In Fig. 31-7, $\mathit{R}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\Omega }$, $\mathit{C}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{70}\mathbf{}\mathit{\mu }\mathit{F}$, and $\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{H}$. The generator provides an emf with rms voltage${\mathit{V}}_{\mathbf{R}\mathbf{C}\mathbf{L}}\mathbf{=}\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{}\mathit{V}$and frequency $\mathit{f}\mathbf{}\mathbf{=}\mathbf{}\mathbf{550}\mathbf{}\mathit{H}\mathit{z}$. (a) What is the rms current? What is the rms voltage across (b) R, (c) C, (d) L, (e) C and L together, and (f) R, C, and L together? At what average rate is energy dissipated by (g) R , (h) C, and (i) L?

Resistance is $R=15.0\Omega$.

Capacitance is$C=4.70\mu F$.

Inductance is$L=25.0mH$.

Rms voltage is$V_{rms}=75.0V$.

Frequency is $f=550Hz$.

Use the concept of RMS current and RMS voltage.Also, use the concept of power.Define the impedance of the circuit. Using the equations that can find the RMS current and RMS voltages across the given components. Using the equation of power, determine the power (rate of energy dissipated) across the given component.

Formulas:

The RMS current is,

${\mathit{I}}_{\mathbf{rms}}\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{rms}}}{\mathbf{Z}}$

The voltage is,

$\mathit{V}\mathbf{=}{\mathit{I}}_{\mathbf{rms}}\mathit{R}$

Here, ${\mathit{I}}_{\mathbf{rms}}$is the RMS current and R is the resistance.

The power is define by,

$\mathit{P}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$

The impedance is given by,

$\mathit{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}{\mathbf{X}}_{\mathbf{L}}\mathbf{-}{\mathbf{X}}_{\mathbf{C}}\mathbf{)}}^{\mathbf{2}}}$

The inductive reactance

${\mathit{X}}_{\mathbf{L}}\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{f}\mathit{L}$

Here, L is the inductor and f is the frequency.

The capacitive reactance.

${\mathit{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi fC}}$

Here, C is the capacitance.

a.

Writethe equation for theimpedance of the circuit:

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

Determine the inductive reactance as below.

$$\begin{gathered} X_L=2\pi fL \\ =2\left(3.14\right)\left(550\right)\left(25\times {10}^{-3}\right) \\ =86.35\Omega \end{gathered}$$

Determine thecapacitive reactance as follow.

$$\begin{gathered} X_C=\frac{1}{2\pi fC} \\ =\frac{1}{2\left(3.14\right)\left(550\right)\left(4.70\times {10}^{-6}\right)} \\ =61.57\Omega \end{gathered}$$

Plugging these values in equation of impedance, you get

$$\begin{gathered} Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ =\sqrt{{15}^2+{\left(86.35-61.57\right)}^2} \\ =28.94\Omega \end{gathered}$$

Plugging this value in equation of RMS current, you have

$$\begin{gathered} I_{rms}=\frac{{\epsilon }_{rms}}{Z} \\ =\frac{75}{28.9} \\ =2.59A \end{gathered}$$

Hence, the RMS current is $I_{rms}=2.59A$.

b.

Write the equation of Ohm’s law as below.

$V_R=I_{rms}R$

$$\begin{gathered} V_R=2.59\left(15\right) \\ =38.8V \end{gathered}$$

Rms voltage across R is $V_R=38.8V$.

c.

Rms voltage across C :

$V_C=I_{rms}{X}_C$

Plugging the values of rms current and capacitive reactance, you obtain

$$\begin{gathered} V_C=2.59\left(61.599\right) \\ =159V \end{gathered}$$

Hence, the RMS voltage across C is $V_C=159V$.

d.

Write the equation for voltage as below.

$V_L=I_{rms}{X}_L$

Plugging the values of rms current and inductive reactance, you get

$$\begin{gathered} V_L=2.59\left(86.35\right) \\ =223.7V \\ =224V \end{gathered}$$

Rms voltage across L is $V_L=224V$.

e.

Rms voltage across C and L together:

We can write

$$\begin{gathered} V_{CL}=\left|V_C-V_L\right| \\ =\left|159-223.7\right| \\ =64.2V \end{gathered}$$

Hence, RMS voltage across C and L together is $V_{CL}=64.2V$.

f.

Rms voltage across R, C, and L together:

We can write

$$\begin{gathered} V_{RCL}=\sqrt{V_R^2+V_{CL}^2} \\ =\sqrt{{38.8}^2+{64.2}^2} \\ =75.0V \end{gathered}$$

Hence the RMS voltage across R, C, and L together is $V_{RCL}=75.0V$.

g.

Usingtheequation of power, you can write

$$\begin{gathered} P_R=\frac{V_R^2}{R} \\ =\frac{{\left(38.8\right)}^2}{15} \\ =100W \end{gathered}$$

Hence, the average rate of energy dissipated by R is $P_R=100W$.

h.

Average rate of energy dissipated by C :

As there is no energy lost in capacitor, no energy is dissipated in capacitor.

i.

Average rate of energy dissipated by L:

As inductor does not cause any energy loss, so the energy dissipated in inductor is zero.