# Complex power calculations formula

• FrankJ777
In summary, the conversation is about complex power calculations and the confusion surrounding the conjugate of I(eff). The formula given in the textbook is S=V(eff)I(eff)*, and it is complex because it takes into account both the real and imaginary components of power. The examples provided in the textbook use the equation S=(1/2)VI*, which is equivalent to S=V(eff)I(eff)*. The conversation also discusses how to calculate real and reactive power, and the relationship between the two. The concept of phasors is also introduced to explain how the conjugate is used in the complex power formula.

#### FrankJ777

I'm confused about a point concerning complex power calculations. The formula my text gives is S=V(eff)I(eff)*. I'm confused about what the conjugate is for I(eff). The way I understand it, I(eff) = I(rms) and I(rms) = I(m)/√2. Since I(m) is a real number, and I(m)/√2 is real, how is there a conjugate.
For example:
i(t)= 20cos(ωt+165)
I(m)=20
I(eff)=20/√2
...so what is I(eff)*?

I've read through the textbook book the only examples that I can find use the equation: S=(1/2)VI*.

I would be grateful for any help you could provide to help me understand this.

Hi FrankJ777! Complex power is supposed to be complex.

For a perfect resistor, the complex power is purely real,

and for a perfect capacitor or inductor, it's purely imaginary …

but for a mixture, it'll be "genuinely complex", but with a constant phase …

(because we assume the current leads the voltage by a constant phase, and so 1/2 VI* subtracts the two phases, and keeps the total phase constant ) …

I think If
V(t) = Vmax exp[jwt + phi] and
I(t) = Imax exp[jwt + phi] then
S = (1/2) V(t) I(t)* is the rms real power.
But even if you use rms values, you still have both volt-amps and watts.

Also...

v(t) = Vmax/ exp[jwt]
i(t) = Imax exp[j(wt + phi)]

Veff (phasor) = v(t)/√2
Ieff (phasor) = i(t)/√2

I*eff = Imax/√2 exp[-j(wt + phi)] (note minus sign)

so (product of phasors)
Veff x I*eff = Vmax/√2 exp[jwt] x Imax/√2 exp[-j(wt + phi)]

Seff = Vmax Imax/2 exp (-jphi)
(plus exp of voltage time and minus exp of current conjugate time cancel, leaving only phi part, that's why conjugate is right for this)

real power P = Real {S} = Veff Ieff cos (-phi)
reactive power Q = Imag{S} = Veff Ieff sin (-phi)
(so 'Reactive Power' is negative for inductive circuits where phi is positive)

the phasors make a vector triangle:
S^2 = P^2 + Q^2

Sorry the formatting is not better. Still, I hope this helps.

Welcome to PF!

Hi Dr.kW! Welcome to PF! (have a phi: φ and an omega: ω and try using the X2 and X2 tags just above the Reply box )

Action replay (of Dr.kW ):-
v(t) = Vmax/ ejωt
i(t) = Imax ej(ωt + phi)

Veff (phasor) = v(t)/√2
Ieff (phasor) = i(t)/√2

I*eff = Imax/√2 e-j(wt + φ) (note minus sign)

so (product of phasors)
Veff x I*eff = Vmax/√2 ejωt x Imax/√2 e-j(ωt + φ)

Seff = Vmax Imax/2 exp-jφ
(plus exp of voltage time and minus exp of current conjugate time cancel, leaving only φ part, that's why conjugate is right for this)

real power: P = Real {S} = Veff Ieff cos (-φ)
reactive power: Q = Imag{S} = Veff Ieff sin (-φ)
(so 'Reactive Power' is negative for inductive circuits where φ is positive)

the phasors make a vector triangle:
S2 = P2 + Q2 ## 1. What is a complex power calculation formula?

A complex power calculation formula is a mathematical equation used in electrical engineering to determine the power in a circuit that contains both reactive and active components. It takes into account both the real power (measured in watts) and the reactive power (measured in volt-amperes reactive).

## 2. Why is it important to calculate complex power?

Calculating complex power is important because it allows us to better understand how energy is being used in a circuit. It also helps engineers design and troubleshoot electrical systems more effectively.

## 3. What are the components of a complex power calculation formula?

The components of a complex power calculation formula include voltage (V), current (I), and impedance (Z). These values are used to calculate the real power (P), reactive power (Q), and apparent power (S) in a circuit.

## 4. How is complex power different from real power?

Complex power takes into account both the real power (P) and the reactive power (Q), while real power only considers the real component of power. Real power is the power that is actually consumed by a circuit, while reactive power is the power that is exchanged between the circuit and the source.

## 5. Can complex power be negative?

Yes, complex power can be negative. This indicates that the circuit is producing more reactive power than real power, and is common in circuits with inductive or capacitive loads. Negative complex power can also occur when power is flowing in the opposite direction to the assumed direction of current flow.