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I need to calculate phase angle between

$$i_1 = -4\sin(377t + 55) \hspace{0.2cm}\quad \mathrm{and}\quad \hspace{0.2cm} i_2 = 5\cos(377t - 65)$$

I need to determine which one is leading too.

My progress so far:

$$\begin{align}i_1 &= -4 \sin(377t + 55)\\ &= 4\cos(377t + 55 + 90)\\ &= 4\cos(377t + 145)\\[1em] i_2 &= 5\cos(377t - 65)\end{align}$$

So, we get phase difference = 145 - (-65) = 210, \$i_1\$ leads \$i_2\$ by 210.

But, $$\begin{align}i_2 &= 5\cos(377t - 65)\\ &= 5\cos(377t - 65 + 360)\\ &= 5\cos(377t + 295)\end{align}$$

Now, we get phase difference = 295 - 145 = 150, \$i_2\$ leads \$i_1\$ by 150.

Are both of these correct ?

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  • 2
    \$\begingroup\$ I recommend collaborating with fellow students and see if they arrived at the same answer. \$\endgroup\$ – TimB Feb 4 '18 at 1:54
  • \$\begingroup\$ Can something lead or lag by more than 180 degrees? How do you define lead and lag? \$\endgroup\$ – jonk Feb 4 '18 at 3:00
  • \$\begingroup\$ I've got this problem in "Fundamentals of electric circuits" book written by Alexander & Sadiku . That gives 210 as answer. Please check this link for how I define lead or lag : imgur.com/a/UEXAy \$\endgroup\$ – Utshaw Feb 4 '18 at 3:07
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    \$\begingroup\$ @Utshaw They plot EXACTLY the same. There is no difference. None at all. You must be plotting in radians and incorrectly adding degrees, or something. Try using \$\operatorname{cos}\left(377 t+\frac{7}{6}\pi\right)\$ and \$\operatorname{cos}\left(377 t-\frac{5}{6}\pi\right)\$. \$\endgroup\$ – jonk Feb 4 '18 at 4:40
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    \$\begingroup\$ Side note (and this is rather my opinion): Matthew Sadiku writes really really really bad textbooks. That being said, and I know the professor assigns certain textbooks for the course so that's out of your control, try to use some outside material when you want to try to learn something. Fundamentals of Circuits by Sadiku is okay at best... Digressing. Sorry... but why would you add 360 degrees? \$\endgroup\$ – KingDuken Feb 4 '18 at 5:52

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