There may be variations, but usually the desired output from a CT is 0-5 amps. Determine the amps of the current passing through the center of the CT, and select the ratio accordongly. For PT, determine the voltage applied to the primary coil, and the output voltage you need. Select a ratio to match.

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When you are connecting to a meter, you also would need the meter's faceplate to be printed with the proper scaled values. On our switchgear, the incoming voltage is nominally 12.5kV and the PT ratio is 12,000:120 or 100:1. This gives a secondary voltage that is well-within the normal insulation range of building wires. Before applying any PT's, it would be well to determine the actual type of connection for the incoming power--Wye (Star) or Delta, because this will also dictate the ratio and connections for the PT's.

Similarly to this, for CT's the same scaling of the meter faceplate is needed. It may be ordered with a 0-5A movement, but scaled so the 5A corresponds to the full-scale primary current of the PT.

CT output is 0 to 5 amps, if you are connecting to a map chart recorder such as a Bristol. The idea is to keep the output from CT to chart recorder at 2.5 amps, so with that in mind read the following...........................Current Transformers
Electronic and mechanical measuring devices monitor the amperage in the
switchboard. To monitor the motor current, it must be reduced to a level that
is easier to measure. If the amperage drawn by the motor is not reduced,
measuring devices have to be large and cumbersome in order to house
components capable of withstanding high currents. Most submersible pump
system measuring devices (motor controllers) are designed to operate from 0
A to 5 A. To reduce or scale-down the actual current in the switchboard to an
acceptable level, current transformers are used.
Current Transformer Ratios
The current transformers should be sized so that under normal operating
conditions, secondary control amperage levels are between 2.5 Aand 3.5 A. This
is done to obtain the best motor controller performance and to have the pen track
in the center of the recording ammeter char t.
The current transformer size calculation is —
current transformer size = expected amperage through the switchboard x 1.6................next part............Example
Motor and Switchboard Information
39 A, 990 V required motor surface voltage; MDFH size 3, 1,500 V switchboard.
Using current transformer size = expected amperage through the switchboard x
1.6,
current transformer size = 39 A x 1.6 = 62 A
Choose a current transformer with a ratio as close to 62:5 as possible. (In this
case, a 75:5 ratio.)
(39/75) x 5 = 2.6 A
hope this helps

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Hi, Could you please tell me where the 1.6 factor (which is applied to the load current in your calculation) comes from? is this just applied to allow for flexibility in full load current? or have i missed something?

My friend, if you don't like the answer, there's not much I can do about it, your question "what is the 1.6 for?" was answered. It is part of the formula I have used for many years without fail. You have asked this forum to which many people reply free of charge to help other people, technical and non-technical to find answers to their technical problems. If you are not happy with the answer or ridicule the person suppling the answer, then don't use this forum, and remain ignorant. All you need to know is that without the "1.6" in the formula you will never get the correct answer you require. So unless you can prove to me you know more than I do on this matter, I have nothing more to say. I am always open to learn, but I suspect you are not, hence your comments.

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The square root of nothing is what you make it!

Dont get me wrong, i appreciate the reply, but as an engineering student, submitting technical information to professional institutes to demonstrate an understanding of basic engineering theory, i am trying to gain an understanding of WHY a factor of 1.6 is applied to a current, not simply "that it must be done". Where this information is sourced is not important, but understanding it is key.

that my friend is a FAR FAR better reply, rather than the previous answer. I thought you might be in college/university, so give me time and will TRY and get a FULLER answer for you. However I would ask this.... why do you apply 3.14 to every equation when calculating the circumference of a circle? Because it's "Pie" and we know it works, same theory, but I will look for answer for you OK?

__________________
The square root of nothing is what you make it!

In reading the exchange you are part of, I believe the answer of the 1.6 factor is embedded in the original answer. IF you are wanting to get the current reading marginally above the midpoint of the meter range (as suggested for keeping a chart recording pen near the middle of its full scale), then the 1.6 is simply a scaling factor for shortening the range of the full-scale reading. That may be a valid approach, depending on the nature of your data needs. Problems with this approach include an increased % reading error and decreased ability to read very low loads.

All-in-all, your approach needs to be determined problem being solved. If accuracy is important, scaling the reading near full-scale is better. If ability to read overloads is important, scaling for the normal load at (say) 60% of full-scale is more reasonable. Whatever you do, make sure that any assumptions and limits to the validity of the readings are included in your documentation. That also is good basic engineering practice.

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