Electronic – Problem understanding CT output

The clamp MUST go around ONE of the two "live wires" only - NOT around the whole cord.

Add 100 Ohms across the output.
Expect 1 Volt per 20A input.
See below.

How can a transformer produce a proportional current if it has no idea about the load? If I connect a 10Mohm resistor across the connections, will I get 10M * 5mA = 50kV across the resistor?

YES it will try to make 50 kV, just as you calculated. But before then you may get arcing, smoke, flames and fun. To limit your fun it probably has back to back zeners rated at about 20V inside.

DO NOT OPERATE WITHOUT EXTERNAL RESISTOR of 100 Ohms or less.

DO NOT

That is a 100A/0.050 A = 2000:1 CT (current transformer). It is designed to have ~~<= 5V at the output with Iin = max rated.
As it makes current YOU must convert this to voltage by adding an output "burden resistor" Rout.
For 5V at 100 A, as this gives 50 mA out
R = V/I = 5V/0.050A = 100 Ohms.
This gives 5V at 100 A in, and eg 1V at 20A in etc for a single turn primary =- wire through core.

As you increase Vout you start to saturate the core. Keeping Vout sensibly low enhances linearity.

Heavyish but useful reading:

SCT 30A CTlower current version of yours.

Family members. Yours is like the one at top left in the table BUT 50 mA output rated. .

The VOLTAGE OUTPUT ones work EXACTLY the same except that the "burbedn resistor" is already included inside the CT.

Yeeha!!!

A CT (current transformer) is an "ordinary transformer" used in an unusual way.

They are usually used with a "one turn primary" which is produced by running a wire through the hole in the core. With "current mode" CTs, with a 1 turn primary they give the stated smaller current at the output when the stated larger current flows in the one turn primary. For 1 100A:50 mA transformer the primary has 1 turn and the secondary has
1 x 100A / 0.050A = 2000 turns.

There is no magic - just brain rearranging.

For an ideal lossless transformer with 1:N turns ratio:
Vout/Vin = N .... 1
Iin/Vout = N .... 2 <- note in and out swapped
Vin x Iin = Vout x Iout .... 3
Iout = Vout / Rload .... 4
Iin = Iout/N = Vout/Rload/N .... 5

If you are not happy with the above 5 formulae either accept them as standard or get out your Google.
Once happy, proceed. We have no trouble believing these equations (perhaps with a little figuring) BUT miss the implications.

We usually set Vin and Vout and let the current adjust as needed.

BUT with the identical transformer lets instead set Iin and Rload and N and see what you can derive.

More later ...

Lets look at this another way. What is the voltage at the terminals of this circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Will the voltage drop? Simple answer: no. Long answer: Well there is no load on it, so no current flowing, so by Ohm's law, no voltage drop across the resistor.

The same is true for your capacitor, there is no current flow, so no voltage drop across it's reactance.

Now what a capacitor will do in an AC circuit if you have a load on it, is to limit the current flowing, which it can do without dissipating large amounts of power like a normal resistor would - it stores and releases charge. That doesn't mean to say it won't heat up, there are losses in it which limit how much power it can safely transfer. It will also change the power factor a long way from unity which can be bad for whatever is supplying your mains voltage (increased losses in transformers and transmission lines, etc.).

Using a capacitor in this way means you would be building a potential divider circuit with whatever you connect, so it won't be a very good regulator. With little load the voltage will go up (no load = open circuit = full mains voltage), and if you put a large load the voltage will drop as more and more voltage is dropped over the capacitor (so high enough rated X series capacitor!).