Calculate Real and Reactive Power and write equation for S = P + jQ

[jaus tail]

Homework Statement

v = 141.4sin(wt + 15)
i = 14.14 cos(wt - 45)

Homework Equations

Draw Phasors.
P = VI cos(theta)
Q = VI sin(theta)
S = P + jQ
theta is angle between V and I

The Attempt at a Solution

[/B]
V = 141.4sin(wt + 15)
So Vrms = 141.4/1.414 sin(wt + 15)
= 100 sin (wt + 15)

I = 14.14 cos(wt - 45)
Now cos (-theta) = cos(theta)
So I = 14.14 cos (45-wt)
So I = 14.14 sin(90-45 + wt)
So I = 14.14 sin(45 + wt)
I rms = 10sin(wt + 45)

Now Vrms = 100sin(wt + 15)
I rms = 10sin(wt + 45)
I leads V by 30 degrees.
theta = 30 degrees.
P = VI cos(theta) = 100 * 10 * cos(30) = 866W
Q = VI sin(theta) = 100 * 10 * sin(30) = 500VAR
So S = P + jQ = 866 + j500

But in answer it is given S = 866 - j500

Where does the minus sign come from?
I understand that it is leading power factor so it gives Q but mathematically sin(30) is positive value. So why should S have negative VAR?

 

[scottdave]

Leading current (current leads voltage) gives phasors pointing down (negative angle). Lagging current (current lags voltage) give phasors pointing up (positive angle).
That is the convention.

 

[cnh1995]

Homework Statement

v = 141.4sin(wt + 15)
i = 14.14 cos(wt - 45)

Homework Equations

Draw Phasors.
P = VI cos(theta)
Q = VI sin(theta)
S = P + jQ
theta is angle between V and I

The Attempt at a Solution

[/B]
V = 141.4sin(wt + 15)
So Vrms = 141.4/1.414 sin(wt + 15)
= 100 sin (wt + 15)

I = 14.14 cos(wt - 45)
Now cos (-theta) = cos(theta)
So I = 14.14 cos (45-wt)
So I = 14.14 sin(90-45 + wt)
So I = 14.14 sin(45 + wt)
I rms = 10sin(wt + 45)

Now Vrms = 100sin(wt + 15)
I rms = 10sin(wt + 45)
I leads V by 30 degrees.
theta = 30 degrees.
P = VI cos(theta) = 100 * 10 * cos(30) = 866W
Q = VI sin(theta) = 100 * 10 * sin(30) = 500VAR
So S = P + jQ = 866 + j500

But in answer it is given S = 866 - j500

Where does the minus sign come from?
I understand that it is leading power factor so it gives Q but mathematically sin(30) is positive value. So why should S have negative VAR?

By convention, leading VARs are negative and lagging VARs are positive, and S=VI*.
(If it were the opposite, you'd have to use S=IV*.)

 

[jaus tail]

So mathematically when we calculate Q then it is V multiplied by I multiplied by the angle by which the voltage leads the current.
Thus here we'll get -theta.

 

[cnh1995]

Take for example, V=20∠30° and I=4∠-30°.
Here, the current is lagging and by the convention S=VI*, you'll get S=P+jQ i.e. positive reactive power.

If you want lagging VARs to be negative (as your own convention), define S=IV*.

https://www.physicsforums.com/threads/power-factor-convention.929532/

 

[jaus tail]

I've also read that if P and Q are in opposite direction then motor/generator is working at leading pf. if P and Q are in same direction then motor/generator is working at lagging pf.

 

[cnh1995]

I've also read that if P and Q are in opposite direction then motor/generator is working at leading pf. if P and Q are in same direction then motor/generator is working at lagging pf.

There is no direction for Q. It is oscillating between the generator and the load.

 

[jaus tail]

Then why do we say that capacitor acts as source of Q and inductor acts as load of Q?