Actual direction of phasor current - Power Calculation - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-07-23T09:46:04Z https://electronics.stackexchange.com/feeds/question/421407 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://electronics.stackexchange.com/q/421407 0 Actual direction of phasor current - Power Calculation Tal J https://electronics.stackexchange.com/users/189653 2019-02-09T19:25:45Z 2019-02-09T19:25:45Z <p>I am trying to calculate the complex power in this circuit. I used nodal analysis to find Vc, and the voltage at the top of the 2H inductor (=3.13&lt;177.8 V_RMS). I did the math for all elements in the circuit, my problem is with the voltage source. Voltage and Current are in phasor form, in RMS.</p> <p>After calculating the complex power of the voltage source using the current direction as in I1, I don't get the correct answer (I summed up all the real and the reactive powers from the elements and the current source and they don't compare).</p> <p>However, if I swap the direction of the current (I just added a minus sign) then the complex power turns out correct.</p> <p>I understand that when using nodal analysis we can arbitrarily specify the current direction, and after doing the math we will find the 'actual' direction of the current - this was easy while doing DC calculations. But now that the current is in complex form (for example 6.48&lt;-78.7 A_RMS) how do I know it's direction? This is critical in calculating the complex power correctly.</p> <p>I asked in class if the current through a voltage-source will always be away from the positive end (as the voltage source pushes power into the circuit) yet my teacher answered that this is not always true, as a powerful source can force another source to accept current through it's positive end.</p> <p>So how can I find the actual direction of current in this and other cases?</p> <p><a href="https://i.stack.imgur.com/xHWlD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xHWlD.png" alt="circuit"></a></p>