Explanation

(b) Ist case From formula
$R = \frac{V^{2}}{P}$
$= \frac{110 \times 110}{330} = \frac{110}{3}\Omega$
Since, current lags the voltage thus, the circuit contains resistance and inductance.
Power factor$\cos{\phi = 0.6}$
$\frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = 0.6$
$\Rightarrow \ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R}^{2} + X_{L}^{2} = \left( \frac{R}{0.6} \right)^{2}\ \ \ $
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X_{L}^{2} = \frac{R^{2}}{(0.6)^{2}} - R^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X_{L}^{2} = \frac{R^{2} \times 0.64}{0.36}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ \ \ X}_{L} = \frac{0.8\ R}{0.6} = \frac{4R}{3}$ …(i)
IInd case
Now $\cos{\phi = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (given)}$
Therefore, circuit is purely resistive, ie, it contains only resistance. This is the condition of resonance in which
${\ \ \ \ \ \ X}_{L} = X_{C}$
$\therefore\ \ \ \ \ X_{C} = \frac{4R}{3} = \frac{4}{3}\ \times \frac{110}{3} = \frac{440}{9}\Omega$ [From Eq. (i)]
$\frac{1}{2\pi fC} = \frac{440}{9}\ \Omega$
$C = \frac{9}{2 \times 3.14 \times 60 \times 440}$
$= 54\ \mu F$