Exam 11: AC Power

Given the following voltage and current on the load, find the power factor, average power, reactive power, complex power, and apparent power. Also, find the equivalent impedance, equivalent resistance, and equivalent inductance of the load. Draw the phasors for P, Q, and S. The voltage and current are peak values. $v ( t ) = 140 \cos \left( 2 \pi 60 t + 30 ^ { \circ } \right) \mathrm { V } , i ( t ) = 12 \cos \left( 2 \pi 60 t + 55 ^ { \circ } \right) \mathrm { A }$

Free

(Essay)

4.8/5

(34)

Question 1

Correct Answer:

Verified

The difference in phase angle is
$\theta = \theta _ { \mathrm { v } } - \theta _ { \mathrm { i } } = 30 ^ { \circ } - 55 ^ { \circ } = - 25 ^ { \circ }$
The power factor is given by
$\mathrm { pf } = \cos ( \theta ) = \cos \left( - 25 ^ { \circ } \right) = 0.906308$
The average power is
$P = \frac { V _ { m } I _ { m } } { 2 } p f = 761.29854 \mathrm {~W}$
The reactive power is
$Q = \frac { V _ { m } I _ { m } } { 2 } \sin ( \theta ) = - 354.99934 \mathrm { VAR }$
The complex power is given by
$\mathbf { S } = \mathrm { P } + \mathrm { jQ } = 761.29854 - \mathrm { j } 354.99934 = 840 \angle - 25 ^ { \circ } \mathrm { VA }$
The apparent power is given by
$| \mathrm { S } | = 840 \mathrm { VA }$ $The difference in phase angle is \theta = \theta _ { \mathrm { v } } - \theta _ { \mathrm { i } } = 30 ^ { \circ } - 55 ^ { \circ } = - 25 ^ { \circ } The power factor is given by \mathrm { pf } = \cos ( \theta ) = \cos \left( - 25 ^ { \circ } \right) = 0.906308 The average power is P = \frac { V _ { m } I _ { m } } { 2 } p f = 761.29854 \mathrm {~W} The reactive power is Q = \frac { V _ { m } I _ { m } } { 2 } \sin ( \theta ) = - 354.99934 \mathrm { VAR } The complex power is given by \mathbf { S } = \mathrm { P } + \mathrm { jQ } = 761.29854 - \mathrm { j } 354.99934 = 840 \angle - 25 ^ { \circ } \mathrm { VA } The apparent power is given by | \mathrm { S } | = 840 \mathrm { VA } \begin{array}{l} \mathbf{Z}=\mathbf{V} / \mathbf{I}=10.5736-\mathrm{j} 4.93055 \Omega \\ \mathrm{R}=10.5736 \Omega \\ \mathrm{X}=-4.93055 \Omega \\ \mathrm{C}=-1 /(\omega \mathrm{X})=537.98954 \mathrm{uF} \end{array}$
$\begin{array}{l}\mathbf{Z}=\mathbf{V} / \mathbf{I}=10.5736-\mathrm{j} 4.93055 \Omega \\\mathrm{R}=10.5736 \Omega \\\mathrm{X}=-4.93055 \Omega \\\mathrm{C}=-1 /(\omega \mathrm{X})=537.98954 \mathrm{uF}\end{array}$

Given the following voltage and current on the load, find the power factor, average power, reactive power, complex power, and apparent power. Also, find the equivalent impedance, equivalent resistance, and equivalent inductance of the load. Draw the phasors for P, Q, and S. The voltage and current are peak values. $v ( t ) = 150 \cos \left( 2 \pi 60 t + 120 ^ { \circ } \right) \mathrm { V } , i ( t ) = 10 \cos \left( 2 \pi 60 t + 85 ^ { \circ } \right) \mathrm { A }$

Free

(Essay)

4.9/5

(35)

Question 2

Correct Answer:

