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Math Help - Current in circuit equation (trigonometry!)

  1. #1
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    Current in circuit equation (trigonometry!)

    Hey this is my first post... been really stuck on this particular question from a paper:

    In a particular circuit the current, I amperes, is given by:


    I=4sinθ - 3cosθ θ>0


    where θ is an angle related to the voltage.


    Given that I=R x sin(θ-a), where R>0 and 0<=a<360degrees,


    a) find the value of R, and the value of a to 1 decimal place.
    b) Hence solve the equation 4sinθ - 3cosθ = 3 to find the values of θ between 0 and 360degrees.
    c) Write down the greatest value for I.
    d) Find the value of θ between 0 and 360degrees at which the greatest value of I occurs.

    I think if someone could show me the steps to answering part a) I could be able to do the rest myself.
    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jaderberg View Post
    Hey this is my first post... been really stuck on this particular question from a paper:

    In a particular circuit the current, I amperes, is given by:


    I=4sinθ - 3cosθ θ>0


    where θ is an angle related to the voltage.


    Given that I=R x sin(θ-a), where R>0 and 0<=a<360degrees,


    a) find the value of R, and the value of a to 1 decimal place.
    b) Hence solve the equation 4sinθ - 3cosθ = 3 to find the values of θ between 0 and 360degrees.
    c) Write down the greatest value for I.
    d) Find the value of θ between 0 and 360degrees at which the greatest value of I occurs.

    I think if someone could show me the steps to answering part a) I could be able to do the rest myself.
    Thanks
    You have:

    I=4 sin(θ) - 3 cos(θ) ..(1)

    and

    I=R sin(θ-a) ..(2)

    where R>0 and θ in [0, 360).

    Applying the trig identity sin(θ-a) = sin(θ) cos(a) - cos(θ) sin(a) and putting it into (2)
    we have:

    I = R sin(θ) cos(a) - R cos(θ) sin(a) ...(3)

    Equating the coefficients of sin(θ) and cos(θ) in (1) and (3) we have:

    R cos(a) = 4 ..(4)

    and

    R sin(a) = 3 ..(5)

    So squaring and adding (4) and (5):

    R^2 cos^2(a) + R^2 sin^2(a) = 4^2 + 3^2 = 25.

    So as cos^2(a) + sin^2(a) = 1, R^2 = 25^2, and so R=5.

    From (4) and (5) we deduce that cos(a)>0 and sin(a)>0, so a is in the first quadrant,
    and dividing (5) by (4) that:

    tan(a) = 3/4,

    so a = arctan(3/4) ~= 36.87 degrees

    RonL
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  3. #3
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    cheers! didnt see the similarities between the two equations after using the addition rule before.
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