[Hamid]

For making of a current injection source with 2000 A rating current, can we use an ordinary current transformer with ratio 2000/1 A, and connect 1 A side to one ordinary low current injection source ( 1 A) for obtaining of 2000 A in other CT side?

For making of a current injection source with 2000 A rating current, can we use an ordinary current transformer with ratio 2000/1 A, and connect 1 A side to one ordinary low current injection source ( 1 A) for obtaining of 2000 A in other CT side?

Topic

[ernairnp]

Assume that on the 2000 A side, the CT has a voltage drop of 1 volts. Then on the 1 A side, to inject 1 A, we require an injection voltage of 2000 volts. The CT secondary side is not designed to withstand this 2000 volts. Hence an attempt to do so will burn out the 1 A winding.

[Hamid]

Creation of 2 KV insulation for current transformers is not difficult, whether are there any other problems? ————————————————————————– Please insert your e-mail address for announce you about new replies. Best Regards Hamid

[ernairnp]

The magnetising current for 5 kv will saturate the iron core and it will be no longer a CT. Regret inability to give email address.

[Hamid]

The 2 KV voltages is the integration of many turn voltage in secondary side. Indeed the linkage magnetizing flux just shall be supported same 1 V for each turn of windings in primary and secondary sides. ————————————————————————— Ok, restful

[Ian Lake]

Practically, you cant do this due to: a) Voltage insulation limitations as mentioned above. b) VA limitations. If you need 1V on the secondary to drive 2000A into the load, the CT needs to be rated at 2000VA. In practice, current injection output transformers with a single turn secondary (as this CT would be) have very large cores to allow operation from a 0-230V input. This is because the transformer has to work at large volts/turn to give the required voltage on the secondary with a single turn. A 2000A injection transformer will typically be 4kVA+ (you need at least 2-3V to drive a few thousand amps down even short output leads). The limiting factor on current is usually the ac impedance of the output leads. Ian Lake

[Hamid]

Ok, Let us assume a real case. Assume switchgear with 2x 100 x 10 mm busbar (2000 A) and relevant current transformer as following characteristics: -Ratio: 2000/1 A – Rated Burden: 25 VA We will have a suitable path for flowing of 2000 A current if 2 meter of mentioned busbar is shorted according to figure in below. Assume we can produce 2000 A current in short circuited paths. In this case the primary short circuit drop voltage can be calculated as follows: V= (2 * 2000) / (56 * 2000) ~ 0.01 V And for consumption burden we can write: S= 0.01 * 2000 = 20 VA < 25 VA ( CT burden ) Is it possible? Can we produce 2000 A in primary side?

[Hamid]

What is your opinion about permitted output stage impedance of current injection source and leakage inductance of 2000 A side?

[R K Mohapatra]

Impedence is always related to total VA rating ( ampere – turns ) , which is anyway persisting in the CT . In a normal Ct , the VA brden being low , this does not play much role . If the core is suitable for operation at 2 kV ( without saturation and without the fear of Ferro Resonance ) and insulation is provided for 2 kV , this is possible .

[Author]

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