Verified

The difference in phase angle is
$\theta = \theta _ { \mathrm { v } } - \theta _ { \mathrm { i } } = 120 ^ { \circ } - 85 ^ { \circ } = 35 ^ { \circ }$
The power factor is given by
$\text { pf } = \cos ( \theta ) = \cos \left( 35 ^ { \circ } \right) = 0.819152$
The average power is
$P = \frac { V _ { m } I _ { m } } { 2 } p f = \frac { 150 \times 10 } { 2 } \times 0.819152 = 614.364 \mathrm {~W}$
The reactive power is
$Q = \frac { V _ { m } I _ { m } } { 2 } \sin ( \theta ) = \frac { 150 \times 10 } { 2 } \sin \left( 35 ^ { \circ } \right) = 430.1823 \mathrm { VAR }$
The complex power is given by
$\mathbf { S } = \mathrm { P } + \mathrm { jQ } = 614.364 + \mathrm { j } 430.1823 = 750 \angle 35 ^ { \circ } \mathrm { VA }$
The apparent power is given by
$| \mathrm { S } | = 750 \mathrm { VA }$
$The difference in phase angle is \theta = \theta _ { \mathrm { v } } - \theta _ { \mathrm { i } } = 120 ^ { \circ } - 85 ^ { \circ } = 35 ^ { \circ } The power factor is given by \text { pf } = \cos ( \theta ) = \cos \left( 35 ^ { \circ } \right) = 0.819152 The average power is P = \frac { V _ { m } I _ { m } } { 2 } p f = \frac { 150 \times 10 } { 2 } \times 0.819152 = 614.364 \mathrm {~W} The reactive power is Q = \frac { V _ { m } I _ { m } } { 2 } \sin ( \theta ) = \frac { 150 \times 10 } { 2 } \sin \left( 35 ^ { \circ } \right) = 430.1823 \mathrm { VAR } The complex power is given by \mathbf { S } = \mathrm { P } + \mathrm { jQ } = 614.364 + \mathrm { j } 430.1823 = 750 \angle 35 ^ { \circ } \mathrm { VA } The apparent power is given by | \mathrm { S } | = 750 \mathrm { VA } \begin{array} { l } \mathbf { Z } = \mathbf { V } / \mathbf { I } = 12.2873 + \mathrm { j } 8.6036 \Omega \\ \mathrm { R } = 12.2873 \Omega \\ \mathrm { X } = 8.6036 \Omega \\ \mathrm { L } = \mathrm { X } / \omega = \mathrm { X } / ( 2 \pi \mathrm { f } ) = 22.8219 \mathrm { mH } \end{array}$
$\begin{array} { l } \mathbf { Z } = \mathbf { V } / \mathbf { I } = 12.2873 + \mathrm { j } 8.6036 \Omega \\\mathrm { R } = 12.2873 \Omega \\\mathrm { X } = 8.6036 \Omega \\\mathrm { L } = \mathrm { X } / \omega = \mathrm { X } / ( 2 \pi \mathrm { f } ) = 22.8219 \mathrm { mH }\end{array}$

A load consists of a parallel connection of paths as shon below. $R _ { 1 } - L _ { 1 } , R _ { 2 } - L _ { 2 } \text {, and } R _ { 3 } - C _ { 1 }$ Assume $\mathrm { R } _ { 1 } = 60 \Omega , \mathrm { R } _ { 2 } = 55 \Omega , \mathrm { R } _ { 3 } = 50 \Omega , \mathrm { L } _ { 1 } = 55 \mathrm { mH } , \mathrm { L } _ { 2 } = 75 \mathrm { mH } , \mathrm { C } _ { 1 } = 200 \mu \mathrm { F } , \left| \mathbf { V } _ { \mathrm { rms } } \right| = 105$ V, f = 60 Hz. Find the complex power, average power, reactive power, apparent power, and power factor of each path. Also, find the complex power, average power, reactive power, apparent power, and power factor of the entire load. $A load consists of a parallel connection of paths as shon below. R _ { 1 } - L _ { 1 } , R _ { 2 } - L _ { 2 } \text {, and } R _ { 3 } - C _ { 1 } Assume \mathrm { R } _ { 1 } = 60 \Omega , \mathrm { R } _ { 2 } = 55 \Omega , \mathrm { R } _ { 3 } = 50 \Omega , \mathrm { L } _ { 1 } = 55 \mathrm { mH } , \mathrm { L } _ { 2 } = 75 \mathrm { mH } , \mathrm { C } _ { 1 } = 200 \mu \mathrm { F } , \left| \mathbf { V } _ { \mathrm { rms } } \right| = 105 V, f = 60 Hz. Find the complex power, average power, reactive power, apparent power, and power factor of each path. Also, find the complex power, average power, reactive power, apparent power, and power factor of the entire load.$

