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 ($\omega$t + 30°)

and

i(t) = 7.07 cos $\omega$t

 

[Astronuc]

AC Power in Phasor Form

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

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

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

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

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 = V_rms² I_rms²

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]

phasor representation

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

 

[rbj]

you might want to explicitly relate V_max to V_rms 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.