# Current Transfomer (CT) ratio setting on energy meter

I am using a smart energy meter which allows me to set the CT ratio among other parameters. I am using 50/5 ratio AC CT's. For the meter to do the kwh calculation correctly, I believe I have to set primary CT ratio to 50 and secondary CT ratio to 5. Now, if I have two rounds of the same 120 volt wire passing through the same CT, then I would set secondary ratio to 100. My question is, will this be equivalent to have just one round of wire through the CT and the primary ratio set to 50 , i.e the kwh calculated by the meter will still be right and I would'nt have to do any manipulations on the values obtained from the meter.

A CT steps down current and a CT specified as stepping down in the ratio 50:5 is the same as stepping down current in the ratio 10:1 so, if you put 10 amps through the primary, you get 1000 mA into your burden resistor (you haven't mentioned this so I assume you know what you are doing or have bought a CT with one inbuilt).

If you put two primary loops through the CT's hole then, in effect you are doubling the ampere turns and if you put 10 amps in you will get 2 A out.

I have no idea what the smart energy meter requires you to enter as values so I can't help you on that one.

I believe I have to set primary CT ratio to 50 and secondary CT ratio to 5.

That's just plain old-fashioned mumbo-jumbo. The primary doesn't have a ratio and neither does the secondary - the primary-in-relation-to-the-secondary is the only thing that has a ratio: -

In the picture above, the primary is 1 (1 wire passing through the hole) and the secondary looks like 10. This means 10 amps in produces 1 amp out (just like your CT).

Now, if I have two rounds of the same 120 volt wire passing through the same CT, then I would set secondary ratio to 100.

Get a grip of this - there is no such concept as a primary ratio (unless there are two primaries and I've never heard of that for a CT).

• I just need to set the CT ratio on the energy meter as I am using ratio CT's (50/5) so that the kwh value displayed by the meter is the actual value so that I dont have to do any math on the value displayed. Feb 2, 2017 at 15:40
• Yes, I understand that but I can't help you with a device that I know NOTHING about. Feb 2, 2017 at 15:48