Free

(Essay)

4.8/5

(40)

Question 3

Correct Answer:

Verified

$\begin{array} { l } \mathrm { Z } _ { \mathrm { L } 1 } = \mathrm { j } \omega \mathrm { L } _ { 1 } = \mathrm { j } 20.7345 \Omega , \mathrm { Z } _ { \mathrm { L } 2 } = \mathrm { j } \omega \mathrm { L } _ { 2 } = \mathrm { j } 28.2743 \Omega , \mathrm { Z } _ { \mathrm { C } 1 } = 1 / \left( \mathrm { j } \omega \mathrm { C } _ { 1 } \right) = - \mathrm { j } 13.2629 \Omega \\\mathbf { Z } _ { 1 } = \mathrm { R } _ { 1 } + \mathrm { Z } _ { \mathrm { L } 1 } = 60 + \mathrm { j } 20.7345 \Omega \\\mathbf { Z } _ { 2 } = \mathrm { R } _ { 2 } + \mathrm { Z } _ { \mathrm { L } 2 } = 55 + \mathrm { j } 28.2743 \Omega \\\mathbf { Z } _ { 3 } = \mathrm { R } _ { 3 } + \mathrm { Z } _ { \mathrm { C } 1 } = 50 - \mathrm { j } 13.2629 \Omega \\\mathbf { S } _ { 1 } = \frac { \left| \boldsymbol { V } _ { \text {rus } } \right| ^ { 2 } \mathbf { Z } _ { 1 } } { \left| \mathbf { Z } _ { 1 } \right| ^ { 2 } } = 164.14718 + \mathrm { j } 56.7251934 = 173.67221 \angle 19.06388 ^ { \circ } \mathrm { VA } \\\mathbf { P } _ { 1 } = 164.14718 \mathrm {~W} , \mathrm { Q } _ { 1 } = 56.7251934 \mathrm { VAR } , \left| \mathbf { S } _ { 1 } \right| = 173.67221 \mathrm { VA } , \mathrm { pf } \mathrm { f } _ { 1 } = \cos \left( 19.06388 ^ { \circ } \right) = 0.9452 \\\mathbf { S } _ { 2 } = \frac { \left| \boldsymbol { V } _ { \text {rms } } \right| ^ { 2 } \mathbf { Z } _ { 2 } } { \left| \mathbf { Z } _ { 2 } \right| ^ { 2 } } = 158.5527 + \mathrm { j } 81.5086 = 178.2768 \angle 27.2067 ^ { \circ } \mathrm { VA } \\\mathrm { P } _ { 2 } = 158.5527 \mathrm {~W} , \mathrm { Q } _ { 2 } = 81.5086 \mathrm { VAR } , \left| \mathbf { S } _ { 2 } \right| = 178.2768 \mathrm { VA } , \mathrm { pf } _ { 2 } = \cos \left( 27.2067 ^ { \circ } \right) = 0.8894 \\\mathbf { S } _ { 3 } = \frac { \left| \boldsymbol { V } _ { \text {rus } } \right| ^ { 2 } \mathbf { Z } _ { 3 } } { \left| \mathbf { Z } _ { 3 } \right| ^ { 2 } } = 206.0051 - \mathrm { j } 54.6445 = 213.1294 \angle - 14.8561 ^ { \circ } \mathrm { VA } \\\mathbf { P } _ { 3 } = 206.0051 \mathrm {~W} , \mathrm { Q } _ { 3 } = - 54.6445 \mathrm { VAR } , \left| \mathbf { S } _ { 3 } \right| = 213.1294 \mathrm { VA } , \mathrm { pf } _ { 3 } = \cos \left( - 14.8561 ^ { \circ } \right) = 0.9666 \\\mathbf { S } = \mathbf { S } _ { 1 } + \mathbf { S } _ { 2 } + \mathbf { S } _ { 3 } = 528.705 + \mathrm { j } 83.5892 = 535.272 \angle 8.9842 ^ { \circ } \mathrm { VA } \\\mathbf { P } _ { 2 } = 705 \mathrm {~W} , \mathrm { Q } = 83.5892 \mathrm { VAR } , | \mathbf { S } | = 535.272 \mathrm { VA } , \mathrm { pf } ^ { 2 } = \cos \left( 8.9842 ^ { \circ } \right) = 0.9877\end{array}$ Alternate method for finding S: $\begin{array} { l } \mathrm { Z } _ { \mathrm { a } } = \mathrm { Z } _ { 1 } \left\| \mathrm { Z } _ { 2 } \right\| \mathrm { Z } _ { 3 } = 20.3443 + \mathrm { j } 3.2165 \Omega \\\mathbf { S } _ { a } = \frac { \left| \boldsymbol { V } _ { r m s } \right| ^ { 2 } \mathbf { Z } _ { a } } { \left| \mathbf { Z } _ { a } \right| ^ { 2 } } = 528.705 + \mathrm { j } 83.5892 = 535.272 \angle 8.9842 ^ { \circ } \mathrm { VA }\end{array}$

