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What method is more accurate to find the turns ratio?

  1. Nov 14, 2011 #1
    Which method of determining turns ratio is more accurate and why?
    1. From the voltage readings in no-load condition
    2. From the current reading in loaded condition

    We used an auto transformer (VARIAC) to supply voltage to a single phase transformer. Supplied voltage was 220 V. Then we measured voltages across the high voltage side and low voltage side in no-load condition of the single phase transformer. V1/V2 = M is used to find the turns ratio.

    Then, we connected a load to the low voltage side and then measured currents and voltages both across the high voltage and low voltage side. I2/I1 = M is used to find the turns ratio.

    Now, voltage ratio gives more accurate result than the current ratio. But why's that?

    In no-load condition there's no current flow in the secondary side which results in a very low current flow on the primary side, that is why voltage drop because of the internal impedance in the source is very small and we get almost same result as the ideal transformer.
    In loaded condition, there's current flow in the secondary as well as in the primary, so the voltage drop is larger than the no-load condition. But what I don't understand is that how we are relating current readings in the determination of turns ratio and why the discrepancy is larger here? I understand that the voltages both on the primary and secondary side are being affected because of the load but how's it affecting the current flow?
    We connected one AC ammeter between the auto transformer and single phase transformer and another between the single phase transformer and the load. That means we are measuring Ip and Is. But the ideal transformer equation is N1/N2 = I2/I1, where I1 is not equal to Ip in practical transformer. I can't understand the situation here.

    The discrepancy is in both voltage and current ratio because of the source impedance, load impedance and losses in the primary and secondary side - is my understanding correct here? If it's right, then why current ratio shows more discrepancy?

    Please be kind enough to explain the operation here. Thank you.
     
  2. jcsd
  3. Nov 14, 2011 #2

    rude man

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    I can think of several reasons, but the first to jump at me is that the primary and secondary winding resistances are out of the picture if you run open-load. Well, the primary carries a small amount of current in no-load, but way less than when the secondary is loaded. The secondary resistance is completely out of the picture.

    Also, leakage inductance (primary & secondary) are similarly unimportant if you run no-load.

    A more accurate picture is formed if you use one of the several transformer equivalent circuits.
     
  4. Nov 14, 2011 #3
    No. The best way is to inject a small alternating volatage signal across the primary and compare it to the secondary signal amplitude. The ratio should be the winds ratio. If you suspect the mutual inductance is not nearly one, then inject the signal into the other winding. If the ratios are in large disgreement, mutual inductance is a factor.
     
  5. Nov 14, 2011 #4

    rude man

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    If the coefficienty of coupling is < 1 it will show up exactly the same way whether you excite the primary or the secondary. You will get gain of kN1/N2 one way and kN2/N1 the other way. That's why it's called 'mutual'.
     
  6. Nov 14, 2011 #5
    I understood the fact that no-load voltage ratio is more accurate because almost all the impedance gets cancelled due to very low current flow.
    I also got the fact that in loaded condition due to current flow significant amount of voltage drops across the impedance of primary and secondary side and so this method gives less accurate result.
    What I don't understand is how I can explain the voltage drop problem in the loaded method in terms of primary and secondary current. In the loaded condition both measured voltage ratio and current ratio give less accurate result. I can explain the reason for discrepancy in terms of voltage but how can I explain it in terms of current?
     
  7. Nov 14, 2011 #6

    rude man

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    The fact of primary resistance R1 and leakage inductance L1 means a portion of your input voltage i1*sqrt(R1^2 + XL1^2) is not 'seen' by the magnetizing inductance. So the secondary sees a lower voltage than expected. And secondary current = magnetizing voltage divided by the sum of your load resistance plus R2 plus secondary leakage inductance. Your current ratio will be correspondingly reduced.
     
  8. Nov 14, 2011 #7

    The Electrician

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    When you measure the voltage ratio of the windings, you operate open circuit; you don't want any current in the secondary.

    To measure the current ratio, you should operate short circuit; you don't want any voltage across the secondary. Having a variac on the primary makes it safe to do.

    Turn the variac all the way down. Short the secondary through a suitable ammeter (able to measure the rated secondary current). Apply the output of the variac through an ammeter to the primary and slowly turn it up until the current reaches some reasonable fraction of rated current. Calculate the ratio of secondary to primary current and compare to your voltage ratio.

    Is this number agreeing with the voltage ratio better than when you used a load resistor?
     
  9. Nov 19, 2011 #8
    I get it. Thanks a lot. I can explain it properly now.
     
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