Answer
(a) $0.962$
(b) increase
(c) $0.998$
Work Step by Step
(a) We know that
$Z=\sqrt{R^2+(2\pi fL-\frac{1}{2\pi fC})^2}$
We plug in the known values to obtain:
$Z=\sqrt{(105\Omega)^2+(2\pi(125Hz)(0.085H)-\frac{1}{2\pi(125Hz)(13.2\times 10^{-6}F)})^2}=109.12\Omega$
The power factor is given as
$cos\phi=\frac{R}{Z}$
We plug in the known values to obtain:
$cos\phi=\frac{105\Omega}{109.12\Omega}=0.962$
Thus, the power factor is $0.962$
(b) As $cos\phi=\frac{R}{\sqrt{R^2+(2\pi fL-\frac{1}{2\pi fC})^2}}$
This simplifies to:
$cos\phi=\frac{1}{\sqrt{1+\frac{1}{R^2}(2\pi fL-\frac{1}{2\pi fC})^2}}$
The above equation shows that the increase in resistance causes a corresponding increase in the power factor.
(c) The impedance is given as
$Z=\sqrt{R^2+(2\pi fL-\frac{1}{2\pi fC})^2}$
We plug in the known values to obtain:
$Z=\sqrt{(525\Omega)^2+(2\pi(125Hz)(0.085H)-\frac{1}{2\pi(125Hz)(13.2\times 10^{-6}F )})^2}$
$Z=525.839\Omega$
Now $cos\pi=\frac{R}{Z}$
We plug in the known values to obtain:
$cos\phi=\frac{525\Omega}{525.839\Omega}=0.998$
