Phasor representation of AC voltage and current

[Astronuc]

Phasor representation of AC voltage and current.

I\,=\,5\angle{0^o}\,=\,5\,+\,j0\,A

V\,=\,100\angle{30^o}\,=\,86.6\,+\,j50\,V


in general

V\,=\,A\angle{\theta^o}\,=\,A cos{\theta}\,+\,jA sin{\theta}\,V

and similarly for I


It is assumed that the angular frequency \omega is the same throughout the system, and it is assumed that the Voltage and Current are RMS values.

For the above phasor values, the voltage and current are:

v(t) = 141.4 cos (\omegat + 30°)

and

i(t) = 7.07 cos \omegat

[Astronuc]

p(t)\,=\,[V_{max}\,cos(\omega{t}+\theta)] \times [I_{max}\,cos(\omega{t}+\phi)]

becomes

p(t)\,=\,\frac{V_{max}I_{max}}{2}[cos(\theta-\phi)\,+\,cos(2\omega{t}+\theta+\phi)]

The average power is

P\,=\,V_{rms}I_{rms}\,cos(\theta-\phi)


In phasor notation,

v\,=\,V_{rms}\angle\theta

i\,=\,I_{rms}\angle\phi

but

P\,\neq\,V_{rms}I_{rms}\angle(\theta+\phi)

Instead

P\,=\,Re\{VI^*\}

and

V\,I^*\,=\,(V_{rms}\angle\theta)\times(I_{rms}\ang le-\phi)

\,=\,V_{rms}I_{rms}\angle(\theta-\phi)

The real part of power is given by

P\,=\,V_{rms}I_{rms}cos(\theta-\phi)

and the reactive or imaginary part of power is

Q\,=\,V_{rms}I_{rms}sin(\theta-\phi)

and the quantity cos(\theta-\phi) is known as the power factor.

The apparent power, S, expressed as volt-amperes (VA) is given by

S (volt-amps) = P (Watts) + jQ (volt-amps-reactive) = VI*

|S|2 = |P|2 + |Q|2 = Vrms2 Irms2

PF = |P|/|S|

VAR is commonly used as a unit for "volt-amperes-reactive"

Some useful background on AC power and phasors.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/phase.html

http://www.physclips.unsw.edu.au/jw/AC.html

http://www.walter-fendt.de/ph11e/accircuit.htm

[infowarrior]

so Phasor representation of an AC voltage is what magnitude? RMS

[rbj]

you might want to explicitly relate Vmax to Vrms and similar for the currents. in fact, Astronuc, i might define the sinusoids as

v(t) \triangleq V_{max} cos(\omega t + \theta) = \sqrt{2} V_{rms} cos(\omega t + \theta)

and

i(t) \triangleq I_{max} cos(\omega t + \phi) = \sqrt{2} I_{rms} cos(\omega t + \phi)

and then crank out the instantaneous and mean power as you did.

i dunno. just a suggestion.