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Using wattmeter with current transformer

  1. Mar 27, 2012 #1
    Hey guys,
    Just looking for some reassurance as to whether I have visually depicted the wattmeter and current transformer correctly in the image I have provided. Needed to step down current to satisfy current ratings of wattmeter. One thing I cannot seem to see is how it detects the voltage difference if it's not physically connected to node A. Have I drawn this correct? Red is transformer, blue is wattmeter.
     

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  2. jcsd
  3. Mar 29, 2012 #2
    You have connected the potential coil of wattmeter wrong. The other point of potential coil should not be connected to the current coil of wattmeter. Connect the across the load.
     
  4. Mar 29, 2012 #3

    psparky

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    Incidentally....how to current transformers lower current without raising voltage? I see these current transformers all the time on single line drawings.

    What am I missing?
     
  5. Mar 29, 2012 #4
    I figured it was something like that.. I could just connect the potential coil of the wattmeter to the top current coil of the current transformer (rather than the bottom one as depicted in the image.
     
  6. Mar 29, 2012 #5

    Averagesupernova

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    The same way a voltage transformer raises voltage without lowering current. Sound impossible? Consider an unloaded voltage transformer. Other than magnetizing current there is no primary current. Increase the windings on the secondary and the voltage will get higher but the current will not go lower because it is already zero.
     
  7. Mar 29, 2012 #6

    psparky

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    Hmmmm....it does sound impossible. Without current flow thru the primary....there is no magnetic field.......how can you change the voltage?

    Explain more how it works for the current tranformer please.
     
  8. Mar 29, 2012 #7

    berkeman

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    A current transformer is often just an N:1 transformer, since you put the wire with the current to be measured through the core (only 1 turn), and you have some number of windings on the core to measure that current. The measuring circuit is usually high impedance to reduce parasitic effects of the measurement on the current in the main wire.

    http://www.pci-pcmcia-express.com/uploads/allimg/1089/1_108902758_1.jpg [Broken]
    http://www.pci-pcmcia-express.com/uploads/allimg/1089/1_108902758_1.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  9. Mar 29, 2012 #8

    psparky

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    Ok.....not quite there yet....but here are my thoughts.

    IN a normal transformer, the voltage and current are exposed from one coil to the other so power in = power out.

    It looks like in the example above....the secondary is getting its power (current) from the rotating magnetic field from the wire on the left. So in this case....it essentially does not sense the voltage from the wire.

    I'm assuming this is how those fancy clip-on amp meters work.

    So if I want to go from 500 to 5 amps.....or 500/5......does this mean they use 100 wraps of wire around the core....or something like that?
     
  10. Mar 29, 2012 #9

    Averagesupernova

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    psparky, in the ideal world, there is no difference between a current and a voltage transformer. In what you call a 'normal' transformer the current and voltage are exposed from one coil to the other so power in = power out, just as you put it. But in a current transformer, guess what? It is the same. The difference is that the voltage is sooooooo low that very little actual power is transferred. And in a potential transformer, one that transforms voltage for measurement purposes, the current is soooooo low that very little actual power is transferred. Current transformers and potential transformers are transformers that are optimized for specific jobs. Power transformers are a sort of cross between them.
     
  11. Mar 29, 2012 #10

    psparky

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    Ok....so power in power out...word....thank god...I was almost going to sue my school...:)

    So you are saying there is very little voltage around the insulation at the wire even with a 5,000 volt wire.....say .1 volt for example...and 500 amps.....so in the current coil....when I drop to 5 amps....the .1 jumps to 10 volts....relatively low?
     
  12. Mar 29, 2012 #11

    jim hardy

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    Sparky -
    I know from many of your posts you are a picture thinker like me.
    And i distinctly remember struggling with exact same questions.

    simplify it in your mind.

    Operate an ideal transformer into a short circuit on its secondary.
    But - limit the primary current to one amp. Use maybe a 120 watt lamp on a 120 volt outlet
    What's secondary current?
    What's voltage across primary winding? (hint -how much of the voltage is dropped across whatever you used to limit limited primary current)
    Transformer equations apply, turns ratio defines both.

    What flux is required to make that primary voltage?

    Now let primary current increase to two amps and repeat above.

