Answer
(a) $29\Omega $
(b) $71\mu F $
(c) decrease
Work Step by Step
(a) We know that
$ Z=\frac{R}{cos\phi}$
We plug in the known values to obtain:
$ Z=\frac{25}{cos30}$
$ Z=29\Omega $
(b) We know that
$ X_C=X_L-(Z^2-R^2)^{\frac{1}{2}}$
We plug in the known values to obtain:
$ X_C=55.3-[(28.86)^2-(25)^2]^{\frac{1}{2}}$
$ X_C=40.89$
Now $ C=\frac{1}{\omega\times 40.89}$
$\implies C=\frac{1}{110\pi\times 40.89}$
$ C=71\mu F $
(c) We know that $ X_L\gt X_C $
We see that $(X_L-X_C=\omega L-\frac{1}{\omega C})$ is greater than 1. If C decreases, $\frac{1}{\omega C}$ increases and as a result the quantity $(\omega L-\frac{1}{\omega C})$ decreases. We know that the impedance is given as $ Z=\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}$. Thus, we conclude that if the value of C is decreased then the impedance of the circuit decreases as well.
