(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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

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