Regarding CT Meter Reading

Quote" Following are the details of Meter : 3x100/5 Amps, 3 x 240 V, 50 Hz, 20 Revolution per KWH having CT of 3x200/5 with a reading of 300."
You usually 1 set of ct's and one set of pt's if needed. I don't know why you have a 3rd set of 200/5. The 100/5= 20 so you would multiply the reading by 20. What is the 200/5 ct's used for? What is the reading of 300? kwh's?

I'm afraid I can't make sense of the CT ratios.

Are the CT's dual ratio with a jumper to select the range?

Your current transformers will give you 5 amps in the secondaries for every 200 amps in the primary. You will need one potential transformer to supply a voltage source which has the same phase rotation as the current to the meter. The meter factor will them depend upon the VT ratio. After that its just arithmetic. Beware that the meter may not accept 10 amps input if it is designed for 100/5 CT's.

As for giving you the answer that would defeat the objective. The next problem you face, you'll be back asking for another answer instead of working it out for yourself.

As the old proverb goes, "give a man a fish and he eats for a day, teach him how to fish he will never go hungry".

I suggest that you find an electrical course to attend that will teach you this... or is this homework?

What did you read...300? then it is 300kWhr. Nothing to bother about the CTs if you have installed the same CT ratio mentioned on the meter.

The OP's problem is that he has a meter designed for 100:5 ratio CTs, and it is fed by 200:5 CTs. He just needs to do a little simple math. What does the change in CT ratio do for the current seen by the meter vs. the actual primary current?

if that is the case then there is something wrong with the whole setup and it needs a competent person to look at it and resolve it.

No amount of advice given here will resolve this particular problem.... safely.

The meter has been manufactured to show the consumption directly in the dial if you connect 3x240v and secondary of ct with ratio 100/5 Amps.

Actually you have connected the meter from a CT having ratio 200/5.

Therfore the MF for this metering arrangement is 2

You have to multiply the consumption shown by the meter by 2 to arrive at the actual consumption

Please see this link first: http://cr4.globalspec.com/thread/36700

Your case is "The energy meter front dial doesn't have any M.F. value and CT ratio is mentioned but the meter is used with CT of different ratio". If the energy meter front dial doesn't have any M.F, consider the Meter M.F is 1. Then it is a new case (probably Case 6).

So, the Actual M.F = 1 x 200/100 = 2.

- MS

The meter is calibrated for 100/5 CT.There will be gear train in the meter for display based on this MF will be indicated in the meter.

Since the CT ratio 200/5 i.e 2 times the calibrated instrument, the readings has to be multiplied by 2.

Supposing you connect 300/5,you have to multiply the reading by 3