Question

In an RLC circuit such as that of Fig. 31-7 assume that $\mathit{R}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\Omega }\mathbf{,}\mathit{L}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{H}\mathbf{,}{\mathit{f}}_{\mathbf{d}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathit{H}\mathit{z}$and ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathit{V}$. For what values of the capacitance would the average rate at which energy is dissipated in the resistance be (a) a maximum and (b) a minimum? What are (c) the maximum dissipation rate and the corresponding (d) phase angle and (e) power factor? What are (f) the minimum dissipation rate and the corresponding (g) phase angle and (h) power factor?

Short answer

(a) The average rate at which energy is dissipated in the resistance would be maximum at $C=1.17\times {10}^{-4}F$.

(b) The average rate at which energy is dissipated in the resistance would be minimum at C = 0F .

(c) The maximum dissipation rate is 90.0W.

(d) The phase angle corresponding to the maximum dissipation rate is $0^{\circ }$.

(e) The power factor corresponding to the maximum dissipation rate is 1.

(f) The minimum dissipation rate is 0.

(g) The phase angle corresponding to the minimum dissipation rate is $\varphi =-{90}^{\circ }$.

(h) The power factor corresponding to the minimum dissipation rate is 0 .

Step-by-step solution

Listing the given quantities:

The resistance $R=5.0\Omega$,

The inductor $L=60.0mH$,

The frequency $f_d=60.0Hz$,

The emf voltage ${\epsilon }_m=30.0V$,

Understanding the concepts of power:

Using the formula for average power, to define the conditions where dissipation rate is minimum and maximum. From this, getthecorresponding capacitance, phase angle, and power factor.

Formulas:

Average power:

$P_{avg}=\frac{{\epsilon }_m^{2}R}{2Z^2}$

Here,the impedance is,

$Z=\sqrt{R^2+\left({\omega }_{d}L-\frac{1}{{\omega }_{d}C}\right)}$

The phase angle is,

$\varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$

(a) Calculation of the value of the capacitance at which average rate energy is dissipated in the resistance is maximum:

Average power is given by,

$P_{avg}=\frac{{\epsilon }_m^{2}R}{2Z^2}$

For maximum power, denominator should be minimum. In this case,

${\omega }_{d}L-\frac{1}{{\omega }_{d}C}=0$ ….. (1)

Therefore, the capacitance will be,

$$\begin{gathered} C=\frac{1}{{{\omega }_d}^{2}L} \\ =\frac{1}{{\left(2\pi f_d\right)}^{2}L} \\ =\frac{1}{{\left(2\times 3.14\times 60\right)}^2\times 0.06} \\ =1.17\times {10}^{-4}F \end{gathered}$$

Hence, the average rate at which energy is dissipated in the resistance would be maximum at $C=1.17\times {10}^{-4}F$.

(b) Calculation of the value of the capacitance at which average rate energy is dissipated in the resistance is minimum:

For minimum power, denominator must have a large value.

Thus, $Z^2$will be maximum when the capacitance C = 0.

Hence, the average rate at which energy is dissipated in the resistance would be minimum at C = 0 F .

(c) Calculation of the maximum dissipation rate:

Rewrite eqiuation (1) as below.

$$\begin{gathered} {\omega }_{d}L-\frac{1}{{\omega }_{d}C}=0 \\ {\omega }_{d}L=\frac{1}{{\omega }_{d}C} \end{gathered}$$

The impedance will becomes.

$$\begin{gathered} Z=\sqrt{R^2+\left({\omega }_{d}L-\frac{1}{{\omega }_{d}C}\right)} \\ =\sqrt{R^2+\left({\omega }_{d}L-{\omega }_{d}L\right)} \\ =R \end{gathered}$$

For maximum power,

$$\begin{gathered} P_{ang}=\frac{{\epsilon }_m^{2}R}{2Z^2} \\ {P}_{\max }=\frac{{\epsilon }_m^2}{2R} \\ =\frac{{\left(30\right)}^2}{2\times 5} \\ =90.0W \end{gathered}$$

Hence, the maximum dissipation rate is 90.0 W.

(d) Calculation of the phase angle corresponding to the maximum dissipation rate:

At maximum power,$X_L=X_C$

Hence, phase angle becomes,

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right) \\ =0 \end{gathered}$$

Therefore, the phase angle corresponding to the maximum dissipation rate is $0^{\circ }$.

(e) Calculation of the power factor corresponding to the maximum dissipation rate:

If $\varphi =0^{\circ }$, power factor becomes

Therefore, the power factor corresponding to the maximum dissipation rate is 1 .

(f) Calculation of the minimum dissipation rate:

Minimum dissipation rate:

$p_{min}=0$

So, the minimum dissipation rate is 0.

(g) Calculation of the phase angle corresponding to the minimum dissipation rate:

If power is minimum,$X_C=\frac{1}{\omega C}$is infinite, which gives

$$\begin{gathered} \varphi ={\tan }^{-1}\left(\frac{X_l-X_c}{R}\right) \\ ={\tan }^{-1}\left(-\infty \right) \\ =-90° \end{gathered}$$

Hence, the phase angle corresponding to the minimum dissipation rate is $\varphi =-90°$.

(h) Calculation of the power factor corresponding to the minimum dissipation rate:

If$\varphi =-90°$

power factor is

$$\begin{gathered} \cos \left(\varphi \right)=\cos \left(-{90}^0\right) \\ =0 \end{gathered}$$

Therefore, the power factor corresponding to the minimum dissipation rate is 0.