# Complex Power Homework: Is My Solution Right?

• Deniz
In summary, the conversation revolves around a specific problem where the individual, Deniz, is seeking help with their solution. They discuss the calculation of total impedance, dividing it with voltage to find current, and using load impedance to find voltage load. They also mention calculating complex power for the load, and the use of AC, which they are not comfortable with yet. They also discuss the issue of converting sine to cosine when looking for power values and the importance of specifying angles in degrees or radians.
Deniz

y = 27

## The Attempt at a Solution

- I calculated the total impedance.
- Divide it with the voltage to get the current.
- Then I use the load impedance to find the voltage load.
- And I calculated the complex power for the load.

I am not comfortable with the AC yet.
Is my solution right?

Hi Deniz,

It's very difficult to make out the details of your handwriting in your image. Helpers are more likely to take time to help if your work is easy to see. Can you provide a larger, clear image, or better still (and highly encouraged), type in the solution so that helpers can reference and quote it line by line? The edit panel gives access to the necessary special characters via the ##\Sigma## icon, or you can use LaTeX syntax to present equations in a very polished manner.

One thing I will mention is, it's not necessary to convert the source function from sine to cosine if you're looking for power values. The angles will sort themselves out as the voltage and current relative phase will be the same regardless of whether the input is a sine or cosine. It's dictated by the impedance angles.

Sorry for the quality gneill, this has a better resolution.

#### Attachments

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I'm looking at your calculation of ##Z_{eq}##, and while the magnitude looks good I'm not happy with the phase angle. What are your rectangular components for ##Z_{load}##?

Are you specifying your angles in degrees or radians? You don't make this clear in your work.

## 1. What is complex power and how is it different from real power?

Complex power is a mathematical representation of the total power in a circuit or system, taking into account both the real power (measured in watts) and the reactive power (measured in volt-amps reactive). Real power represents the actual energy being used or transferred, while reactive power represents the energy that is stored and released by inductive and capacitive elements. Complex power is expressed as a complex number in the form P + jQ, where P is the real power and Q is the reactive power.

## 2. How is complex power calculated?

Complex power can be calculated using the formula S = P + jQ, where S is the complex power, P is the real power, and Q is the reactive power. Real power can be calculated by multiplying the voltage and current in a circuit, while reactive power can be calculated by multiplying the voltage and current that are out of phase with each other.

## 3. What is the significance of finding the complex power in a circuit?

Finding the complex power in a circuit allows us to understand the total power being used or transferred, as well as the balance between real and reactive power. This information is important for designing and analyzing efficient and stable electrical systems.

## 4. How can I check if my solution for complex power homework is correct?

To check if your solution for complex power homework is correct, you can use the formula S = P + jQ and plug in the values you calculated for real power and reactive power. The resulting complex power should be the same as the one you initially calculated. Additionally, you can compare your solution with others or ask your instructor for feedback.

## 5. Are there any common mistakes to avoid when solving for complex power?

Common mistakes when solving for complex power include forgetting to include the j term or using the wrong units for voltage and current. It is also important to make sure that the voltage and current values used in the calculations are for the same point in the circuit. Additionally, it is important to be mindful of the direction of current flow and the sign convention used for reactive power. Double-checking your calculations and units can help avoid these mistakes.

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