Exam 11: AC Power

\[\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: \[\begin{array} { l } \mathrm { Z } _ { \mathrm { a } } = \mathrm { Z } _ { 1 } \left\| \mathrm { Z } _ { 2 } \right\| \mathrm { Z } _ { 3 } = 20.3443 + \mathrm { j } 3.2165 \Omega \\ \mathbf { S } _ { a } = \frac { \left| \boldsymbol { V } _ { r m s } \right| ^ { 2 } \mathbf { Z } _ { a } } { \left| \mathbf { Z } _ { a } \right| ^ { 2 } } = 528.705 + \mathrm { j } 83.5892 = 535.272 \angle 8.9842 ^ { \circ } \mathrm { VA } \end{array}\]

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 \(\begin{array}{l} \mathbf{Z}=\mathbf{V} / \mathbf{I}=10.5736-\mathrm{j} 4.93055 \Omega \\ \mathrm{R}=10.5736 \Omega \\ \mathrm{X}=-4.93055 \Omega \\ \mathrm{C}=-1 /(\omega \mathrm{X})=537.98954 \mathrm{uF} \end{array}\)