# How both current and voltage values obtained at secondary of CT are in range of milli-volt/ ampere?

I have a power distribution unit (PDU) used in servers . It has current rating of 32 A. I want to measure this current and convert it into proportional voltage using CT (current transformer). The current of 32 A passing through live cable of PDU . This wire is passed through 1 CT . Its turns ratio is 1:2500. I will connect a burden resistor of 7.5 ohm accros CT and measure voltage across burden resistor .

As per theory of transformers, V1/V2=N1/N2 =I2/I1 . Suppose V1=Vin =230 , I1=Iin=32A , N1=1, N2=2500 . Here , So V2 should be 230*2500= 575000V and I2=Iout should be I1/2500=0.0128A .When I performed simulation of the above schematic, I am getting same I2 value but V2 is 96mV? I think they done following calculation V2=Vout=I2*Rburdon= 0.0128*7.5=96mV .

But I don't understand how it is possible ? Why simulation results are showing V2=96mV instead of 575000V?How both current and voltage values obtained at secondary of CT are in range of milli-volt/ ampere ?What about transformer turns ratio relationship?

The voltage on the primary of the current transformer is the reflected impedance of the 7.5 Ohm burden resistor times the current . So the primary voltage is very small, not 230V. Most of the voltage is dropped across the load. (If the voltage across the CT were 230V, you would have zero volts across your load by KVL, right? You need a very small CT primary impedance for small insertion loss and low power dissipation.)

Reflected impedance in a transformer results from the transformer equations you wrote- The primary impedance will be equal to the secondary impedance (7.5 Ohms) divided by the turns ratio squared. You can derive that easily from the equations. (In a perfectly ideal transformer.) Since the transformer is not ideal there are other factors to take into account which are detailed in the linked document.

Here's a good discussion of how CTs work: Current Transformer Theory

Equivalent circuit excerpted from link above:

• Instead of re-creating all the math I edited my answer to add a link to a really good discussion of how a CT works. Jan 12 '17 at 17:57

As per theory of transformers, V1/V2=N1/N2 =I2/I1 . Suppose V1=Vin =230 , I1=Iin=32A , N1=1, N2=2500 . Here , So V2 should be 230*2500= 575000V and I2=Iout should be I1/2500=0.0128A .

No, no that's not how it works. The voltage seen "across" the primary of the CT is a millivolt at best - it's the slight volt-drop across a small section of your 32 amp cable that is passing through the hole in the centre of the CT. It isn't the rated AC voltage of the 32 amp circuit!

Your burden is 7 ohms - this is where you begin to analyse things. So, 7 ohms attached to a 1:1 transformer looks like 7 ohms on the other side of the transformer but, when you use a transformer with a none 1:1 turns ratio, the impedance of 7 ohms gets stepped-up or stepped-down. For a CT, the 7 ohms gets stepped down.

The ratio it gets stepped down is the turns ratio squared so, 7 ohms on the secondary looks like: -

$\dfrac{7}{2500^2}$ ohms on the primary winding.

That's 1.12 micro ohms. Yes it sounds incredible but it has to be this way for a CT to work and to avoid core saturation. This means that with 32 amps in the primary circuit, the effective primary voltage is about 36 uV RMS.