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Compound angles

  1. Mar 21, 2007 #1
    i am just about to start an electrical course at college and ive been reading up on some mathematics i will need when i start my problem is i dont know much about compound angles and trig and i want to complete some examples within a book the following problem is the one i need help with

    the power P in an electrical circuit is given by P=V²/R

    find the power in terms of V, R and cos2t when v= Vcost

    any help would be appreciated

  2. jcsd
  3. Mar 21, 2007 #2


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    First, clarify your notation. Is the "V" in P= V2r the "V" or "v" in v= V cos t?

    If it is really P= V2/R, there is nothing to be done, P is already in terms of V and R (and cos 2t is irrelevant).

    If it is really P= v2/R, then v= Vcos t so v2= V2 cos2 t and you can use the trigonometric identity cos 2t= cos2 t- sin2 t= cos2 t- (1- cos2 t)= 2cos2 t- 1 so that 2cos2 t= cos 2t+ 1 and then
    cos2t= (cos(2t)+ 1)/2.

    v2= V2cos2 t= V2(cos(2t)+ 1)/2.

    Now, if P= v2/R, then P= V2(cos(2t)+ 1)/(2R).

    A third, though unlikely, possibility is that P= V2/R and v= Vcos t but you meant "in terms of v, R, and cos 2t. In that case, V= v/cos t so cos2 t belongs in the denominator.
    In that case, P= 2v2/(R(cos(2t)+ 1)).
  4. Mar 21, 2007 #3


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    Yes Chis your notation is a little confused, but I suppose that’s to be expect since you are also a little confused by the area you’re studying. Still what about some punctuation, there’s no excuse for that!

    The general statement for power in a resistive load is that the instantaneous power is,

    [tex]P = v^2 / R[/tex]

    For the “DC case” where the voltage is a constant, [tex]v = V[/tex], so [tex]P=V^2/R[/tex]

    For the “AC case” where the voltage is sinusoidal, [tex]v = V cos(t)[/tex], so

    [tex]P(t) = V^2 cos^2(t) / R[/tex]

    Now often we are interested primarily in just the average value of the power. In this case it is convenient to use the trig identity,

    [tex]2 \cos^2(t) = 1 + \cos(2t)[/tex],

    to write the AC power as,

    [tex]P(t) = \frac{V^2}{2R} \left( 1 + \cos(2t) \right)[/tex]

    Note here that the cos(2t) part has zero average. Hence for the AC case the average power is,

    [tex]P_{av} = V^2 / 2R[/tex]
    Last edited: Mar 21, 2007
  5. Mar 24, 2007 #4
    Thankyou Hallsofivy

    I did think i should be using trig identities:surprised . I was a little confused:confused: because the question was taken directly from a book it is written exactly as i wrote it in my request.

    P.S sorry about my grammer as i was rushing before work.

    Thanks again

    Chris :approve:
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