Let the voltage across a load be $\mathbf { V } _ { \mathrm { rms } } = 530 \angle 60 ^ { \circ } \mathrm { V }$ $\mathbf { I } _ { \text {rms } }$ . Find the coplex power, apparent power, average power, reactive power, and $= 8.5 \angle 30 ^ { \circ } \mathrm { A }$ power factor. Also, find the equivalent impedance, equivalent resistance, and equivalent inductance of the load. Draw the phasors for $P , Q , \text { and } S . f = 60 \mathrm {~Hz}$

(Essay)

5.0/5

(37)

Question 4

A load consists of a parallel connection of paths as shon below. $\mathrm { R } _ { 1 } - \mathrm { L } _ { 1 } , \mathrm { R } _ { 2 } - \mathrm { L } _ { 2 } \text {, and } \mathrm { R } _ { 3 } - \mathrm { C } _ { 1 }$ Assume , $\mathrm { R } _ { 1 } = 80 \Omega , \mathrm { R } _ { 2 } = 95 \Omega , \mathrm { R } _ { 3 } = 75 \Omega , \mathrm { L } _ { 1 } = 65 \mathrm { mH } , \mathrm { L } _ { 2 } = 85 \mathrm { mH } , \mathrm { C } _ { 1 } = 250 \mu \mathrm { F } , \left| \mathbf { I } _ { \mathrm { rms } } \right| = 20 \mathrm {~A}$ = 60 Hz. Find the complex power, average power, reactive power, apparent power, and power factor of each path. Also, find the complex power, average power, reactive power, apparent power, and power factor of the entire load. $A load consists of a parallel connection of paths as shon below. \mathrm { R } _ { 1 } - \mathrm { L } _ { 1 } , \mathrm { R } _ { 2 } - \mathrm { L } _ { 2 } \text {, and } \mathrm { R } _ { 3 } - \mathrm { C } _ { 1 } Assume , \mathrm { R } _ { 1 } = 80 \Omega , \mathrm { R } _ { 2 } = 95 \Omega , \mathrm { R } _ { 3 } = 75 \Omega , \mathrm { L } _ { 1 } = 65 \mathrm { mH } , \mathrm { L } _ { 2 } = 85 \mathrm { mH } , \mathrm { C } _ { 1 } = 250 \mu \mathrm { F } , \left| \mathbf { I } _ { \mathrm { rms } } \right| = 20 \mathrm {~A} = 60 Hz. Find the complex power, average power, reactive power, apparent power, and power factor of each path. Also, find the complex power, average power, reactive power, apparent power, and power factor of the entire load.$

(Essay)

4.8/5

(29)

Question 5

Showing 1 - 5 of 5

Voltage, Current, Power, and Sources

Circuit Laws

Circuit Analysis Methods

Circuit Theorems

Operational Amplifier Circuits

Capacitors and Inductors

RL and RC Circuits

RLC Circuits

Phasors and Impedances

Analysis of Phasor Transformed Circuits

Three-Phase Systems

Magnetically Coupled Circuits

The Laplace Transform

Circuits Analysis in the S-Domain

Filters