Chapter 21, Problem 60P

(a)

To determine

The value of the unknown capacitor.

Answer to Problem 60P

The value of the unknown capacitor is $1.7\text{μF}$.

Explanation of Solution

Write the expression for phase angle.

  $\tan \varphi =\frac{X_L-X_C}{R}$                                                                      (I)

Here, $\varphi$ is the phase angle, $X_L$ is the inductive reactance, $X_C$ is the capacitive reactance, and $R$ is the resistance.

Write the expression for capacitive reactance.

  $X_C=\frac{1}{\omega C}$                                                                              (II)

Here, $\omega$ is the angular frequency and $C$ is the capacitance.

Write the expression for inductive reactance.

  $X_L=\omega L$                                                                               (III)

Here,  is the angular frequency and $L$ is the inductance

Conclusion:

Substitute the equation (III) and (II) in the equation (I).

  $\begin{array}{c}\tan \varphi =\frac{\omega L-\frac{1}{\omega C}}{R}\\ R\tan \varphi -\omega L=-\frac{1}{\omega C}\end{array}$

Substitute $2\pi f$ for $\omega$ and rearrange the above equation for capacitance.

  $\begin{array}{c}\omega C=\frac{1}{\omega L-R\tan \varphi }\\ C=\frac{1}{{\omega }^{2}L-\omega R\tan \varphi }\\ =\frac{1}{4{\pi }^2{f}^{2}L-2\pi fR\tan \varphi }\end{array}$

Here, $f$ is the frequency of the ac source.

Substitute $440.0\text{Hz}$ for $f$, $0.750\text{H}$ for $L$, $4.00\text{k}\Omega$ for $R$, and      for $\varphi$ to find $C$.

  $\begin{array}{c}C=\frac{1}{4{\left(3.14\right)}^2{\left(440.0\text{Hz}\right)}^2\left(0.750\text{H}\right)-2\left(3.14\right)\left(440.0\text{Hz}\right)\left(4.00\text{k}\Omega \right)\left(\frac{{10}^3\Omega }{1\text{k}\Omega }\right)\tan 25.0°}\\ =1.7\times {10}^{-6}\text{F}\left(\frac{{10}^6\text{μF}}{1\text{F}}\right)\\ =1.7\text{μF}\end{array}$

Therefore, the value of the unknown capacitor is $1.7\text{μF}$.

(b)

To determine

If person F should connect a second capacitor in parallel across the first capacitor or in series in the circuit.

Answer to Problem 60P

Person F needs to add another capacitor in series to reduce the equivalent capacitance.

Explanation of Solution

Write the expression for the power proportional to the cosine of the phase angle.

  $\cos \varphi =\frac{R}{Z}$                                                                            (IV)

Here, $R$ is the resistance and $Z$ is the impedance.

The impedance of the $RLC$ circuit is.

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

Since $X_L-X_C$ must be made as small as possible.

Conclusion:

Substitute $2\pi f$ for $\omega$ in the equation (II).

  $X_C=\frac{1}{2\pi fC}$                                                                          (VI)

Substitute $2\pi f$ for $\omega$ in the equation (III).

  $X_L=2\pi fL$                                                                           (VII)

Substitute $440.0\text{Hz}$ for $f$ and $1.7\text{μF}$ for $C$ in equation (VI).

  $\begin{array}{c}{X}_C=\frac{1}{2\left(3.14\right)\left(440.0\text{Hz}\right)\left(1.7\text{μF}\right)\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)}\\ =0.213\times {10}^3\Omega \left(\frac{{10}^{-3}\text{k}\Omega }{1\Omega }\right)\\ =0.2\text{k}\Omega \end{array}$

Substitute $440.0\text{Hz}$ for $f$ and $0.750\text{H}$ for $L$ in equation (VI).

  $\begin{array}{c}{X}_L=2\left(3.14\right)\left(440.0\text{Hz}\right)\left(0.750\text{H}\right)\\ =2.1\times {10}^3\Omega \left(\frac{{10}^{-3}\text{k}\Omega }{1\Omega }\right)\\ =2.1\text{k}\Omega \end{array}$

Since the capacitive reactance must be increased. Thus,$X_C\propto \frac{1}{C}$, the capacitance must be increased to decrease the capacitive reactance. By placing a capacitor in series will decrease the equivalent capacitance of the circuit.

The capacitive reactance is too small, so person F needs to add another capacitor in series to reduce the equivalent capacitance, thereby increasing the capacitive reactance.

(c)

To determine

The value of the capacitor requited to get maximum power.

Answer to Problem 60P

The value of the capacitor requited to get maximum power is $0.19\text{μF}$.

Explanation of Solution

For the maximum power, the capacitive reactance should be equal to the inductive reactance. Since $X_L=X_C$.

The required capacitance of the additional capacitor is.

  $\omega L=\frac{1}{\omega C_{\text{equ}}}$

Here, $C_{\text{equ}}$ is the equivalent capacitor.

Replace $1/C_{\text{equ}}$ for $1/C+1/C_2$ and rearrange the above equation for $C_2$.

  $\begin{array}{c}\omega L=\frac{1}{\omega }\left(\frac{1}{C}+\frac{1}{C_2}\right)\\ {\omega }^{2}L=\left(\frac{1}{C}+\frac{1}{C_2}\right)\\ C_2={\left({\omega }^{2}L-\frac{1}{C}\right)}^{-1}\end{array}$

Here, $C_2$ is the additional capacitor.

Conclusion:

Substitute $2\pi f$ for $\omega$ in the above equation.

  $C_2={\left(4{\pi }^2{f}^{2}L-\frac{1}{C}\right)}^{-1}$

Substitute $440.0\text{Hz}$ for $f$, $0.750\text{H}$ for $L$, and $1.7\text{μF}$ for $C$ to find $C_2$.

  $\begin{array}{c}{C}_2={\left(4{\left(3.14\right)}^2{\left(440.0\text{Hz}\right)}^2\left(0.750\text{H}\right)-\frac{1}{1.7\text{μF}\times \frac{{10}^{-6}\text{F}}{1\text{μF}}}\right)}^{-1}\\ =1.9\times {10}^{-7}\text{F}\\ =0.19\times {10}^{-6}\text{F}\left(\frac{{10}^6\text{μF}}{1\text{F}}\right)\\ =0.19\text{μF}\end{array}$

Therefore, the maximum power needed for the value of capacitor is $0.19\text{μF}$.