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

## 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
Homework Helper
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.

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jaus tail
cnh1995
Homework Helper
Gold Member

## 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.

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
Homework Helper
Gold Member
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.

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