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

AI Thread Summary
The discussion revolves around solving a circuit problem involving a resistor-inductor (RL) circuit with one voltage source and one current source. The user initially attempted to apply Kirchhoff's Voltage Law (KVL) but encountered two unknowns, leading to confusion. They then assumed the current through the circuit equals the current from the source and calculated the voltage of the current source to be approximately 482.25V. Using the root mean square (Irms) current and total impedance, they derived the apparent power as 1414W and determined the real power delivered to the resistor as 1.282kW, with reactive power to the inductor at 597.6 VAR. The conclusion drawn is that all power is supplied by the current source, with no contribution from the voltage source, which can be verified by checking the power balance in the circuit.
asim1701
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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:
 

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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.
 
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