Sinusoid Analysis

Discussion in 'Homework Help' started by Digit0001, Apr 14, 2011.

  1. Digit0001

    Thread Starter Member

    Mar 28, 2010

    i got a few question i am not sure how should i go about it:

    View attachment tut5qns.doc
    This is what i have done so far:
    Question 1

    So i simplify the circuit i get j5+10

    now to change to phasor

    11.18 phasor 0.46 degrees

    Question 2

    Z1 = 1+j

    Z2 = j

    Z3 = Z1 || Z2

    [text]Z3 = \frac{(1+j)j}{(1+j+j)} = \frac{j-1}{1+2j} = * (1-2j) = \frac{1+3j}{-15}[/text]

    [text]\frac{1}{-15} - \frac{3}{-15}[/text]

    Find V0 = used voltage division = (\frac{Z1}{Z1+Z2})*10 phasor 0 degree
  2. NISM1906

    New Member

    Dec 27, 2010

    - AC analysis can be very difficult and tedious especially when dealing with phase shifts in sine and cosines. To evaluate the first problem you need to determine what the frequency is, based on the information that the problem gives you. For an AC waveform the equation is A*cos(ω*t) where ω = 2πfreq (Not getting into all the details). From this you can find the frequency which, since ω = 200, is 200/2π = 31.83Hz. Now you can calculate the reactance of the capacitor and the inductor while remembering to include the phase angle -90 degrees for capacitors and +90 degrees for inductors. You should come up with very simple numbers i.e. 4Ω@90degrees and 1Ω@-90degrees which when added up equals 3Ω@90degrees. Next add in the 10Ω resistor @ 0 degrees and you are left with 10+3j (in rectangular notation). From here convert to polar notation (look this up) and you will get ~10.44 Ω@16.699degrees (looks familiar?). Now calculate the current by taking the voltage and dividing by the impedance i.e. 50V@ 0 degrees/10.44 Ω@16.699degrees (Note that there is no phase shift for the voltage so the angle is 0 degrees) and the answer is 4.79A@-16.7degrees or 4.79cos(200t-16.7)A. Apply the same principle the second problem and you should be ok. Let me know if you have any problems with this.

  3. Digit0001

    Thread Starter Member

    Mar 28, 2010
    In the second question i get the following, however according to the answers it is incorrect.

    2-j = 2.236@-26.565degrees

    I = 10 / 2.236@-26.565degrees = 4.47@26.565

    V = 4.47@26.565 / 1@45 degrees = 4.47cos(t-18.43)
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    You have correctly calculated both the total impedance and source current.

    The unknown voltage will in fact be the source current multiplied by the parallel branch impedance of (1-j)Ω.

    So v(ω)=4.47@26.565° * (1-j) = 6.325@-18.43°