sqrt{3}}$
$\implies X_L=\frac{R}{\sqrt{3}}+X_C$
The inductive reactance is given as $X_L=\omega L$
and the capacitance in terms of frequency and capacitance is $X_C=\frac{1}{\omega C}$
Now $X_L=\frac{R}{\sqrt{3}}+X_C$
As $X_L\gt X_C$
$\implies \omega _L\gt \frac{1}{\omega C}$
$\implies \omega \gt \sqrt{\frac{1}{LC}}$
We know that for resonance frequency, the inductive reactance and capacitive reactance are equal in magnitude. Thus, we have $\omega=\sqrt{\frac{1}{LC}}$ and we conclude that the driving frequency is greater than the resonance frequency of the circuit.
