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Transformer (Power For Load)

  1. Jun 25, 2014 #1

    dwn

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

    Please see the image attached. We are asked to find the average power (P) delivered to ZL.

    2. Relevant Questions

    (My observation of the circuit) The circuit on the left is as follows: V + I1*(5 + jwL) = 0.
    For the circuit on the right: V = I2*(Z + jωL)

    Is a= N1/N2 or N2/N1?

    3. The attempt at a solution

    (a) Given that there is no source in the left circuit, we use the RE values from the right circuit:
    Code (Text):
    [B]I[/B][SUB]2(max)[/SUB] = 2A    [B]I[/B][SUB]2(rms)[/SUB] = 2/2 = 1A       [B]P[/B] = 1[SUP]2[/SUP] *60 = 60 W

    (b) Since we are given a V1(rms)= 100∠60° V we can calculate V2 by ( a = V1/V2 ).
    Code (Text):
    a = N1/N2 = 0.5       V[SUB]2[/SUB]= 100/0.5 = 200V[SUB](rms)[/SUB]
    [B]P[/B] = V[SUP]2[/SUP]/Z    →   200[SUP]2[/SUP]/60 = 666.67 W
    (c)
    Code (Text):
    V[SUB]S(max)[/SUB] = 120∠0°   By finding V[SUB]1[/SUB], we are able to find V[SUB]2[/SUB].    V[SUB]S[/SUB]= I[SUB]1[/SUB]*5 + jωL
    I thought this was the correct way of solving this since, jωL = V/I but we are not given the frequency ω. So obviously this is not the correct approach.

    Can someone please critique and advise on everything. I do have a couple other questions regarding this circuit, but I will wait until the problem is completed. Thank you very much any help!
     

    Attached Files:

  2. jcsd
  3. Jun 25, 2014 #2

    rude man

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    Homework Helper
    Gold Member



    OK

    No. Your V is right but your Z is not. Use the correct |Z| to get the current, then do I^2 R where R = 60 ohms.
    What is the impedance of the load Z looking at the transformer input? Call that Z'. You have an ideal transformer. What is Z' as a function of Z and a? BTW a = N2/N1 = 2.

    Then you can put R=5 ohms in series with the transformer input to compute I1, and you know I2/I1 = N1/N2. Then use |I2| to compute load power as in (b).

    EDIT:
    It may be simpler to write KVL for the inut and output circuits, realizing V2 = aV1 and I2=I1/a. So you get 2 equations in 2 unknowns: in I2 and V2. Then again P = |I2|^2 R, R = 60.
     
    Last edited: Jun 25, 2014
  4. Jun 25, 2014 #3

    dwn

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    Part B:
    Code (Text):
    Z[SUB]L[/SUB]= 60+j80        I[SUB]2[/SUB]= 200/(60+j80)  = 1.2 - j1.6  
    Magnitude of I[SUB]2[/SUB]= 2A
    P = 2[SUP]2[/SUP]*60 = 240 W
    Part C:
    Code (Text):
    a[SUP]2[/SUP]= Z[SUB]s[/SUB]/Z[SUB]p[/SUB]   Z[SUB]p[/SUB] = 15+j20

    I[SUB]1[/SUB]= 120/(20+j20) = 3+j3       Magnitude of I[SUB]1[/SUB] = 4.24A
    I[SUB]2[/SUB]/I[SUB]1[/SUB] = a        2(4.24) = 8.48A = I[SUB]2[/SUB]
    [B]P[/B]=8.48[SUP]2[/SUP]*60 = 4314.62W
    How does it look now?

    Thank you a lot for the guidance. I forgot about the Zp : Zs relationship.
     
  5. Jun 26, 2014 #4

    rude man

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    Homework Helper
    Gold Member

    Getting there.

    For (c) you first need to reduce the 120V by sqrt(2).
    That reduces your |I1|.
    Then, it's a = I1/I2, not the other way around. Remember, a = N2/N1 = 2 here.
    That reduces |I2| by a lot.
    Then P = |I2|^2 * 60.
     
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