Apparent power (S) is the complex conjugate of the complex power (P+jQ), where P is the real power and Q is the reactive power. To convert between real power, reactive power, and apparent power, you can use the following equations:

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## convert from real power and reactive power to apparent power

S = P + jQ

The equation S = P + jQ is used to convert from real power and reactive power to apparent power. Let me explain each term in the equation and provide an example:

- S: Apparent Power, measured in volt-amperes (VA), represents the total power that is supplied to a circuit, including both real power and reactive power.
- P: Real Power, measured in watts (W), represents the power consumed or produced by a circuit, which is used to perform useful work.
- Q: Reactive Power, measured in volt-amperes reactive (VAR), represents the power that is stored and released by reactive components, such as inductors and capacitors. Reactive power does not perform any useful work but is required for the circuit to function properly.

Let’s say that a circuit has a real power of 1000 W and a reactive power of 500 VAR. To find the apparent power, we can use the equation S = P + jQ:

S = 1000 W + j(500 VAR)

We can simplify the equation by multiplying the imaginary term by “j” to get:

S = 1000 W + j(500 VAR) = 1000 + j500 VA

Therefore, the apparent power of the circuit is 1000 + j500 VA.

Note that in a purely resistive circuit (no reactive components), the reactive power is zero, and the apparent power is equal to the real power.

In a purely reactive circuit (no resistive components), the real power is zero, and the apparent power is equal to the reactive power.

In a circuit with both resistive and reactive components, the apparent power is always greater than the real power due to the presence of reactive power.

## convert from apparent power and real power to reactive power

Q = sqrt(S^2 – P^2)

Let’s say we have a circuit with an apparent power of 1000 VA and a real power of 800 W. We can use the equation Q = sqrt(S^2 – P^2) to find the reactive power of the circuit.

Q = sqrt(S^2 – P^2) = sqrt((1000 VA)^2 – (800 W)^2) = sqrt(1,000,000 VA^2 – 640,000 W^2) = sqrt(360,000) VAR = 600 VAR

Therefore, the reactive power of the circuit is 600 VAR.

## convert from apparent power and reactive power to real power

P = sqrt(S^2 – Q^2)

In the above equations, “j” represents the imaginary unit and “sqrt” denotes the square root. Note that the units of real power, reactive power, and apparent power are watts (W), volt-amperes reactive (VAR), and volt-amperes (VA), respectively.

**Let’s consider **a circuit with an apparent power of 1200 VA and a reactive power of 500 VAR. We can use the equation P = sqrt(S^2 – Q^2) to find the real power of the circuit.

P = sqrt(S^2 – Q^2) = sqrt((1200 VA)^2 – (500 VAR)^2) = sqrt(1,440,000 VA^2 – 250,000 VAR^2) = sqrt(1,190,000) W = 1090.87 W (rounded to two decimal places)

Therefore, the real power of the circuit is approximately 1090.87 W.

## Using the power factor formula

It is also possible to convert between real power, reactive power, and apparent power using the power factor formula.

The power factor (PF) is the ratio of real power (P) to apparent power (S), expressed as a decimal or percentage:

**PF = P / S**

Using this formula, we can solve for any of the three quantities given the other two. Here are the equations for converting between real power (P), reactive power (Q), and apparent power (S) using the power factor formula:

- To find real power (P) given apparent power (S) and power factor (PF):
**P = S x PF**

Let’s consider a circuit with an apparent power of 1000 VA and a power factor of 0.8. We can use the equation P = S x PF to find the real power of the circuit.

P = S x PF = 1000 VA x 0.8 = 800 W

Therefore, the real power of the circuit is 800 W.

- To find apparent power (S) given real power (P) and power factor (PF):
**S = P / PF**

Let’s consider a circuit with a real power of 2000 W and a power factor of 0.9. We can use the equation S = P / PF to find the apparent power of the circuit.

S = P / PF = 2000 W / 0.9 = 2222.22 VA (rounded to two decimal places)

Therefore, the apparent power of the circuit is approximately 2222.22 VA.

- To find reactive power (Q) given real power (P) and power factor (PF):
**Q = P x tan(acos(PF))**

Let’s say that a circuit has a real power of 1000 W and an apparent power of 1200 VA. We can use the power factor formula to find the power factor:

PF = P / S = 1000 W / 1200 VA = 0.8333

Now, let’s say that we want to find the reactive power of the circuit. We can use equation 3 above:

Q = P x tan(acos(PF)) = 1000 W x tan(acos(0.8333)) = 384 VAR

Therefore, the circuit has a reactive power of 384 VAR.