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Homework Help: Alternating current problem.

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Complex current of I = (1+j) on frequency of 50 Hz. Find maximum current Im, effective current Ief, and current at time 1s, i(1)

    2. Relevant equations


    3. The attempt at a solution
    Effective current is length of that hypotenuse on complex current, that is, graph, so Ief^2 = 1^2 + 1^2 = sqrt(2)
    Maximum current is Im=Ief*sqrt(2)=sqrt(2)*sqrt(2)=2, and ω=2πf=314.ψ = arctan(1/1)=π/4 so i(t)=2*sin(314t+π/4) A.
    i(1)=2*sin(314*1+π/4)? Now how should I solve this? Just take sine from it, or rearrange it somehow?
  2. jcsd
  3. Apr 21, 2013 #2


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    Gold Member

    The actual current is (in the conventions I'm used to using) the real part of the time dependent complex current which is your I=1+j times [itex] (\cos(\omega t) + j \sin(\omega t)[/itex]. (Check with your instructor or texts to see the convention assumed.)

    Something doesn't look right to me about your maximum current. As the actual current is the real part of the complex (time dependent) current which is a "length of the hypotenuse" long complex phasor which rotates around 0 in the complex plane. At some point it points totally in the real direction so should give a maximum current equal to its length.

    Another note, be absolutely sure your calculator is in radian mode!!! That 314 is big and so might lead you to think it is in degrees. But that's 314 radians per second (since it rotates 50 whole cycles per second).

    Let me see .... you're using a phase factor so using the sine vs cosine is just a matter of different phase but you should be sure you're using the appropriate phi. I am interpreting your I=(1+j) as the t=0 complex current.

    There are some choices of convention but the standards I'm used to express current as:
    [itex] I(t) = \Re( I_0 e^{j\omega t}) = \Re( I_0 \cdot [\cos(\omega t) + j\sin(\omega t)])[/itex] with [itex]\Re[/itex] meaning "take the real part of".

    The phase comes in as [itex]I_0[/itex] is the initial complex current which has a polar form: [itex] I_m e^{j\phi} = I_m[\cos(\phi) + j\sin(\phi)][/itex] where [itex]I_m[/itex] is (a real number) the magnitude of the complex current. (I think that is what you're using it as.)

    Multiplying gives you:
    [itex] I(t) =\Re(I_me^{j\phi} e^{j\omega t}) =\Re(I_m e^{j(\omega t + \phi)}) = I_m \cos(\omega t + \phi)[/itex]

    This is the conventional way I'm used to expressing it. Your use of sin instead of cos might imply an equivalent way (multiplying through by [itex]\pm j[/itex] in the equations) which simply adds or subtracts a 90 degree phase factor. It is only a problem if you mix two conventions.

    I would suggest you work out the full complex current at t=1 and then find from that your real current at t=1.

    Keep in mind that the complex representation is supposed to make things easier (provided you're up on your complex arithmetic). Trust this and use it.
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