## Phasor representation of AC voltage and current

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
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 Admin $$p(t)\,=\,[V_{max}\,cos(\omega{t}+\theta)] \times [I_{max}\,cos(\omega{t}+\phi)]$$ 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 = 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...ric/phase.html http://www.physclips.unsw.edu.au/jw/AC.html http://www.walter-fendt.de/ph11e/accircuit.htm
 so Phasor representation of an AC voltage is what magnitude? RMS

## Phasor representation of AC voltage and current

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.