1 V source, 1 I source, RL circuit, find power

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 5K views
asim1701
Messages
2
Reaction score
0

Homework Statement




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, I am so confused:confused::confused::confused:
 

Attachments

  • elecq.jpg
    elecq.jpg
    16.1 KB · Views: 602
Physics news on Phys.org
ok I've 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.

Correct?
 
Well you can check the solution by summing up all the forces since summation of real and apparent power must be equal to 0.