91Ó°ÊÓ

Resistance is $R=16.0\mathrm{Î\copyright }$.

Capacitance is $C=31.2\mathrm{μ}F$.

Inductance is$L=9.20mH$.

Emf is ${\mathrm{Î\mu }}_m={\mathrm{Î\mu }}_m\sin \left({\mathrm{Ó\neg }}_{d}t\right)$.

The magnitude of emf is ${\mathrm{Î\mu }}_m=45.0V$.

Angular frequency is ${\mathrm{Ó\neg }}_d=3000rad/s$.

Time is $t=0.442ms$.

Use the concept of power in generator, capacitor, inductor, and resistor.For generator,use power related to current and emf. Using the given equations and the equations of power, calculate the rate of energy supplied or changing across the given element.

Formulas:

The power is define by,

$$\begin{gathered} \mathit{P}\mathbf{=}\mathit{I}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{I}\left(IR\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R} \end{gathered}$$

Here, Vis the voltage, R is the resistance, and I is the current.

The impedance is given by,

$\mathit{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_C\right)}^{\mathbf{2}}}$

The inductive reactance

$$\begin{gathered} {\mathit{X}}_{\mathbf{L}}\mathbf{=}{\mathit{Ó\neg }}_{\mathbf{d}}\mathit{L} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathit{Ï€}\mathit{f}\mathit{L} \end{gathered}$$

Here, ${\mathit{Ó\neg }}_{\mathbf{d}}$is the angular frequency, L is the inductor and f is the frequency.

The capacitive reactance.

$$\begin{gathered} {\mathit{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Ó\neg }}_{\mathbf{d}}\mathbf{C}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}\mathbf{f}\mathbf{C}} \end{gathered}$$

Here, C is the capacitance.

The current is define by,

$\mathit{I}\mathbf{=}\frac{{\mathbf{Î\mu }}_{\mathbf{m}}}{\mathbf{Z}}$

The phase angle is define by,

$\mathit{Ï•}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{X_L-X_C}{R}\right)$

The power is,

${\mathit{P}}_{\mathbf{L}}\mathbf{=}\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}$.

a.

The rate $P_g$at which energy is being supplied by the generator:

Power across generator is

$$\begin{gathered} P_g=i\left(t\right)×\mathrm{Î\mu }\left(t\right) \\ =Isin\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)×{\mathrm{Î\mu }}_{m}sin{\mathrm{Ó\neg }}_{d}t \end{gathered}$$

Here, current I and the phase angle are unknown; so calculate them using the equation of current as given below.

$I=\frac{{\mathrm{Î\mu }}_m}{Z}$

Write the equation for the impedance of the circuit:

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

Determine the inductive reactance as below.

$$\begin{gathered} X_L={\mathrm{Ó\neg }}_{d}L \\ =\left(3000\right)\left(9.20×{10}^{-3}\right) \\ =27.6\mathrm{Î\copyright } \end{gathered}$$

Determine thecapacitive reactance as follow.

$$\begin{gathered} X_C=\frac{1}{{\mathrm{Ó\neg }}_{d}C} \\ =\frac{1}{\left(3000\right)\left(31.2×{10}^{-6}\right)} \\ =10.6837\mathrm{Î\copyright } \end{gathered}$$

Plugging these values in equation of impedance, you get

$$\begin{gathered} Z=\sqrt{{\left(16.0\right)}^2+{\left(27.6-10.6837\right)}^2} \\ =\sqrt{256+286.161} \\ =\sqrt{542.161} \\ =23.2843\mathrm{Î\copyright } \end{gathered}$$

Plugging this value in equation of RMS current, you have

$$\begin{gathered} I=\frac{{\mathrm{Î\mu }}_m}{Z} \\ =\frac{45.0}{23.2843} \\ =1.93A \end{gathered}$$

Now we can find $\mathrm{Ï•}$using the equation

$$\begin{gathered} \mathrm{ϕ}={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right) \\ ={\tan }^{-1}\left(\frac{27.6-10.6837}{16}\right) \\ ={\tan }^{-1}\left(1.057\right) \\ =46.5° \end{gathered}$$

Plugging these values in the equation of power of generator, you have

$P_g=Isin\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)×{\mathrm{Î\mu }}_{m}sin{\mathrm{Ó\neg }}_{d}t$

Substitute all known numerical values ion the above equation.

$$\begin{gathered} P_g=1.93×sin\left(3000\left(0.442×{10}^{-3}\right)-46.5°\right)×\left(45.0\right)×\text{sin}(3000\left(0.442×{10}^{-3})\right) \\ =1.93×\sin \left(1.326rad×\left(\frac{57.3°}{rad}\right)-46.5°\right)\left(45.0\right)\sin \left(1.326rad×\left(\frac{{57.3}^0}{rad}\right)\right) \\ =1.93×\sin \left(29.4798°\right)×\left(45.0\right)×\sin \left(75.9798°\right) \\ =1.93×0.4921×45.0×0.9702 \end{gathered}$$$P_g=41.4W$

The rate at which energy is being supplied by the generator is $P_g=41.4W$.

b.

