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Application to AC circuits

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that current ##I = I_0\sin{\omega t}## can be rewritten as ##I = I_0 e^{i\omega t}##

    2. Relevant equations
    Euler's formula ##re^{i\theta} = r(\cos{\theta}+i\sin{\theta})##

    3. The attempt at a solution
    If ##r = I_0## and ##\theta = \omega t ## then ##re^{i\theta} = r(\cos{\theta}+i\sin{\theta}) ## will be ##I = I_0e^{i\omega t} = I_0(\cos{\omega t}+i\sin{\omega t})##. The left hand side ##I = I_0 e^{i\omega t}## is correct, but the right hand side is not equal to ##I = I_0\sin{\omega t}##
    Thank you
     
  2. jcsd
  3. Oct 19, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    I think what you're looking at is the introduction of phasors to your toolkit. As you have spotted, eiωt is not formally congruent to sinωt. However, if one considers a phasor to be a rotating vector (a vector that rotates in direction around the origin with a constant angular frequency ω), then a projection of that vector on the imaginary axis does indeed follow sinωt. Similarly, its projection on the real axis follows cosωt. Now, eiωt can be considered to be a rotating vector, as t is a time variable. So, for example,##I_0 e^{i\omega t}## would be a phasor of magnitude ##I_0## rotating at a rate of ##\omega## radians per second.

    Typically (but not always) all the phasors in a given circuit have the same angular frequency, and the convention is to drop the common ##e^{i \omega t}## from the notation. Declare that ##I## is a phasor and the ##e^{i \omega t}## is implied.

    Does that help?
     
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