    There was a very interesting thread on CT a week or two ago - i learned a lot. See https://www.physicsforums.com/showthread.php?t=586722
    They can be made to have an effective non-integer turns ratio.



    Current transformers have extremely good cores so they're close to ideal when operated at modest voltage..
    If you ever get a chance lay your hands on one of those huge transformer bushing CT's they make a terrific arc welder for your home workshop. I once had one ~ two foot diameter . Gave two volts per turn of welding cable.
     
    Last edited: Mar 29, 2012
  13. Mar 29, 2012 #12
    Start with Faraday's Law:
    [tex] \oint E\space d\ell= - \frac{d}{dt}\int_{A}^{}B\cdot n \space dA [/tex]
    The induced voltage in an N-turn loop around the perimeter of a cross-sectional area Ao is equal to minus the time derivative of the perpendicular magnetic field integrated over the area inside the loop. If B(t) = Bosin(ωt) at frequency ω, then the peak output voltage is
    [tex] V= \omega NA_0 B_0 [/tex]
    Using B0 = 1.5 Tesla (max peak field without saturating iron), the maximum voltage output for a N=10 turn secondary at 60 Hz is V = 377·10·1.5 = 5700 volts per square meter.
     
    Last edited: Mar 29, 2012
  14. Mar 30, 2012 #13

    jim hardy

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    PS -

    we get so accustomed to using transformers for power conversion

    that we forget the only fundamental difference between a current transformer and a power transformer is how it's connected in the circuit.

    In a power transformer the current is set by load connected in parallel with secondary.
    In a current transformer the current is set by load connected in series with primary.
     
  15. Mar 30, 2012 #14
    It is also important to understand that in a power transformer, the input is a voltage source, while in the current transformer (CT), the input is a current source.

    In a CT, if there is no secondary, the current in the primary will drive the iron into saturation. This is also true if the secondary termination resistance R1 (in Berkeman's attachment, post #7) is too high.
     
    Last edited: Mar 30, 2012
  16. Mar 30, 2012 #15

    psparky

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    Thanks guys....I'll try to take a good look at it over the weekend and reply with my latest revelation.....lol.
     
  17. Mar 30, 2012 #16

    psparky

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    Ok....I think bells just started ringing........

    Lets take a simple clip on ammeter. It is obviously using this CT principle.

    The single conductor going thru the clip on meter is the primary. Like you guys said...the winding of the primary.....is the wire you are measuring. The clip on meter is the secondary. The secondary has 50 windings or whatever the ratio you want. The mysterious voltage that generates in the primary I belive to be the voltage drop in the area of the wire you are measuring......the resistance is quite small....so the voltage is quite small. Whatever the ratio is....the voltage will pump up accordingly.....but since it's so small....it is insignifigant. But yes, power in = power out.

    Now....if I take the wire thru the ampmeter....and wrap it around the amp meter clip 5 times.....I just made the primary winding 5 times as big.....therefore lowering the amps on my meter by a factor of 5.

    In additon....in regards to the resistance in the secondary being too high.....I think a Norton equivalent is a good way to look at this. If you take a 5 amp current and throw it thru a 1 ohm resistor....no biggy.....5 volts. Now....take that same 5 amp current and throw it thru a 1,000 ohm resistor.....and whalla~......5,000 volts!!!

    Am I getting warmer?
     
  18. Mar 30, 2012 #17

    psparky

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    Wow Bob!!! You are certainly brilliant!!!

    However, reading this makes me feel like an idiot. Picture a hamster on a wheel who is passed out for a week. Nobody home.

    I would probably have a clearer understanding of what you said if you wrote it in Chinese.
    Thanks though.....I'm sure somebody understood it!!!

    Who's this Faraday guy?
     
    Last edited: Mar 30, 2012
  19. Mar 30, 2012 #18

    jim hardy

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    Indeed !
    a CT should never be asked to produce very many volts on its secondary else it becomes inaccurate.

    Low secondary volts measns low primary volts, so the insertion loss in circuit whose current you're measuring is small.

    With a one turn pass thru the voltage induced in that primary wire going thru donut hole is miniscule. Call it counter-emf if you like.

    Dont you love it when something clicks ? I still do (no innuendo intended) .
     
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