eatsleep 5 Posted July 2, 2019 Report Share Posted July 2, 2019 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 Quote Link to post Share on other sites

FelizENG PE 18 Posted July 2, 2019 Report Share Posted July 2, 2019 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. ThanksCould see the prob? I have never seen current conj used to obtain the real power Quote Link to post Share on other sites

Zach Stone, P.E. 116 Posted July 3, 2019 Report Share Posted July 3, 2019 20 hours ago, eatsleep said: 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. Quote Link to post Share on other sites

omniware19 4 Posted July 3, 2019 Report Share Posted July 3, 2019 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 Quote Link to post Share on other sites

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