# time domain to phasor



## waterboy22 (Jun 4, 2010)

I have one question regarding question no. 1.1 in power system analysis by john grainger......

in this question voltage is given as 141.4 sin (wt + 30) and current as 11.31 cos(wt -30)

my solution...

first convert voltage in cosine form... which is 141.4 cos(wt +30-90) and current as 11.31 cos (wt - 30)

now phasor values become V = 100&lt;-60deg and I = 8&lt;-30deg

but we need to take voltage as ref. so V = 100&lt;0deg so I = 8&lt;30deg. and therefore as I leads its an capacitive circuit.... IS THIS CORRECT...

OR ANOTHER METHOD TO CONVERT CURRENT TO SINE DOMAIN...

V = 100&lt;30 I = 8&lt;60deg....

and so ... if volt is ref. V=100&lt;0deg and I = 8&lt;30deg... and again the ans stays same I leads so its capacitive....

Am I doing this correct... plz correct me if i am wrong... thank you ...


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## pelaw (Jun 4, 2010)

EDIT: I thought you were electrical. Here is the basics.

For v(t) = Vmax(wt + A), then

1) if Current i(t) = Imax sin [(wt + A) + D], then it is leading,

2) if Current i(t) = Imax sin [(wt + A) - D], then it is lagging.

Since, current in the problem is in terms of cosine: i(t) = Imax cos(wt - C), then it must be converted to sine.

So, i(t) = Imax cos (wt - C) = Imax sin(wt - C + 90)

11.31 sin (wt - 30 + 90) = 11.31 sin (wt + 60) = 11.31 sin (wt + 30 + 30).

Since v(t) = 141sin(wt+30), and i(t)=11.31sin[(wt+30) + 30], then 1) above applies, and the circuit is leading.


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