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  1. #1
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    How to satisfy ohm law in power transformer using nameplate? Ohm's Law justification

    Suppose there is a power step-down transformer of ratings. [Name-Plate Attached].

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    100KVA

    HV Side specifications:-

    volts-> 11KV
    current-> 5.25

    LV Side specifications:-
    volts-> 0.415KV
    current-> 139.1


    P=VI

    p=11x10^3 * 5.25 = 57750

    P=0.415x10^3 * 139.1 = 57726


    Can any one help me about calculation? I believes that power should be 100KVA as mentioned in the transformer's nameplate?






    Another Question (Power is same):-

    If a man accidentally contacts with the transformer either LV side or HV side which one is dangerous?

    I want to say that high current is harmful or high voltage is harmful?

    From which side man get more shock (LV or HV) as power is same on both sides?



    Consider another case (My Confusion):-

    The overhead cables have the high voltage but little current it is dangerous and may cause Arc Flash. I believe that if a man contact with them (if path completes) he dies due to high voltage.

    We can limit the current by using RCD, MCB etc

    If we limit the current of very high voltage say 33KV to 3A then my question is that if someone contact with them he gets shock or not?


    How to justify the LV and HV side of transformer by ANALOGY OF ELECTRICITY (Flow of Water in Garden Hose) figure attached?

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    Say from HV Side

    cable used is -> 3Cx300mm2


    From LV Side

    cable use is -> 3Cx300mm2


    Both LV and HV have the same cable used so that means have the same resistance but the voltage & current of either side of Voltages is different.

    How to justify it.

    Same pipe area water pressure is high and water flow is slow etc.....
    Last edited by tipu_sultan; 22nd December 2013 at 14:05.

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  2. #2
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    Re: How to satisfy ohm law in power transformer using nameplate? Ohm's Law justifica

    What Ohm's law?

    Quote Originally Posted by tipu_sultan View Post
    ......
    P=VI
    p=11x10^3 * 5.25 = 57750
    P=0.415x10^3 * 139.1 = 57726

    Can any one help me about calculation? I believes that power should be 100KVA as mentioned in the transformer's nameplate?
    That's a three-phase transformer.
    If you have the line to line voltage and the phase current the power will be:
    P = √3*Vl*Ip = 1.73*11x10^3*5.25 = 100kVA


    If a man accidentally contacts with the transformer either LV side or HV side which one is dangerous?
    I want to say that high current is harmful or high voltage is harmful?
    From which side man get more shock (LV or HV) as power is same on both sides?
    From HV side.


    Consider another case (My Confusion):-
    The overhead cables have the high voltage but little current it is dangerous and may cause Arc Flash. I believe that if a man contact with them (if path completes) he dies due to high voltage.
    We can limit the current by using RCD, MCB etc
    If we limit the current of very high voltage say 33KV to 3A then my question is that if someone contact with them he gets shock or not?
    With 3A your eyes will blow.

    http://hyperphysics.phy-astr.gsu.edu...ric/shock.html

    http://en.wikipedia.org/wiki/Electric_shock


    How to justify the LV and HV side of transformer by ANALOGY OF ELECTRICITY(Flow of Water in Garden Hose) figure attached?
    ? ? ? A pressuse booster may be an analogy of transformer, not a garden hose.


    Say from HV Side
    cable used is -> 3Cx300mm2

    From LV Side
    cable use is -> 3Cx300mm2

    Both LV and HV have the same cable used so that means have the same resistance but the voltage & current of either side of Voltages is different.

    How to justify it.
    You are trying to establish a relation between the pig and the speed of light.



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    Re: How to satisfy ohm law in power transformer using nameplate? Ohm's Law justifica

    Quote Originally Posted by tipu_sultan View Post
    Suppose there is a power step-down transformer of ratings.

    100KVA

    HV Side specifications:-

    volts-> 11KV
    current-> 5.25

    LV Side specifications:-
    volts-> 0.415KV
    current-> 139.1


    P=VI

    p=11x10^3 * 5.25 = 57750

    P=0.415x10^3 * 139.1 = 57726


    Can any one help me about calculation? I believes that power should be 100KVA as mentioned in the transformer's nameplate?
    Maybe the current and voltage ratings are for each phase, but 100KW is the total.

    Quote Originally Posted by tipu_sultan View Post
    Another Question (Power is same):-

    If a man accidentally contacts with the transformer either LV side or HV side which one is dangerous?

    From which side man get more shock (LV or HV) as power is same on both sides?
    Both are dangerous, but HV is worse. Touching the LV may elecrocute and kill you. Touching the HV will kill you and burn or cook your body.

    Ohms law: I=V/R
    So the higher voltage will cause more current to flow through the resistance of your body.

    Quote Originally Posted by tipu_sultan View Post
    Consider another case (My Confusion):-

    The overhead cables have the high voltage but little current it is dangerous and may cause Arc Flash. I believe that if a man contact with them (if path completes) he dies due to high voltage.

    We can limit the current by using RCD, MCB etc

    If we limit the current of very high voltage say 33KV to 3A then my question is that if someone contact with them he gets shock or not?
    Less than 100mA is enough to Kill you.



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    Re: How to satisfy ohm law in power transformer using nameplate? Ohm's Law justifica

    Dear,

    I m satisfied your answer about the Power transformer's name-plate. But i have a still confusion about the LV and HV.

    Current is the flow of electron.
    Voltage is actually the force which push electrons and tends them to move depends upon the magnitude of force.
    Resistance is the hindrance which depends upon the wire area and length.


    By considering above definitions we came across the scenario.

    HV side of transformer have a very low current and very high voltage and the opposite is applicable for LV.

    By Ohm Law we know

    I=V/R

    If there is little resistance (say R approaches to zero) then there will be infinite current.
    The above example is true if a person is in-contact with the wires.


    Consider in general that we have voltage from the fixed source is 220V and little current 2A (Power->0.44KVA), the current & voltage is fixed (Limited) means current will not exceed from 2A.

    when we step it down to 10V then the current will become 44A (Power 440V).

    According to me (correct me if i m wrong) how matter how the magnitude (Say 1000A) of current is but there is little voltage (say 2V) it is not dangerous.

    But if the magnitude is voltage is high but current is very small say approaches to zero it is dangerous.


    I will be very thankful to you if you clear my confusion.



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    Re: How to satisfy ohm law in power transformer using nameplate? Ohm's Law justifica

    Nameplates convey ratings.

    Not all ratings can be supported simultaneously in all
    cases.

    Until you understand the basis of them there's no point
    in electrical lawyering, Ohm's or otherwise.

    You seem not to understand the nature of the human
    body as load, in the context of electrocution. You make
    handwaving assumptions about this and it confuses
    your argument. Such as the stuff about 2V and 1000A.
    Not gonna happen, so why try to prove some point
    with such speculation?



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