In a current transformer, the primary current causes a magnetic field in the core, which in turn generates a current in the secondary. Fine.

How come then that a power transformer outputs a voltage, and not a current? Isn't it the same principle?


3 Answers 3


A transformer is a transformer whether intended for current sensing use or power conversion use. All transformers work on the same principle.

However, there is considerable latitude in various parameters when designing a transformer. These different tradeoffs give the transformer different characteristics and therefore make it suitable for different applications.

A current sense transformer is optimized to have small primary impedance so as to minimize the voltage drop in the line it is intended to measure current in. The secondary is also intended to be connected to a low resistance. This reflects a lower impedance to the primary. The transformer is run primarily in short-circuit output mode. Note that little power is transferred thru the transformer. Energy is taken from the magnetic field by the secondary almost as soon as it is put there by the primary. As a result, the core can be small since it never has to hold much energy at any one time.

A power transformer has a different purpose, which is to transfer power from the primary to the secondary. Sometimes they are just for isolation, but often it is also to get a different combination of voltage and current on the output than the input. To get power, you need both voltage and current, which means the transformer needs to be operated somehwere between short circuit output where there is no voltage and open circuit output where there is no current. Generally power transformers are designed so that the secondary looks reasonably low impedance and therefore it's voltage doesn't sag too much at the rated power output. They also have to behave reasonably with light load or no load, meaning the open circuit case. Again you want low impedance so that the voltage in the light load case is not too different from the full load case. This type of transformer has to be able to handle larger energy in the magnetic field. This means a physically larger and therfore heavier core.


The difference is not in the phisical principle, just in the usage.

The power transformer is used to convert voltage using the number of windings in the two coils as ratio, while the current transformer is just an inductor placed around a wire to sense the magnetic field caused by the changing current. So you use it to measure the (AC) current without breaking the circuit.

But both the transformers output a voltage, basically, which is given by Faraday's induction law. The difference is that the power transformer is voltage driven, and the current is determined by the load on the other winding.

Update for comment

The transformer's principle is that a changing current will induce a magnetic field, and the magnetic field induces a voltage. Then there is Ohm's law, which implies that for a voltage applied to a load, you have a current proportional to the load's resistance.

If you put them together, you have an infinite loop in which the current in the load has an influence on the magnetic field that generates the voltage on the load itself. That's how the current in the primary of the Power Transformer is determined.

About the current transformer, you want the biggest possible load to avoid a significant current flowing in it, because it generates that feedback effect.

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    \$\begingroup\$ But why doesn't the CT show a voltage proportional to the number of windings, even when it wouldn't be terminated? After all the primary's winding number is 1. \$\endgroup\$ May 2, 2012 at 12:39
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    \$\begingroup\$ I'm going to edit the answer, it's simpler. And the winding number is not even 1, unless you wrap the wire around the toroid. \$\endgroup\$
    – clabacchio
    May 2, 2012 at 12:44
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    \$\begingroup\$ @Federico: Actually a current transformer will produce a open circuit voltage that is the primary voltage times the turns ratio. Consider what the primary voltage is though. The primary is often just a single turn (or less!) of wire, so there isn't going to be much voltage accross it. \$\endgroup\$ May 2, 2012 at 12:48

Simple summary:

A current transformer is a "normal" (voltage in):(voltage out) transformer that is optimised for a special task.

A current transformer is ALWAYS operated with a defined load resistor.

A constant K is able to be calculated based on load resistor and turns ratio such that
Iin = Vout x k. See below for details.
So Iin can be determined by measuring Vout.

Despite the name, a current transformer works according to the standard transformer related equations (ignoring non idealities such as winding resistance). The primary is usually effectively a single turn, produced by running a wire carrying the circuit to be measured through the core. :

  • Vout = Vin x Turns_Out/Turns_In ...... (1)

Turns in = primary turn or turns.
Turns out = secondary turns. Define Turns ratio = TR = Turns_out/Turns_in

  • Vin x Iin = Vout x Iout ...... (2)

  • Iin = Iout x Vout/Vin ...... (3) = rearrangement of (2)

BUT if we have a resistive load = Rout then

  • Iout = Vout/Rload ...... (4)


  • Iin = Vout/Rload x Turns_out/Turns_in ...... (5) - combination of 1, 3, 4 above. or
  • Iin = Vout x TR/RLoad ...... (5b)

    (So Vout = Iin x Rl / TR) ...... (5c)

For a given Rload and given turns ratio TR/Rload is a constant = K say so

- Iin = Vout x K ...... (6) <- target result

So for a given load we can determine Iin from Vout multiplied by a constant.

Some current transformers have Rout included as part of the assembly.
Some CTs need Rout added.
Failure to add an Rout gives Vout = very very very big, but usually not for long.

Usually the input "winding" is a single turn or a wire passing through the core. Using multiple turns or looping a wire carrying the target current through the core several times decreases the turns ratio - so (see 5c) Vout drops.

Iout as to be such that the core does not saturate and operates as linearly as possible to Rl and thus Vout may not be "too large". Max Rl and/or Vout are specified by the manufacturer.

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    \$\begingroup\$ In a CT, Vin will depend on Iin and Iout. Equation 5c would imply that Vout approaches infinity as Rload approaches infinity, but with the secondary open-circuit, Vout = M.dIp/dt where M is approximately Ls/(turns ratio). In general, Vout is a linear combination of Iin and dIin/dt \$\endgroup\$
    – MikeJ-UK
    May 2, 2012 at 14:13
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    \$\begingroup\$ @MikeJ-UK - As I'm sure you know, in a CT Vin is of very little interest in most cases and is a byproduct of the process. You are correct about what 5c implies and right enough about what happens in the very very non ideal case of running a CT secondary O/C. But in practice, as I noted, when a CT is run as a CT the output is ALWAYS terminated by a lowish value resistor to keep Vout low and the core in the linear region. An unterminated CT in operation is NOT a CT. It may be a fireworks display, spark generator, simulated eruption or harrowing experience :-). \$\endgroup\$
    – Russell McMahon
    May 2, 2012 at 14:21
  • \$\begingroup\$ Perhaps my first sentence was misleading. Ideally a CT should be terminated with a short circuit (not often useful in practice - granted!). My main point was that an unterminated CT will not necessarily produce a high output voltage. \$\endgroup\$
    – MikeJ-UK
    May 2, 2012 at 14:43

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