Chapter 29, Problem 69P

(a)

To determine

RMS voltage between the points A and B.

Answer to Problem 69P

  $V_{AB}=80.344\text{V}$

Explanation of Solution

Given:

Rms voltage of the generator, $\epsilon =115\text{V}$

Frequency of the generator, $f=60\text{Hz}$

Inductance, $L=137\text{mH}$

Resistance, $R=50\Omega$

Capacitance, $C=25\text{μF}$

Formula used:

  $\begin{array}{l}{I}_{rms}=\frac{\epsilon }{Z}\\ Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ X_L=2\pi fL\\ X_C=\frac{1}{2\pi fC}\end{array}$

Calculation:

The rms voltage between points A and B is,

  $\begin{array}{l}{V}_{AB}=I_{rms}{X}_L\\ V_{AB}=\frac{\epsilon }{Z}{X}_L\\ V_{AB}=\frac{\epsilon }{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}{X}_L\\ V_{AB}=\frac{\epsilon }{\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}}2\pi fL\\ V_{AB}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(\left(2\pi \times 60\text{Hz}\times \text{137mH}\right)-\frac{1}{\left(2\pi \times 60\text{Hz}\times 25\text{μF}\right)}\right)}^2}}\left(2\pi \times 60\text{Hz}\times \text{137mH}\right)\\ V_{AB}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(51.648\Omega -106.10\Omega \right)}^2}}51.648\Omega \\ V_{AB}=80.344\text{V}\end{array}$

Conclusion:

The rms voltage through points A and B where the inductor is present is, $V_{AB}=80.344\text{V}$ .

(b)

To determine

RMS voltage between B and C.

Answer to Problem 69P

  $V_{BC}=77.78\text{V}$

Explanation of Solution

Given:

Rms voltage of the generator, $\epsilon =115\text{V}$

Frequency of the generator, $f=60\text{Hz}$

Inductance, $L=137\text{mH}$

Resistance, $R=50\Omega$

Capacitance, $C=25\text{μF}$

Formula used:

  $\begin{array}{l}{I}_{rms}=\frac{\epsilon }{Z}\\ Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ X_L=2\pi fL\\ X_C=\frac{1}{2\pi fC}\end{array}$

Calculation:

The rms voltage between points B and C is,

  $\begin{array}{l}{V}_{BC}=I_{rms}R\\ V_{BC}=\frac{\epsilon }{Z}R\\ V_{BC}=\frac{\epsilon }{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}R\\ V_{BC}=\frac{\epsilon }{\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}}R\\ V_{BC}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(\left(2\pi \times 60\text{Hz}\times \text{137mH}\right)-\frac{1}{\left(2\pi \times 60\text{Hz}\times 25\text{μF}\right)}\right)}^2}}50\Omega \\ V_{BC}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(51.648\Omega -106.10\Omega \right)}^2}}50\Omega \\ V_{BC}=77.78\text{V}\end{array}$

Conclusion:

Across the resistor in between the points B and C, the rms voltage is $V_{BC}=77.78\text{V}$ .

(c)

To determine

RMS voltage between points C and D.

Answer to Problem 69P

  $V_{CD}=165.05\text{V}$

Explanation of Solution

Given:

Rms voltage of the generator, $\epsilon =115\text{V}$

Frequency of the generator, $f=60\text{Hz}$

Inductance, $L=137\text{mH}$

Resistance, $R=50\Omega$

Capacitance, $C=25\text{μF}$

Formula used:

  $\begin{array}{l}{I}_{rms}=\frac{\epsilon }{Z}\\ Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ X_L=2\pi fL\\ X_C=\frac{1}{2\pi fC}\end{array}$

Calculation:

The rms voltage between points D and C is

  $\begin{array}{l}{V}_{CD}=I_{rms}{X}_C\\ V_{CD}=\frac{\epsilon }{Z}{X}_C\\ V_{CD}=\frac{\epsilon }{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}{X}_C\\ V_{CD}=\frac{\epsilon }{\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}}{X}_C\\ V_{CD}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(\left(2\pi \times 60\text{Hz}\times \text{137mH}\right)-\frac{1}{\left(2\pi \times 60\text{Hz}\times 25\text{μF}\right)}\right)}^2}}2\pi \times 60\text{Hz}\times 25\text{μF}\\ V_{CD}=\frac{115\text{V}}{\sqrt{{\left(50\Omega \right)}^2+{\left(51.648\Omega -106.10\Omega \right)}^2}}106.10\Omega \\ V_{CD}=165.05\text{V}\end{array}$

Conclusion:

Rms voltage across the capacitance is $V_{CD}=165.05\text{V}$ .

(d)

To determine

The rms voltage between the points A and C.

Answer to Problem 69P

  $V_{AC}=111.58\text{V}$

Explanation of Solution

Given:

Rms voltage of the generator, $\epsilon =115\text{V}$

Frequency of the generator, $f=60\text{Hz}$

Inductance, $L=137\text{mH}$

Resistance, $R=50\Omega$

Capacitance, $C=25\text{μF}$

Rms voltage between AB, $V_{AB}=80.344\text{V}$

Rms voltage between BC, $V_{BC}=77.78\text{V}$

Calculation:

The voltage across AC is calculated using the Pythagoras theorem as the voltage in the inductor leads the voltage in the resistor according to the phasor diagrams,

  $\begin{array}{c}{V}_{AC}=\sqrt{V_{AB}{}^2+V_{BC}{}^2}\\ =\sqrt{80{\text{V}}^2+77.78{\text{V}}^2}\\ =111.58\text{V}\end{array}$

Conclusion:

The voltage across AC is $V_{AC}=111.58\text{V}$ .

(e)

To determine

The rms voltage across BD.

Answer to Problem 69P

  $V_{BD}=182.46\text{V}$

Explanation of Solution

Given:

Rms voltage of the generator, $\epsilon =115\text{V}$

Frequency of the generator, $f=60\text{Hz}$

Inductance, $L=137\text{mH}$

Resistance, $R=50\Omega$

Capacitance, $C=25\text{μF}$

Rms voltage between AB, $V_{AB}=80.344\text{V}$

Rms voltage between DC, $V_{CD}=165.05\text{V}$

Calculation:

The voltage across CD is such that the voltage across the resistor leads the capacitor voltage according to the phasor diagram,

  $\begin{array}{c}{V}_{BD}=\sqrt{V_{CD}{}^2+V_{BC}{}^2}\\ V_{BD}=\sqrt{\text{165}{\text{.05V}}^2+77.78{\text{V}}^2}\\ V_{BD}=182.46\text{V}\end{array}$

Conclusion:

Across BD the rms voltage is $V_{BD}=182.46\text{V}$ .