46 - Power in AC Circuits 3 | Power Triangle - Apparent, Real and Reactive Power

[@ManulaRodrigo-ve4hw]

Can't we directly get the apparent power by adding S1,S2, and S3

[@ebodonkoh6636]

Good day sir With the C part of the question If we assume that the power factor 0.95 leads the load of 1.2KVA how do u solve for the total P and Q

[@SkanCityAcademy_SirJohn]

You will do it like specified in the video just that in finding the total Q, the Q with pf leading becomes negative, so for instance, if pf for c is leading, Qt = Q1+Q2+(-Q3). Hope you get it.

[@ebodonkoh6636]

@@SkanCityAcademy_SirJohn I get it But will it also affect the total real power???

[@SkanCityAcademy_SirJohn]

No, it doesn't affect, P.

[@ebodonkoh6636]

Noted Gracias 🙏

[@ephe8949]

Can you divide the polar form by a scalar quantity? E.g. finding the rms value of a current in polar form.

[@SkanCityAcademy_SirJohn]

Yes. You can. The scalar will have a zero argument

[@SkanCityAcademy_SirJohn]

So 5 is just 5

[@ephe8949]

@@SkanCityAcademy_SirJohn Thank you.

[@user-ChrisB]

Please how would you know the final power would be lagging or leading?

[@SkanCityAcademy_SirJohn]

for an rc circuit, which is capacitive, current leads voltage, for an rl circuit, which is inductive, current lags voltage, but for an rlc circuit, check the phase angles of the current and voltage of the circuit. Assuming v has phase angle of 0 and I has a phase angle of 60. it means current lead voltage by 60 hence pf is leading. this video will help kzbin.info/www/bejne/Z2mcdI2gn5V9eJo

[@DesmondDugrah-jk2ti]

Please considering the question i, since we have the S total to be in KW why didn't we multiply it by by 10 exponent 3 before dividing by V to find the I total in A Or better still have our final answer( I total in KA) since we didn't get rid of K?

[@SkanCityAcademy_SirJohn]

the reason is that first, we want to avoid having huge values that is why we worked in K's, notice that at the end, 240v is converted to 0.24 which is is kv, hence k cancels out, that is why you have current in A. in actual sense you would have had 3540/240 which is still 14.75A.