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enter image description herethat's my first question here. I'm studying fundamentals of electric ciruits and I faced this question:

Determine which one leads and by how much: $$v1(t) = 4*\cos(377t + 10)$$ $$v2(t) = -20*\cos (377t)$$

So I converted

$$v2 = -20*\cos (377t)$$ to $$v2 = +20*\cos (377t+180)$$

and graphed the two sinusoids on the unit circuit but I'm confused about which one is leading the other and by how much

Thanks in advance!! =)

EDIT!!:

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    \$\begingroup\$ Show the graph you produced and explain how you are confused by it. It's quite possible you have done this incorrectly and that is what is confusing you but without it being seen, nobody can tell. \$\endgroup\$
    – Andy aka
    Commented Dec 10, 2016 at 12:10
  • \$\begingroup\$ I have attached the graph now \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 13:44

4 Answers 4

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You have got these two equations: $$v1(t) = 4*cos(377t + 10)$$ and $$v2(t) = 20*cos (377t+180)$$ If you substitute t=0, you will get \$v1(0) = 4*cos(10)\$ and \$v2(0) = 20*cos(180)\$. From the values you can conclude that at t =0, v2 has a value which cosine does at the end of its period and v1 has a value which cosine does at the beginning of its period.
So, we can say v2 is leading v1.

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  • \$\begingroup\$ Is that really a convenient way of solving that question? does it apply in all cases? \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 16:16
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Represent both in terms of sine before finding which one leads/lags.

\$v1(t) = 4*cos(377t + 10)\$ becomes \$4*sin(377t + 10 + 90) = 4*sin(377t +100)\$

\$v2(t) = 20*cos (377t+180)\$ becomes \$20*sin(377t + 180 + 90) = 20*sin(377t + 270)\$

Normalize both. So you get normalized \$v1 = sin(377t +100)\$ and normalized \$v2 = sin(377t + 270)\$. Now you can see v2 leads v1 by 170 degrees.

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  • \$\begingroup\$ Do I have to represent in terms of sine in all cases? or this is just a special case? \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 13:45
  • \$\begingroup\$ If both are in cosine form, then at time t = 0 you will know which one leads which one. I just said it so that you follow it as a convention. I mean the cases where one is sine and another is cosine. \$\endgroup\$ Commented Dec 10, 2016 at 14:00
  • \$\begingroup\$ It's written in the handbook that both voltages or currents must be a cosine or both must be a sine \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 14:02
  • \$\begingroup\$ Fundamentals of electric circuits 5th edition by sakidu... there is two things to make sure before solving these types of questions (1)both must be in terms of sine or both must in terms of cosine (2)both must be negative or both must be positive \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 14:10
  • \$\begingroup\$ Yes, that is what I also meant. Read above comments. You convert both to sine if in case one of the two is cosine (you can convert both to cosine as well but. I just said a convention). \$\endgroup\$ Commented Dec 10, 2016 at 14:14
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In this case you already represent both in the same terms (cosine) so you can see which one is leading by seeing which phase is greater. In this case \$v1(t) = 4*\cos(377t + 10)\$ and \$v2 = +20*cos (377t+180)\$. You can see that v2 has greater phase (180) than v1 (10). Now you know that V2 is leading V1 by 170 (180-10).

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  • \$\begingroup\$ -"so you can see which one is leading by seeing which phase is greater".. Does that work in all cases? \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 14:04
  • \$\begingroup\$ that should work in all cases since if you plot the graph you should be able to see the wave with greater phase is leading. \$\endgroup\$
    – shafiyyah
    Commented Dec 10, 2016 at 14:27
  • \$\begingroup\$ and don't forget that before comparing you have to make sure the amplitude have to be both positive or both negative \$\endgroup\$
    – shafiyyah
    Commented Dec 10, 2016 at 14:32
  • \$\begingroup\$ I'm not allowed to use a grapher in exams do you mean unit circle graph as I did? \$\endgroup\$
    – Seraj
    Commented Dec 10, 2016 at 16:17
  • \$\begingroup\$ The graph that i mean is the waveform, it only a means to proof that your answer is correct. So you don't actually have to use it. \$\endgroup\$
    – shafiyyah
    Commented Dec 10, 2016 at 17:26
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This looks like a "split single phase" except unbalanced (v1 << v2)

  • regardless of inbalance, v1(t) is clearly leading +10

  • if offset was 0 you would have a complementary sinewaves or a "split-single phase " outputs. e.g. centre-tapped transformer output. with v1 much lower amplitude.

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