Answer
(a) $0.602$
(b) $0.106W$
Work Step by Step
(a) We know that
$Z=\sqrt{R^2+\omega^2L^2}$
We plug in the known values to obtain:
$Z=\sqrt{(2.00K\Omega)^2+(2\pi(1.34)(315mH))^2}=3.32\Omega$
Now $cos\phi=\frac{R}{Z}$
$cos\phi=\frac{2.00K\Omega}{3.32K\Omega}=0.602$
(b) We can find the average power as
$P_{avg}=\frac{(V_{rms})^2}{Z}R$
We plug in the known values to obtain:
$P_{avg}=(\frac{24.2}{3.32K\Omega})^2(2.00K\Omega)=0.106W$
