Solving Circuits with James: Finding I, Power, & Phase Angle

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SUMMARY

The discussion focuses on calculating the phasor current (I), power, reactive power, and apparent power in an electrical circuit involving a capacitor and a resistor. The user, James, initially replaced the capacitor with its complex impedance of -j300.3 ohms and calculated the total impedance as 186.48 ohms. However, his final current calculation of 10.08 A was incorrect, as the expected result was 15.11 ∠ 20.66 degrees. The key takeaway is the importance of treating capacitive reactance as a complex quantity during impedance calculations.

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  • Basic trigonometry for phase angle calculations
  • Understanding of reactive and apparent power
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James889
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Hi,

I have the following circuit
[PLAIN]http://img707.imageshack.us/img707/3024/upgp580.png

And i need to find the phasor current I, the power, reactive power and apparent power delivered by the source and state whether it's lagging or leading. But I am kind of stuck at the first step.

First i replace the capacitor with it's complex impedance
[tex]-j\frac{1}{333\cdot10\cdot10^{-6}} = -J300.3[/tex]

And that is parallel with the 115ohm resistor, yielding:

[tex]\frac{1}{(1/115)+(1/-300)} = 186.48\ohm[/tex]

And the phase angle

[tex]arctan\bigg(\frac{1/-300}{1/115}\bigg)= -20.93[/tex]

So the current I is given by
[tex]\frac{1585\sqrt{2}}{186.48} \cdot cos(12-(-21)) = 10.08[/tex]

Which is as usual incorrect.
According to the key the current I is [tex]15.11 \angle 20.66[/tex]

Please help
James
 
Last edited by a moderator:
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You should consider capacitive reactance as a complex quantity while calculating total impedance.
 

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