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Ideal Transformer

  1. May 18, 2010 #1


    related picture: 11760pt.jpg

    Above link, says: "If the voltage is increased, then the current is decreased by the same factor."

    I didn't understand. Is this means, if I increase Vp, Ip will decrease? Or means Is will decrease?

    One more... Suppose an ideal transformer. Vp/Vs , Np/Ns and Is/Ip ratios are equal(like given pic). If I increase Vp then Vs will too because Np/Ns ratio is constant. So, Vp/Vs ratio will be equal to Np/Np everytime. Let's suppose, load resistance is variable(potentiometer) and resistance is increasing;

    1. Vs won't change. Because Vp, Np, Ns didn't change.
    2. Is will decrease. Because Vs = Is * R(Ohm Law).
    3. Ip will decrease like Is.

    As a result: If transformer is ideal and ONLY load(resistance) decrease, neither Vp nor Vs don't change. Only both Ip and Is decrease. Surely if load(resistance) increase both circuits decrease but both voltages stay the same.

    Am I right?
    Last edited: May 18, 2010
  2. jcsd
  3. May 18, 2010 #2
    The best way to look at this is to realise that transformers transfer power.
    The effect of changing the voltage or current or impedance by the turns ratio is a very convenient and useful side effect.
    We actually distinguish (select) transformer types according to whether we want to change the voltage, current or impedance.
    Having chosen what property we wish to change the othere follow like sheep, from Joules and Ohms laws.

    Now power = volts x amps for both the primary and the secondary

    The power stays constant for an ideal transformer since the efficiency = 100%

    [tex]{P_s} = {I_s}{V_s} = {P_p} = {I_p}{V_p}[/tex]

    Does this help?
  4. May 18, 2010 #3
    Yes, i know that equation. And I suppose am right. But asked to be sure. If load resistance change, both currents change but voltages not, isn't it?
  5. May 18, 2010 #4


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  6. May 18, 2010 #5
    thank you :)
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