# Homework Help: Calculate Real and Reactive Power and write equation for S = P + jQ

1. Nov 30, 2017

### jaus tail

1. The problem statement, all variables and given/known data
v = 141.4sin(wt + 15)
i = 14.14 cos(wt - 45)

2. Relevant equations
Draw Phasors.
P = VI cos(theta)
Q = VI sin(theta)
S = P + jQ
theta is angle between V and I
3. The attempt at a solution

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?

2. Nov 30, 2017

### 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.

3. Dec 1, 2017

### cnh1995

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*.)

4. Dec 1, 2017

### 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.

5. Dec 1, 2017

### 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*.

6. Dec 2, 2017

### 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.

7. Dec 2, 2017

### cnh1995

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

8. Dec 2, 2017

### jaus tail

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