AC Power Analysis

Instantaneous Power

The instantaneous power changes with time and it is therefore difficult to measure. The average power is more convenient to measure. For sinusoidal signals or any other periodic signal we compute average power over one period using the following relation:

• Product of time-domain voltage 

      and time-domain current
           p(t) = v(t) i(t)

• Determine maximum value
• Transients
• Steady-state

ƒ

• v(t) = Vmcos(ωt+θ) and i(t) = Imcos(ωt +ϕ)

ƒ

• p(t) = Vm Imcos(ωt+θ) cos(ωt +ϕ) =

            = ½ Vm Imcos(θ-ϕ) + ½ Vm Imcos(2ωt + θ+ ϕ)

where:

p(t) = instantaneousƒ,    two parts,     ƒaverage,     ƒperiodic

 ½ Vm Imcos(θ-ϕ)  =  constant,     ƒindependent of t,     ƒthe "average",     ƒwantedƒ,    active or real

+ ½ Vm Imcos(2ωt + θ+ ϕ) = periodicƒ,     period is ½T,     ƒaverage is zero,     ƒunwanted

EXAMPLE:

1. Voltage across impedance (V= 100∠25V and Z= 50∠55Ω). Determine active power absorbed.

I   =V / Z
    = 2 ∠25-55°
    = 2 ∠-30°
I   = 1.7 - j A

Z= 28.7 + j41.9 Ω

P   = ½ 2228.7
     = ½ (1.72+12) 28.7
P   = 57.4 W

EFFECTIVE OR RMS VALUE

Effective or RMS Value

The RMS Value is a constant itself which depending on the shape of the function i(t).

The effective value of a periodic current is equal to dc current that delivers the same average power to a resistor as the periodic current.

The RMS Value of a sinusoid i(t) = Imcos(wt) is:

                                                              I rms =  Im 

                                                                           √2

                                             

                                                           P = Irms Vrms cos(θv-θi)

• Take note: Supply is always Vrms, unless stated that it is maximum voltage.
  
  

   

   Apparent Power and Power Factor

  
  
  • Apparent Power S, is theproduct of the RMS values of voltage and current.
  • It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts.
  
  
  
  
                                             P = Vrms Irms cos(θv-θi) = S cos(θv-θi)
                                  where:
                                             S - the apparent power
                                             (θv-θi) - the power factor, PF 
            
                                                         PF = P/S = (θv-θi)
• PF is the cosine of the phase difference between the voltage and current. Also, angle of the load impedance. 

                                                                 P = S 0.9

                                                   where: 

                                                             P - active power (W)

                                                             S - apparent power (VA)

                                                            0.9 - reactive power (VAR)

Purely resistive load (R) - θv-θi = 0

                                        - PF = 1

                                        - P = S

Purely reactive load (L or C) - θv-θi = ±90

                                              - PF =0

                                               - P = 0

Resistive and Inductive load (R and L) - θv-θi > 0

                                                                - P = S x PF

                                                                - 0 < PF < 1, lag

Resistive and Capacitive load (R and C) - θv-θi < 0

                                                                  - 0 < PF < 1, lead

                     

                                                 

REVIEW:

• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0deg and 90deg.

COMPLEX POWER

S = 1/2 VI* = Vrms Irms ∠ (θv-θi)

         

                  V = Vm∠θv           I = Im∠θpf

S =  Vrms Irms cos(θv-θi) + jVrms Irms sin(θv-θi)

S = P +jQ

P: is the average power in watts delivered to a load and it is the only useful power.

Q: is the reactive power exchange between the source and the reactive part of the load. It is measured in VAR.

• Q = 0, for resistive loads (unity PF)
• Q < 0, for capacitive loads (leading PF)
• Q > 0, for inductive loads (lagging PF)