It measures alternating current flowing through a conductor. Since it is used to measure current, a current transformer is often classified as a type of instrument transformer. One could measure the voltage drop across a known resistor.
This is okay for low current applications but is often impractical for high current applications. The resistor consumes a lot of power (lowering efficiency) unless the resistor is very low in value, in which case there may be very little voltage to measure. The resistor could be excessively large. The resistor’s heat may affect the resistor value, thereby reducing the accuracy of the measurement. A current transformer can accurately measure the alternating current and put out a reasonable voltage, which is proportional to the current, but without as much heat and size that an appropriate resistor would require.
The current transformer can perform its function with very little insertion loss into the conductor current being measured. The current transformer also provides voltage isolation between the conductor and the measuring circuitry. Proper function of a current transformer requires use of a load resistor. The load resistor is often referred to as a “burden resistor”. In the typical current transformer application, the primary winding consists of one to a few turns of wire. The primary wire size is much larger than the secondary wire size. The number of secondary winding turns is a selected multiple of the primary turns. Figure 1 gives a circuit schematic of a current transformer application. The current transformer shown represents an ideal transformer. The ideal transformer has infinite no-load input impedance, 100% magnetic coupling between transformer windings ( hence no leakage inductance), zero winding resistance, zero core losses, and no capacitance. ( Capacitance, leakage inductance, winding resistance, and core losses are considered to be parasitic components. ) The output voltage is exactly proportional to the primary voltage times the turns' ratio. There is no regulation drop. There are no losses. Since there are no parasitic components the ideal current transformer is 100% accurate. The conservation of energy requires that the output power equals the input power, hence Vp x Ip must equal Vs x Is. Since Vs = Vp x Ns / Np, it can be shown that Is = Ip x Np / Ns. Is = Vs / RL, hence Ip = Ns x Vs / ( RL x Np ). With an ideal current transformer there is no phase shift ( except 180 degrees depending on the choice of output connections ).