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

1. Jun 10, 2007

### asim1701

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

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2. Jun 12, 2007

### asim1701

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.

Correct?

3. Jul 9, 2007

### l46kok

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