Chapter 29, Problem 70P

(a)

To determine

The average power supplied to the circuit.

Answer to Problem 70P

  $P_{av}=933.38\text{W}$

Explanation of Solution

Given:

RMS voltage, $V_{rms}=120\text{V}$

Frequency, $f=60\text{Hz}$

RMS current, $I_{rms}=11\text{A}$

Phase difference between the current and the voltage, $\varphi ={45}^{\text{o}}$

Calculation:

The average power is

  $\begin{array}{l}{P}_{av}=V_{rms}{I}_{rms}\cos \varphi \\ P_{av}=120\text{V}\times \text{11A}\times {\text{cos 45}}^{\text{o}}\\ P_{av}=933.38\text{W}\end{array}$

Conclusion:

The average power supplied to the circuit is $P_{av}=933.38\text{W}$ .

(b)

To determine

The resistance of the circuit.

Answer to Problem 70P

  $R=7.71\Omega$

Explanation of Solution

Given:

RMS voltage, $V_{rms}=120\text{V}$

Frequency, $f=60\text{Hz}$

RMS current, $I_{rms}=11\text{A}$

Phase difference between the current and the voltage, $\varphi ={45}^{\text{o}}$

Average power, $P_{av}=933.38\text{W}$

Calculation:

The resistance is,

  $\begin{array}{l}{P}_{av}=I_{rms}{}^{2}R\\ R=\frac{P_{av}}{I_{rms}{}^2}\\ R=\frac{933.38\text{W}}{{\left(11\text{A}\right)}^2}\\ R=7.71\Omega \end{array}$

Conclusion:

  $R=7.71\Omega$

This is the resistance of the circuit.

(c)

To determine

The capacitance of the circuit.

Answer to Problem 70P

  $C=2.4\times {10}^{-4}\text{F}$

Explanation of Solution

Given:

Rms voltage, $V_{rms}=120\text{V}$

Frequency, $f=60\text{Hz}$

Rms current, $I_{rms}=11\text{A}$

Phase difference between the current and the voltage, $\varphi ={45}^{\text{o}}$

Average power, $P_{av}=933.38\text{W}$

Inductance, $L=50\text{mH}$

Resistance, $R=7.71\Omega$

Formula used:

Capacitive reactance, $X_C=\frac{1}{2\pi fC}$

Impedance, ${\text{Z}}^2=R^2+{\left(X_L-X_C\right)}^2=\frac{{\epsilon }^2}{I_{rms}{}^2}$

Calculation:

The capacitive impedance is,

  $\begin{array}{l}{\text{Z}}^2=R^2+{\left(X_L-X_C\right)}^2\\ \frac{{\epsilon }^2}{I_{rms}{}^2}=R^2+{\left(2\pi fL-X_C\right)}^2\\ {\left(\frac{120\text{V}}{11\text{A}}\right)}^2=7.71{\Omega }^2+{\left(2\pi \times 60\text{Hz}\times \text{50}\times {\text{10}}^{\text{-3}}-X_C\right)}^2\\ 119.01=59.44+{\left(18.85-X_C\right)}^2\\ 18.85-X_C=\sqrt{119.01-59.44}\\ X_C=18.85-7.71\\ X_C=11.14\Omega \end{array}$

The capacitance is derived as,

  $\begin{array}{l}C=\frac{1}{2\pi f{X}_C}\\ C=\frac{1}{2\pi \times 60\text{Hz}\times 11.14\Omega }\\ C=2.4\times {10}^{-4}\text{F}\end{array}$

Conclusion:

The capacitance is $C=2.4\times {10}^{-4}\text{F}$ .

(d)

To determine

The change in the capacitance to make the power factor equal to 1.

Answer to Problem 70P

  $\Delta C=1\times {10}^{-4}\text{F}$

Explanation of Solution

Given:

Rms voltage, $V_{rms}=120\text{V}$

Frequency, $f=60\text{Hz}$

Rms current, $I_{rms}=11\text{A}$

Phase difference between the current and the voltage, $\varphi ={45}^{\text{o}}$

Average power, $P_{av}=933.38\text{W}$

Inductance, $L=50\text{mH}$

Capacitance, $C=2.4\times {10}^{-4}\text{F}$

Formula used:

Capacitive reactance, $X_C=\frac{1}{2\pi fC}$

Frequency, $2\pi f=\frac{1}{\sqrt{LC}}$

Calculation:

The capacitance at power factor 1 is $C_{pf=1}$ .The frequency is independent on the power factor and the inductance is remaining constant as per the conditions. So, $C_{pf=1}$ is establish as,

  $\begin{array}{l}{C}_{P_{f=1}}=\frac{1}{{\left(2\pi f\right)}^{2}L}\\ C_{P_{f=1}}=\frac{1}{{\left(2\pi \times 60\text{Hz}\right)}^2\times 0.05\text{H}}\\ C_{P_{f=1}}=1.4\times {10}^{-4}\text{F}\end{array}$

The change in the capacitance for which the power factor is 1 is,

  $\begin{array}{l}\Delta C=C-C_{P_{f=1}}\\ \Delta C=\left(2.4\times {10}^{-4}-1.4\times {10}^{-4}\right)\text{F}\\ \Delta C=1\times {10}^{-4}\text{F}\end{array}$

Conclusion:

The capacitance must change $\Delta C=1\times {10}^{-4}\text{F}$

(e)

To determine

The change in inductance to make the power factor 1.

Answer to Problem 70P

  $\Delta L=0.02\text{H}$

Explanation of Solution

Given:

RMS voltage, $V_{rms}=120\text{V}$

Frequency, $f=60\text{Hz}$

RMS current, $I_{rms}=11\text{A}$

Phase difference between the current and the voltage, $\varphi ={45}^{\text{o}}$

Average power, $P_{av}=933.38\text{W}$

Inductance, $L=50\text{mH=0}\text{.05H}$

Capacitance, $C=2.4\times {10}^{-4}\text{F}$

Formula used:

Capacitive reactance, $X_C=\frac{1}{2\pi fC}$

Frequency, $2\pi f=\frac{1}{\sqrt{LC}}$

Calculation:

The capacitance at power factor 1 is $L_{pf=1}$ . The frequency is independent of the power factor and the capacitance is remains constant as per the conditions. So, $L_{pf=1}$ is established as:

  $\begin{array}{l}{L}_{P_{f=1}}=\frac{1}{{\left(2\pi f\right)}^{2}C}\\ L_{P_{f=1}}=\frac{1}{{\left(2\pi \times 60\text{Hz}\right)}^2\times 2.4\times {10}^{-4}\text{F}}\\ L_{P_{f=1}}=0.03\text{H}\end{array}$

The change in the inductance for which the power factor is 1 is

  $\begin{array}{l}\Delta L=L-L_{P_{f=1}}\\ \Delta L=\left(0.05-0.03\right)\text{H}\\ \Delta L=0.02\text{H}\end{array}$

Conclusion:

The change in inductance to make power factor 1 is $\Delta L=0.02\text{H}$ .