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

In summary, the conversation discusses a question from a past exam that involves using KVL and superposition to solve for unknown values in a circuit. The individual asking for help receives assistance in solving the problem and confirms their solution by checking the summation of real and apparent power. The final conclusion is that the voltage source does not provide any power in the circuit.
  • #1
asim1701
2
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:
 

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  • #2
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?
 
  • #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.
 

1. What is a "1 V source" and "1 I source" in an RL circuit?

A "1 V source" refers to a voltage source that has a potential difference of 1 volt, while a "1 I source" refers to a current source that has an intensity of 1 ampere. In an RL circuit, these sources are used to provide the electrical energy needed for the circuit to function.

2. How is power calculated in an RL circuit?

Power in an RL circuit can be calculated using the formula P = VI, where V is the voltage and I is the current. In this circuit, the voltage and current are dependent on each other and can change over time, so the power output will also vary.

3. What is the significance of the RL circuit in electrical engineering?

The RL circuit is a common type of circuit used in electrical engineering, as it allows for the control of current and voltage in a circuit. It is often used in filters, amplifiers, and other electronic devices.

4. How does the resistance (RL) affect the power in an RL circuit?

The resistance (RL) in an RL circuit affects the power output by influencing the amount of current that can flow through the circuit. A higher resistance will result in a lower current and therefore a lower power output.

5. Can the power output in an RL circuit be increased?

Yes, the power output in an RL circuit can be increased by either increasing the voltage or the current. This can be achieved by using a higher voltage or current source, or by using a transformer to step up the voltage or current.

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