Transformer Question

Discussion in 'Homework Help' started by exidez, Sep 29, 2008.

  1. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    I have the following Circuit with the information below:
    It is an ideal transformer.
    [​IMG]

    i need to find the number of turn ratios so that maximum power is transferred to the 10 Ohm resistor.

    I used (V^2)/4R
    (50^2)/(4*1000) = 625 mW

    then 625 mW = (V^2)/(4*10)
    V = 2.5V

    50/2.5 = 20

    ratio 1: 20

    Is this correct?

    part c)
    Calculate I1, I2 and V2

    I have no idea how to do this.
    What is the resistance of the Tansformer?
     
  2. S_lannan

    Active Member

    Jun 20, 2007
    247
    2
    First i would take the impedance ratio 100 : 1
    To find the turns ratio needed i would take the square root of the impedance ratio 10 : 1

    Take in mind i chose that ratio so the secondary side will reflect an impedance of 1000ohms, maximum power is transferred then.

    The only mistake you made was in assuming the full 50V would appear on the primary side. Thats why your turns ratio is wrong.
     
  3. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    that seems simple enough and easy to remember. Thanks
    So how about working out the current?
    I can work out the secondary side if i knew the primary input voltage but because there is a voltage drop before it goes to the transformer i don't know it. I could work out I1 but i need total impedance, but what is the impedance of the transformer? Or do i neglect that?
    is I1 simply (50 < 0) / 1000 ?
     
  4. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    exidez,

    If the transformer is ideal, and the primary reactance is much greater than R1, and the secondary reactance if much greater than R2, then the primary impedance of the transformer is R2/N^2 = 1000, where R2 = 10 and N = 1/10 .

    Ratch
     
  5. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    however in this case i will be neglecting the impedance of the transformer yeah?

    so would it be safe to assume that input current it 50 V / 2000 Ω = 25mA ?
    meaning a voltage of 0.025 * 1000 = 25 Volts

    therefore output Voltage will be 25(1/10) = 2.5V
    and output current is 25(10/1) = 250 mA


    is this what is meant by an ideal transformer?
     
  6. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    exidez,

    Your calculations are correct.

    Sentences in English start with a capital letter.

    An ideal transformer has 100% coupling, and there is negligible leakage inductance and leakage flux. An ideal transformer has zero losses. This means negligible dc winding resistance. It also means negligible ac winding resistance and core losses at the frequencies of interest. In an ideal transformer, either the primary and/or the secondary winding reactance is very large relative to any impedance in the secondary loop. The power put into the primary winding will equal the power dissipated in the load.

    Ratch
     
  7. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    Thanks, you help is much appreciated. That cleared up my doubts.
     
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