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A load consists of a parallel connection of paths as shon below. R1−L1,R2−L2, and R3−C1R _ { 1 } - L _ { 1 } , R _ { 2 } - L _ { 2 } \text {, and } R _ { 3 } - C _ { 1 } Assume R1=60Ω,R2=55Ω,R3=50Ω,L1=55mH,L2=75mH,C1=200μF,∣Vrms∣=105\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.

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\[\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 Vrms=530∠60∘V\mathbf { V } _ { \mathrm { rms } } = 530 \angle 60 ^ { \circ } \mathrm { V } Irms \mathbf { I } _ { \text {rms } } . Find the coplex power, apparent power, average power, reactive power, and =8.5∠30∘A= 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, and S.f=60 HzP , Q , \text { and } S . f = 60 \mathrm {~Hz}

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\[\begin{array} { l }
\mathbf { S } = \...

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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)=140cos⁡(2π60t+30∘)V,i(t)=12cos⁡(2π60t+55∘)Av ( 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 }

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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 \[| \mathrm { S } | = 840 \mathrm { VA }\] 11ecc078_8cb0_107f_9ec6_ab37c162ecc5_TB8726_11 \(\begin{array}{l} \mathbf{Z}=\mathbf{V} / \mathbf{I}=10.5736-\mathrm{j} 4.93055 \Omega \\ \mathrm{R}=10.5736 \Omega \\ \mathrm{X}=-4.93055 \Omega \\ \mathrm{C}=-1 /(\omega \mathrm{X})=537.98954 \mathrm{uF} \end{array}\)

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)=150cos⁡(2π60t+120∘)V,i(t)=10cos⁡(2π60t+85∘)Av ( 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 }

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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 \[| \mathrm { S } | = 750 \mathrm { VA }\] 11ecc078_82ce_283e_9ec6_a7536bd224ee_TB8726_11 \[\begin{array} { l } \mathbf { Z } = \mathbf { V } / \mathbf { I } = 12.2873 + \mathrm { j } 8.6036 \Omega \\ \mathrm { R } = 12.2873 \Omega \\ \mathrm { X } = 8.6036 \Omega \\ \mathrm { L } = \mathrm { X } / \omega = \mathrm { X } / ( 2 \pi \mathrm { f } ) = 22.8219 \mathrm { mH } \end{array}\]

A load consists of a parallel connection of paths as shon below. R1−L1,R2−L2, and R3−C1\mathrm { R } _ { 1 } - \mathrm { L } _ { 1 } , \mathrm { R } _ { 2 } - \mathrm { L } _ { 2 } \text {, and } \mathrm { R } _ { 3 } - \mathrm { C } _ { 1 } Assume , R1=80Ω,R2=95Ω,R3=75Ω,L1=65mH,L2=85mH,C1=250μF,∣Irms∣=20 A\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.

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\[\begin{array} { l }
\mathrm { Z } _ {...

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