Relation between power factor and phase angle

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Vriska
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Homework Statement


apparently your effective power is V_maxI_max * cos(x) where x is the phase angle. So I must consider only real power

Homework Equations


power = VI
current = I sin(wt)

The Attempt at a Solution



Let impedence be = e^ix . I'll write current as Ie^i(wt) consideronly the real part, so V = e^(ix) *e^(iwt), sooo power is P = VI e^(ix)*e^(2iwt). Sooooooo...now what? take the real part? I don't know and ignore that 2iwt? I'm not quite comfortable with this notation.

random question :

is a phasors axes supposed to be a V vs I plot, I think there's something wrong with by textbook
 
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First consider a totally resistive situation: current and voltage in-phase. If you multiply the instantaneous values of the sine waves, what will the "power wave" look like? What would be the frequency of this wave, compared to the voltage and current waves?

Hint, in situation I described, you could rewrite power, instead of V*I, you could have I²R, or V²/R.
 
scottdave said:
First consider a totally resistive situation: current and voltage in-phase. If you multiply the instantaneous values of the sine waves, what will the "power wave" look like? What would be the frequency of this wave, compared to the voltage and current waves?

Hint, in situation I described, you could rewrite power, instead of V*I, you could have I²R, or V²/R.

aright the power wave will look like road bumps. frequency will be half of that of voltage and current waves

p =v^2 /r sin^2(wt). so max power = v^2/r.. how do I go from here?
 
scottdave said:
Can you clarify what you mean? Can you take a picture of the textbook example that you are asking about?

they drew a v versus I plot, with Voltage on the y-axis and drew a line on the plot with its y coordinate of its end being v_i - v_c and the x coordinate being v_r. The angle between the line and the x-axis was the phase angle. not quite sure what they're doing here :|