Solving Complex Power with a 400-V, 100-Hz Source and a Load of 400+j30 VA

In summary, the problem involves a 400-V, 100-Hz power source supplying power to a load that absorbs 400+j30 VA of power. The magnitude of current drawn by the load can be found by taking the square root of the average power squared plus the reactive power squared. The apparent power absorbed by the load can be found by taking the square root of the average power squared plus the reactive power squared. The power factor of the load can be found by dividing the average power by the apparent power. The impedance of the load is not given and cannot be determined without additional information. A power triangle can be sketched and labeled to represent the relationship between average power, reactive power, and apparent power.
  • #1
Khesahn
5
0

Homework Statement



1. A 400-V, 100-Hz source supplies power to a load. If the load absorbs 400+j30 VA of power, find the following: (2 pts each)
a. The magnitude of current drawn by the load.
b. The apparent power absorbed by the load.
c. The power factor of the load.
d. The impedance of the load.
e. Sketch and label the power triangle.

Homework Equations



none given

The Attempt at a Solution



I just wanted to make sure I was starting this off in the right direction. I am looking at this problem as the complex power = 400+j30 VA. This would give me the average power is 200W, and reactive power is 30 VAR correct? I could then find the apparent power by taking √[(Avg. Power)^2 + (Reactive Power)^2].

I could then find the power factor by taking the (average power) / (apparent power). I was wondering if you guys could let me know if this is heading in the right direction so far? Thanks.
 
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  • #2
How did you determine that the average power is 200W?
 
  • #3
For the Apparent Power, check you decimals
for Power factor, remember arc-cos
 

Related to Solving Complex Power with a 400-V, 100-Hz Source and a Load of 400+j30 VA

1. What is complex power with loads?

Complex power with loads is a measure of the total power consumed by a circuit that has both reactive components (such as inductors and capacitors) and resistive components (such as resistors). It is represented by a complex number that has both a real part (representing the active or real power) and an imaginary part (representing the reactive power).

2. How is complex power with loads calculated?

Complex power with loads is calculated using the equation S = P + jQ, where S is the complex power, P is the real power, and Q is the reactive power. The real power is calculated as the product of the voltage and current in the circuit, while the reactive power is calculated as the product of the voltage and the reactive component of the current.

3. What is the difference between complex power with loads and apparent power?

Complex power with loads takes into account the reactive components of a circuit, while apparent power only accounts for the total amount of power (both active and reactive) that is drawn from the circuit. Apparent power is represented by a real number, while complex power is represented by a complex number.

4. How is complex power with loads used in practical applications?

Complex power with loads is used in electrical engineering to analyze and design circuits, particularly those with reactive components. It is also used in power systems and renewable energy systems to optimize energy efficiency and reduce power losses.

5. Can complex power with loads be negative?

Yes, complex power with loads can be negative. This indicates that the reactive power is flowing in the opposite direction of the active power. Negative complex power is often seen in circuits with inductors or capacitors, as they can store and release energy back into the circuit.

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