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Utility Metering Current Transformer

  1. Aug 12, 2015 #1


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    The electric utility has a digital wattmeter that seems to be measuring the power usage at the high voltage side of our distribution transformer (11 kV - 420 V 3 phase, 250 kVA) via a Current Transformer.

    How exactly does this work? Specifically, what I'm confused about is that transformers generally obey a VxI=const. relationship (barring losses). So is this CT stepping down the current or is it stepping down the voltage?

    Wikipedia says this:

    "When current in a circuit is too high to apply directly to measuring instruments, a current transformer produces a reduced current accurately proportional to the current in the circuit, which can be conveniently connected to measuring and recording instruments. A current transformer isolates the measuring instruments from what may be very high voltage in the monitored circuit. "​

    This sounds to me like they are saying it is reducing the current as well as reducing the Voltage. How can a transformer do both at once? What gives?
  2. jcsd
  3. Aug 12, 2015 #2
    The voltage of the conductor referenced to earth is 6500 V. But the voltage between any two close points along the conductor is (almost) zero

    The primary of the CT is the HV conductor itself, which is essentially one turn. The Secondary has many turns.

    The CT is placed along the conductor (the primary), so what it is doing is 'isolating' the secondary small wiring from the high voltages of the circuit.

    CTs are also used in LV metering where the current is too high (>100A) for direct metering.

    It works just like your clip-on ammeter.
    Last edited: Aug 12, 2015
  4. Aug 12, 2015 #3


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    Thanks @William White!

    So my confusion was I was assuming the CT-primary sees the rated primary potential whereas in reality it only sees a small potential just the drop across the conductor for however large the CT it.

    In any case, given that the primary is one turn (in effect) and the secondary many turns, the CT will be a step-up TX in terms of Voltage? i.e. Current reduces?

    i.e. If the current of the distribution TX primary were 10 A then the CT will actually see less than 10 A etc. depending on how many turns?
  5. Aug 12, 2015 #4
    most CTs for power systems protection are rated at 1A or 5A.

    So you will see written on the side of the CT (or on the protection relay) 100:5 ; which means 100A of primary current give 5A secondary current

    so you have

    Is = Ip Np / Ns

    where Is is secondary current, Ip is primary current Np is primary tunrs and Ns is secondary turns

    the voltage across the secondary circuit is just ohms law
    Vs = Is Rs

    because Is is small (just 1A or 5A) and Rs (the resistance of the secondary circuit) is small (milli ohms) the voltage is small.

    This is where it is essential not to open circuit the secondary! If you do do that the CT becomes a voltage transformer and you get huge voltages across the CT secondary.
    If your primary voltage is 6400V and your CT has 100 turns on the secondary, then if the open circuit CT voltage:
    Vs = 6.4kV x 100 = 640 kV!
    Last edited: Aug 12, 2015
  6. Aug 12, 2015 #5


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    Hmm....I don't think I get this part.

    Like you explained in your earlier post shouldn't it act like a step-up but the primary voltage shouldn't be the full 6400 V but only " the voltage between any two close points along the conductor"? After all the CT primary winding isn't seeing the conductor-to-earth full PD? Only the primary Distribution Transformer Winding sees this?
  7. Aug 12, 2015 #6
    If you open circuit the secondary there is a huge increase in the magnetising flux in the secodary winding which induces a voltage in the secondary winding. In theory, an open circuit has resistance of infinity (it doesn't in practice because the air is a conductor, so there is no true purely open circuit - even the vacuum in a VCB is not an infinite resistance); so you might think the voltage would increase without limit. What happens is the flux increases until the secondary is saturated or the thing blows up. It is one reason why you should never ever put any protection (fuses/breakers etc) in the secondary circuit. If the protection operated, the circuit is open and you get very high voltages.
    Last edited: Aug 12, 2015
  8. Aug 12, 2015 #7


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    The secondary of a CT is loaded by a low valued resistor.

    Say that N1 = 1 , I1 = 100A , N2 = 1000 , then I2 will be 100A/1000 = 100 mA. Now load the secondary with a 10Ω resistor and you will get a voltage = 1V. That's the one to be measured.

    You may never install a CT without having the secondary properly loaded due to very high no-load voltage.

    The core of the CT will not be saturated because N1*I1 ≈ N2*I2.
    Last edited: Aug 12, 2015
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