Answer
Please see the work below.
Work Step by Step
(a) The reactance of the capacitor is given as
$X_C=\frac{1}{\omega_C}$
$\implies X_C=frac{1}{300\pi \times 0.35\times 10^{-6}}$
$X_C=3030\Omega$
Now the impedance is given as
$Z=\sqrt{R^2+X_C^2}$
We plug in the known values to obtain:
$Z=\sqrt{(4000)^2+(3030)^2}$
$Z=5018\Omega$
The power is $cos\phi=\frac{R}{Z}$
$cos\phi=\frac{4000}{5018}$
$cos\phi=0.797$
(b) We know that the power factor is inversely proportional to the impedance of the circuit. But when frequency increases, the impedance decreases and as a result the power factor increases.
