Question

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?

Short answer

a. RMS current is $I_{rms}=2.59A$.

b. RMS voltage across R is $V_R=38.8V$.

c. RMS voltage across C is $V_C=159V$.

d. RMS voltage across L is $V_L=224V$.

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

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

g. The average rate of energy dissipated by R is $P_R=100W$.

h. The average rate of energy dissipated by C is zero.

i. The average rate of energy dissipated by L is zero.

Step-by-step solution

Listing the given quantities

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 $f=550Hz$.

Understanding the concepts of power and rms value

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.

Calculation of the RMS current

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 $I_{rms}=2.59A$.

Calculation of the Rms voltage across R

b.

Write the equation of Ohm’s law as below.

$V_R=I_{rms}R$

Using the value of RMS current and resistance, you get

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

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

Calculation of the Rms voltage across C

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 $V_C=159V$.

Calculation of the Rms voltage across  

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 $V_L=224V$.

Calculation of the Rms voltage across C and L together

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 $V_{CL}=64.2V$.

Calculation of the Rms voltage across R, C, and L together

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 $V_{RCL}=75.0V$.

Calculation of the average rate of energy dissipated by R:

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 $P_R=100W$.

Calculation of the average rate of energy dissipated by C :

h.

Average rate of energy dissipated by C :

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

Calculation of the average rate of energy dissipated by L 

i.

Average rate of energy dissipated by L:

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