Complex power calculations 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.

 

[tiny-tim]

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

 

[Bob S]

If
V(t) = V_max exp[jwt + phi] and
I(t) = I_max 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.

 

[Dr.kW]

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.

 

[tiny-tim]

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) = V_max/ e^jωt
i(t) = I_max e^{j(ωt + phi)}

V_eff (phasor) = v(t)/√2
I_eff (phasor) = i(t)/√2

I*_eff = I_max/√2 e^{-j(wt + φ)} (note minus sign)

so (product of phasors)
V_eff x I*_eff = V_max/√2 e^jωt x I_max/√2 e^{-j(ωt + φ)}

S_eff = V_max I_max/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} = V_eff I_eff cos (-φ)
reactive power: Q = Imag{S} = V_eff I_eff sin (-φ)
(so 'Reactive Power' is negative for inductive circuits where φ is positive)

the phasors make a vector triangle:
S² = P² + Q²