The rate $P_C$at which energy in the capacitor is changing:

We know voltage across the capacitor is

$$\begin{gathered} V_c\left(t\right)=V_c\text{sin}\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}-\frac{\mathrm{Ï€}}{2}\right) \\ =-V_c\text{cos}\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right) \end{gathered}$$

Also $V_c=\frac{I}{{\mathrm{Ó\neg }}_{d}C}$.

Now, the rate at which energy in capacitor changes is

$$\begin{gathered} P_c=\frac{d}{dt}\left(\frac{q^2}{2C}\right) \\ =\frac{iq}{C} \\ =i×V_c \end{gathered}$$

Plugging the values, you obtain

$$\begin{gathered} P_c=-Isin\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)×\left(\frac{I}{{\mathrm{Ó\neg }}_{d}C}\right)\text{cos}\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right) \\ =-\frac{I^2}{2{\mathrm{Ó\neg }}_{d}C}\text{sin}\left[2\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)\right] \\ =-\frac{{\left(1.93\right)}^2}{2\left(3000\right)\left(31.2×{10}^{-6}\right)}\text{sin}\left[2\left(3000\left(0.442×{10}^{-3}\right)-46.5°\right)\right] \\ =-\frac{{\left(1.93\right)}^2}{2\left(3000\right)\left(31.2×{10}^{-6}\right)}\text{sin}\left[2\left(1.326rad×\left(\frac{{57.3}^0}{rad}\right)-46.5°\right)\right] \end{gathered}$$

$$\begin{gathered} P_c=-\frac{3.7249}{0.1872}\text{sin}\left[2\left(75.9798°-46.5°\right)\right] \\ =-19.8979×\text{sin}\left(58.9596°\right) \\ =-19.8979×0.8568 \\ =-17.0W \end{gathered}$$

Hence, the rate $P_C$at which energy in the capacitor is changing is $P_c=-17.0W$.

c.

The rate at which energy changes in inductor is,

$$\begin{gathered} P_L=\frac{d}{dt}\left(\frac{1}{2}L{i}^2\right) \\ =Li\frac{di}{dt} \end{gathered}$$

Plugging the value of current, you can write

$P_L=LIsin\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)\frac{d\left[Isin\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)\right]}{dt}$

Therefore,

$$\begin{gathered} P_L=\frac{1}{2}{\mathrm{Ó\neg }}_{d}L{I}^2\text{sin}\left[2\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right)\right] \\ =\frac{1}{2}\left(3000\right)\left(9.20×{10}^{-3}\right)\left({1.93}^2\right)\text{sin}\left[2\left(3000\left(0.442×{10}^{-3}\right)-46.5°\right)\right] \\ =\frac{1}{2}\left(3000\right)\left(9.20×{10}^{-3}\right)\left({1.93}^2\right)\text{sin}\left[2\left(3000\left(0.442×{10}^{-3}\right)-46.5°\right)\right] \end{gathered}$$

$$\begin{gathered} P_L=51.4036×\text{sin}\left[2\left(1.326rad\left(\frac{57.3°}{rad}\right)-46.5°\right)\right] \\ =51.4036×\text{sin}\left(58.959°\right) \\ =44.0W \end{gathered}$$

The rate $P_L$at which energy in the inductor is changing is $44.0W$.

d.

Using equation

$$\begin{gathered} P_R=I^{2}R \\ =I^{2}Rsi{n}^2\left({\mathrm{Ó\neg }}_{d}t-\mathrm{Ï•}\right) \end{gathered}$$

$$\begin{gathered} P_R={\left(1.93\right)}^2\left(16.0\right)si{n}^2\left(1.326rad\left(\frac{57.3°}{rad}\right)-46.5°\right) \\ ={\left(1.93\right)}^2\left(16.0\right)si{n}^2\left(29.4798°\right) \\ =14.4W \end{gathered}$$

Therefore, the rate $P_R$at which energy is being dissipated in the resistor is $P_R=14.4W$.

e.

Define $P_g$as follow.

$$\begin{gathered} P_L+P_R+P_C=44.0-17.0+14.4 \\ =41.4W \\ =P_g \end{gathered}$$

Hence, the sum of $P_C$, $P_L$, $P_R$and is equal to $P_g$.