P=V*I

[eatsleep]

I have a pretty basic question here but its tripping me up. A practice problem i was doing used P=V*I (thats the conjugate phase angle of V). Can someone explain this equation? Is it valid? I understand how and why we use S=VI* but i am uncertain of P=V*I.

Thanks

 

[FelizEng PE]

I have a pretty basic question here but its tripping me up. A practice problem i was doing used P=V*I (thats the conjugate phase angle of V). Can someone explain this equation? Is it valid? I understand how and why we use S=VI* but i am uncertain of P=V*I.
Thanks

Could see the prob? I have never seen current conj used to obtain the real power

 

[Zach Stone P.E.]

I have a pretty basic question here but its tripping me up. A practice problem i was doing used P=V*I (thats the conjugate phase angle of V). Can someone explain this equation? Is it valid? I understand how and why we use S=VI* but i am uncertain of P=V*I.

Thanks

Is it possible that the asterisk is meant to just be multiplication and the load is purely resistive (I.e. no need to multiply by power factor)? 

Or is it possible that the book accidentally put the conjugate in front of the I by mistake? 

It would be helpful to see the full problem and solution.

 

[omniware19]

Proof:

1) S=P+jQ = VI*

if P=V*I then,

2) S=V*I + jQ = VI*

3) V*I + jQ = P+jQ

4) From like term in 3), P=V*I.

Verify,

let V=v*exp(a), I = i*exp(b)

S=VI* = vi*exp(a-b) = P+jQ

S*=V*I = vi*exp[-(a-b)] = P-jQ

S+S* = 2P = 2vicos(a-b)

P=vi*cos(a-b) = vi * R{exp(a-b)]  = Re[vexp(a)* iexp(-b)]

note that P is real so that P=P*

P = Re{ vexp(-a)* iexp(b)] = V*I