|Jun10-07, 08:45 AM||#1|
1 V source, 1 I source, RL circuit, find power
1. The problem statement, all variables and given/known data
Hey guys, this is a question from one of the past exams. I've got my exams coming up and I was looking at this question, and I couldn't figure out how to start. I initially used KVL, but I then got two unknowns, the Voltage of the current source and the current of the voltage source.
Could anybody give me a jump start on this question, im so confused
|Jun12-07, 02:34 AM||#2|
ok ive tried working this out, but not with superposition. I assume that the current flowing THROUGHOUT the circuit to be the same as the current source, ie 14.14A(angle 170). Then,
(Voltage of Voltage Source - Voltage of Current Source)/Total Impedance = A cuurent of 14.14A
Solving for this, I get Vc (Voltage of Current Source) as 482.25V (approx. no angle).
Next, I use Irms^2Z to find the apparent power delivered to the Impedance. Irms=10A(angle 170) and Z=14.14(angle 45)
I get apparent power S = 1414W(angle of 25)
then from this information, we get that real power delivered (ie power to the resistor) = 1.282kW, and Reactive power to Inductor = 597.6 VAR.
but since all of the current was provided by the current source, that means it provides a real power of 1.282kW and a reactive power of 597.6VAR. Therefore, no power is provided by the Voltage source.
|Jul9-07, 10:04 PM||#3|
Well you can check the solution by summing up all the forces since summation of real and apparent power must be equal to 0.
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