How to find True power dissipated and apparent power?

Well, the input impedance is given by:

$$underline{text{Z}}_{spacetext{in}}=frac{1}{frac{1}{text{R}}+frac{1}{text{j}omegatext{L}}+text{j}omegatext{C}}tag1$$

The true power is given by:

$$text{P}=text{V}_{text{in}}^*cdottext{I}_{text{in}}^*cdotcosleft(varphiright)tag2$$

Where $x^*$ is the RMS value.

In your case, $text{V}_{text{in}}^*=240$ is given. So, we get:

• $$text{I}_{text{in}}^*=frac{1}{sqrt{2}}cdotfrac{240}{left|underline{text{Z}}_{spacetext{in}}right|}tag3$$
• $$varphi=argleft(underline{text{Z}}_{spacetext{in}}right)tag4$$

It is not hard, to show that:

$$underline{text{Z}}_{spacetext{in}}=frac{1}{68 left(left(frac{1}{5 pi }-frac{57 pi }{10000}right)^2+frac{1}{4624}right)}+frac{2890000 pi left(57 pi ^2-2000right)}{59642000 pi ^2-938961 pi ^4-1156000000}cdottext{j}tag5$$

So:

• $$text{I}_{text{in}}^*=frac{3 sqrt{578000000-29821000 pi ^2+frac{938961 pi ^4}{2}}}{2125 pi }tag6$$
• $$varphi=-arctanleft(frac{68}{5pi}-frac{969pi}{2500}right)tag7$$

The answer, gives:

$$text{P}=frac{7200 sqrt{2}}{17}approx598.961spacetext{W}tag8$$