I need to find the "phasor difference" between two voltages of the same value & differing in phase by 60deg?
I am unsure if that means the phasor difference between the two voltages themselves or the two voltages in relation to a 0 reference line or neither?
To be honest i am not sure about this as i have only done phasor addition etc.
The only information i have is that the "phasor difference" is the unknown side of a parallelogram developed from one side and the diagonal?.
I have taken a stab at it in the attachment below, i first worked out the resultant (Vtotal) of the two voltages as you'll see & then added what i think the phasor difference is (in this case 10deg) but this seems to be in relation to the 0 ref line--no idea?
I have attached a couple of phasor drawings that i think are what you mentioned.
I am a little--well a lot confused about this Phasor Difference.
The Voltages V1 & V2 = 240V & the result from the method mentioned (if i have it correct) & the scale i am using also is 240V so i am not sure if i have done this correct or if the Phasor Difference is supposed to be in Degrees of the resultant angle which is 60deg but this is the original phase angle?
I understand what you mean about the question being vague & the meaning of Phasor Difference.
I thought the previous attachment was phasor subtraction which is throwing me off.
I don't mean to be a pain in the butt & i certainly don't want to waste peoples time but can we look at this in a different way just for a moment.
I have attached three diagrams on one sheet, the first diagram is the original attachment No1, the second is for (V1 NOT KNOWN) so to find V1 & the Phasor Difference in respect to 0 Ref.
The third diagram is (V2 NOT KNOWN) so to find V2 & the Phaser Difference in respect of 0 Ref.
The question says, find the phasor differences of two voltages, could this mean "individually" in respect of 0 Ref as i have tried to do in the attachment.
Are my diagrams correct for finding the "unknown values mentioned & the phasor difference" in respect to 0 Ref, if in fact that is what Phasor difference means, or am i way out?
You'll have to shoot me, as i find this interesting Haha!
Thank You
Hi.
To make math between out-of-phase phasors, you need to project each phasor to the bases.
For example:
You know that the both phasors are dephased by 60º. So it doesn't hurt to make V1 = 0° and V2 = -60°.
But what is V1?
V1 is V1*cos(0°)i + V1*sen(0°)j. -> That means: V1*cos(0°) on base "i" and V1*sen(0º) on base "j".
And V2?
V2 is V2*cos(-60°)i + V2*sen(-60°)j.
Which takes us to V1 - V2 = (V1 - V2*cos(-60°))i + (0 - V2*sen(-60°))j
To find the magnitude again, just use the Phytagoras theorem.
Yes, i guess your right i have been trying to decipher Phasor Difference a bit, i just thought i had better understand what it actually meant before trying to answer it if you get my drift.
Unfortunately i don't have all the related material from the course as i have just a copy of the questions from a friend of the family who is doing the actual course full time, because i am to young to get the course proper & they wouldn't let me tackle it.
Once i complete this though i will try to get enrolled again showing what i have done from the paper work i have recieved & that may change there mind & let me do it properly.
At the moment it's all study & trying to understand things so i can complete a trade when i am of age.
I will go & ask if there is any related paperwork regarding this & see if i can get a copy & post it.
Thanks very much for your time.
Hayato
Thanks for the equations regarding this they will be helpful, i don't mean to sound ignorant but i am not sure what "sen" is?
Did you mean sin or is there a sen as well ?
My bad, it was a sin to write sen instead of sin. =D.
If you have doubts, always decompose your vectors to the base and do the operations there. And remember that bases are always orthogonal, in a scalar multiplication that means:
(Ai + Bj) . (Ci + Dj) = A.Ci + B.Dj
the terms (A.D)i.j , (B.C)i.j are 0, because i.j = 0
My bad, it was a sin to write sen instead of sin. =D.
If you have doubts, always decompose your vectors to the base and do the operations there. And remember that bases are always orthogonal, in a scalar multiplication that means:
(Ai + Bj) . (Ci + Dj) = A.Ci + B.Dj
the terms (A.D)i.j , (B.C)i.j are 0, because i.j = 0
this is my take. The resultant of a subtraction is drawn between the short diagonal of the parallelogram. This is the same as tail to tail of phasors which compares to head to tail for addition.