Answer
(a) $0.6$
(b) $85\mu F; 300\mu F$
Work Step by Step
(a) We know that the power factor of the motor is given as
$cos\phi=\frac{R}{Z}$
We plug in the known values to obtain:
$cos\phi=\frac{15\Omega}{25\Omega}$
$cos\phi=0.6$
(b) We know that
$Z=\sqrt{R^2+(X_L-X_C)^2}$
We plug in the known values to obtain:
$Z=\sqrt{(15\Omega)^2+(20\Omega-\frac{1}{120\pi C})^2}$
We also know that
$cos\phi=\frac{R}{Z}$
$\implies 0.8=\frac{15\Omega}{\sqrt{(15\Omega)^2+(20\Omega-\frac{1}{120\pi C})^2}}$
$\implies (20\Omega-\frac{1}{120\pi C})^2=126.5625$
$\implies 20\Omega-\frac{1}{120\pi C}=\pm 11.25$
This simplifies to:
$C=85\mu F$ and $C=300\mu F$
