91Ó°ÊÓ

The angle between the voltage and current phasors is given by the equation

$\mathbf{³Ù²¹²ÔÏ•}\mathbf{=}\left(\frac{X_L-X_C}{R}\right)$

The inductive reactance of the coil can be calculated by ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{Ó\neg ³¢}$ and capacitive reactance by ${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{Ó\neg °ä}}$. Impedance is given by $\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_C\right)}^{\mathbf{2}}}$. The expression for rms voltage is${\mathbf{V}}_{\mathbf{rms}}\mathbf{=}{\mathbf{I}}_{\mathbf{rms}}\mathbf{Z}$ and the expression for average power is${\mathbf{P}}_{\mathbf{avg}}\mathbf{=}{\mathbf{V}}_{\mathbf{rms}}{\mathbf{I}}_{\mathbf{rms}}\mathbf{³¦´Ç²õÏ•}$ and also ${\mathbf{P}}_{\mathbf{avg}}\mathbf{=}{\left(I_{\mathrm{rms}}\right)}^{\mathbf{2}}\mathbf{R}$.

Plug the values for L and to get inductive reactance and also C and $\mathrm{Ó\neg }$to get capacitive reactance.

$$\begin{gathered} {\mathrm{X}}_{\mathrm{L}}=2\mathrm{Ï„Ï„´Ú³¢} \\ =2\mathrm{Ï„Ï„}\left(400\mathrm{Hz}\right)\left(0.120\mathrm{H}\right) \\ =301.6\mathrm{Î\copyright } \end{gathered}$$

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{1}{2\mathrm{Ï„Ï„´Ú°ä}} \\ =\frac{1}{2\mathrm{Ï„Ï„}\left(400\mathrm{Hz}\right)\left(7.3×{10}^{-6}\mathrm{F}\right)} \\ =54.5\mathrm{Î\copyright } \end{gathered}$$

$$\begin{gathered} \tan \mathrm{Ï•}=\left(\frac{301.6-54.5}{240}\right) \\ =1.028 \end{gathered}$$

Hence, the angle is $\mathrm{ϕ}=+45.8°$and the power factor$\cos \mathrm{ϕ}=0.697$

$$\begin{gathered} \mathrm{Z}=\sqrt{\left({240}^2\right)+{\left(301.6-54.5\right)}^2}\mathrm{Î\copyright } \\ =344\mathrm{Î\copyright } \end{gathered}$$

Hence, the value of impedance is 344Ω.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{rms}}=\left(0.450\mathrm{A}\right)\left(344\mathrm{Î\copyright }\right) \\ =155\mathrm{V} \end{gathered}$$

Therefore, the rms voltage is 155V.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\left(155\mathrm{V}\right)\left(0.450\mathrm{A}\right)\left(0.697\right) \\ =48.6\mathrm{W} \end{gathered}$$

Therefore, the average power delivered by the source is 48.6W.

The average rate here represents the average power dissipated into the resistor, so we use equation ${\mathrm{P}}_{\mathrm{avg}}={\left({\mathrm{I}}_{\mathrm{rms}}\right)}^2\mathrm{R}$.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{avg}}=\left(155\mathrm{V}\right)\left(0.450\mathrm{A}\right)\left(0.697\right) \\ =48.6\mathrm{W} \end{gathered}$$

It is the same result as of part (d).

So, as shown in the parts (d) and (e), all generated power is dissipated in the resistor R only. Hence, no power is consumed in the capacitor and inductor.

This occurs because the capacitor charges and discharges the energy when it is connected to ac source.

Hence, the answers for all the parts are:

(a)$\mathrm{Ï•}=+45.8^{°}$ and$\mathrm{³¦´Ç²õÏ•}=0.697$

(b)$Z=344\mathrm{Î\copyright }$

(c) ${\mathrm{V}}_{\mathrm{rms}}=155\mathrm{V}$

(d) ${\mathrm{P}}_{\mathrm{avg}}=48.6\mathrm{W}$,(e)${\mathrm{P}}_{\mathrm{avg}}=48.6\mathrm{W}$

(f)${\mathrm{P}}_{\mathrm{avg}}=0$,(g)${\mathrm{P}}_{\mathrm{avg}}=0$