Been trying to get out two matching power factors (p.f) from the following circuit Calculation,
What am I doing wrong? Been working on it for 8 hours in total. Used matlab to minimize my error and so on. But I keep getting a differnece...
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\$\begingroup\$ The frequency is 60 Hz. \$\endgroup\$– Slowpoke-coderCommented Jan 24, 2018 at 5:17
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\$\begingroup\$ For \$60\: \text{Hz}\$, I get \$100\:\Omega\mid\mid\left(5\:\Omega+10\:\text{mH}\right) = 4.88451729 + 3.41501831j= 5.95993785\: \angle 34.959385^\circ\$. \$\endgroup\$– jonkCommented Jan 24, 2018 at 5:30
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\$\begingroup\$ Even for such a small differnece the end result between the two p.f dosen't match unfourtunetly. \$\endgroup\$– Slowpoke-coderCommented Jan 24, 2018 at 5:50
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1\$\begingroup\$ Please rotate your photos. \$\endgroup\$– DanielCommented Jan 24, 2018 at 5:55
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\$\begingroup\$ @Zeclot What is the power and power factor for \$R_1\$? (It should be obvious.) The voltage across \$R_1\$ is the same as the voltage across the \$R_2+L\$ leg. But the voltage across both these two legs is \$120.0\:\text{V}_\text{RMS}\:\angle0^\circ\$, isn't it? \$\endgroup\$– jonkCommented Jan 24, 2018 at 6:48
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1 Answer
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I solved the problem. I used P=VI instead of VI' ('= conjugate). So the angle became wrong and gave extra errors.
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\$\begingroup\$ You can edit your question. Please delete this because it is not answer and edit your question with these pics. \$\endgroup\$– HazemCommented Jan 24, 2018 at 10:54
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\$\begingroup\$ How is this an "Answer" to your question? \$\endgroup\$– TylerCommented Jan 24, 2018 at 13:15
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\$\begingroup\$ I did a comment on the problem. But when I added the new pictures it turned in to an answer. But solved the problem. Had a conjugate problem when calculating the power \$\endgroup\$ Commented Jan 25, 2018 at 23:04