91Ó°ÊÓ

The resistance $R=5.0\mathrm{Î\copyright }$,

The inductor $L=60.0mH$,

The frequency $f_d=60.0Hz$,

The emf voltage ${\mathrm{Î\mu }}_m=30.0V$,

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:

$P_{avg}=\frac{{\mathrm{Î\mu }}_m^{2}R}{2Z^2}$

Here,the impedance is,

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

The phase angle is,

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

Average power is given by,

$P_{avg}=\frac{{\mathrm{Î\mu }}_m^{2}R}{2Z^2}$

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

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

Therefore, the capacitance will be,

$$\begin{gathered} C=\frac{1}{{{\mathrm{Ó\neg }}_d}^{2}L} \\ =\frac{1}{{\left(2\mathrm{Ï€}{f}_d\right)}^{2}L} \\ =\frac{1}{{\left(2×3.14×60\right)}^2×0.06} \\ =1.17×{10}^{-4}F \end{gathered}$$

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

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 .

Rewrite eqiuation (1) as below.

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

The impedance will becomes.

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

For maximum power,

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

Hence, the maximum dissipation rate is 90.0 W.

At maximum power,$X_L=X_C$

Hence, phase angle becomes,

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

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

If $\mathrm{ϕ}=0^{∘}$, power factor becomes

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

Minimum dissipation rate:

$p_{min}=0$

So, the minimum dissipation rate is 0.

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

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

Hence, the phase angle corresponding to the minimum dissipation rate is $\mathrm{ϕ}=-90°$.

If$\mathrm{ϕ}=-90°$

power factor is

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

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