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.

Question

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.

Quizplus · Accepted Answer

The Answer is

\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 } ight) = - \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 } _ { ext {rus } } ight| ^ { 2 } \mathbf { Z } _ { 1 } } { \left| \mathbf { Z } _ { 1 } ight| ^ { 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 } ight| = 173.67221 \mathrm { VA } , \mathrm { pf } \mathrm { f } _ { 1 } = \cos \left( 19.06388 ^ { \circ } ight) = 0.9452 \\mathbf { S } _ { 2 } = \frac { \left| \boldsymbol { V } _ { ext {rms } } ight| ^ { 2 } \mathbf { Z } _ { 2 } } { \left| \mathbf { Z } _ { 2 } ight| ^ { 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 } ight| = 178.2768 \mathrm { VA } , \mathrm { pf } _ { 2 } = \cos \left( 27.2067 ^ { \circ } ight) = 0.8894 \\mathbf { S } _ { 3 } = \frac { \left| \boldsymbol { V } _ { ext {rus } } ight| ^ { 2 } \mathbf { Z } _ { 3 } } { \left| \mathbf { Z } _ { 3 } ight| ^ { 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 } ight| = 213.1294 \mathrm { VA } , \mathrm { pf } _ { 3 } = \cos \left( - 14.8561 ^ { \circ } ight) = 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 } ight) = 0.9877\end{array}

Alternate method for finding S:

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

A Load Consists of a Parallel Connection of Paths as Shon