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Current and Parallel Impedance

  1. Oct 31, 2014 #1

    Zondrina

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    Homework Helper

    1. The problem statement, all variables and given/known data

    Find ##i(t)## in the following circuit:

    Screen Shot 2014-10-31 at 12.46.32 PM.png

    2. Relevant equations

    ##Z = \frac{V}{I} \Rightarrow I = \frac{V}{Z}##

    3. The attempt at a solution

    I've solved this, but I'm wondering why my answer is different than the book's answer. The book lists the answer as ##i(t) = 3.88 cos(377t - 39.2^o) \space A##.

    Writing the given voltage as a phasor: ##v(t) = 120 \angle 60^o \space V##.

    Calculating the impedances of each component:

    ##Z_R = 20 \Omega##
    ##Z_L = j \omega L = (377)(40 \times 10^{-3})j = (15.08j) \Omega##
    ##Z_C = - \frac{j}{\omega C} = - \frac{j}{(377)(50 \times 10^{-6})} = -(53.05j) \Omega##

    Calculating the required equivalent impedances:

    ##Z_{eq_1} = Z_R + Z_L = (20 + 15.08j) \Omega##

    ##Z_{eq_2} = (\frac{1}{Z_C} + \frac{1}{Z_{eq_1}})^{-1} = (- \frac{1}{53.05 j} + \frac{1}{20 + 15.08j})^{-1} = (30.56 + 4.97j) = 30.96 \angle 9.24^o \space \Omega##

    Finding the current phasor:

    ##I = \frac{V}{Z} = \frac{120 \angle 60^o}{30.96 \angle 9.24^o} = 3.88 \angle 50.8^o \space A##

    ##i(t) = 3.88 cos(377t + 50.8^o) \space A##

    My answer differs from the books answer by exactly ##90^o## (50.8 + 39.2 = 90). Why is this? Have I done something wrong?
     
  2. jcsd
  3. Oct 31, 2014 #2

    gneill

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    Staff: Mentor

    Your method and answer look good to me.
     
  4. Oct 31, 2014 #3

    Zondrina

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    That's good to know. Although I'm a little perplexed as to why my answer differs by an angle of ##\frac{\pi}{2}##.
     
  5. Oct 31, 2014 #4

    gneill

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    Staff: Mentor

    Text books have been known to have incorrect solutions on occasion.
     
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