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I read that surge arresters in power stations suppress the high incoming voltage and the output passes on to the current transformer.

If surge arrester already reduced the voltage then why is the current transformer needed?

The reason I am asking this question because Ohm's Law states that: V∝I.

enter image description here

Surge arrester.

enter image description here

Current transformer.

EDIT 1: Why surge arresters are used in electric power distribution plants before current transformer?

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    \$\begingroup\$ What else did the source say? \$\endgroup\$
    – Solar Mike
    Nov 15 '20 at 8:15
  • \$\begingroup\$ I’m voting to close this question because lack of research. \$\endgroup\$
    – winny
    Nov 15 '20 at 9:36
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I read that surge arresters in power stations suppress the high incoming voltage and the output passes on to the current transformer.

If surge arrester already reduced the voltage then why is the current transformer needed?

The two devices are not related.

  • The surge arrester protects equipment from over-voltage due to lightning strike or system failure. The spark gap distance is set so that the breakdown occurs at the required voltage and this limits the voltage seen by the equipment.
  • Current transformers have two main functions.
    • Over-current monitoring which will signal the circuit-breakers to trip.
    • Energy metering for billing purposes.

The reason I am asking this question because Ohm's Law states that: V∝I.

If the surge is not arrested then bad things will happen:

  • The insulation (including air insulation) may break down and destroy the equipment.
  • High currents may damage cables and switchgear due to thermal effects.

As V∝I according to Ohm's Law, the surge arrester reduces the voltage. I want to know if the depreciation in the voltage causes current to decrease or not.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A rough equivalent circuit.

You have to remember that the impedance of the transformer will only remain high if the core does not go into saturation. If the core does saturate then the impedance will drop to a low value and huge currents can flow. The way to do that is to limit the voltage applied to the transformer.

The spark gap acts as a potential divider with two modes of operation.

  • In normal mode there is no arc and it's impedance is infinite. The transformer sees the line voltage.
  • In over-voltage such as a lightning strike the arc is struck and the spark gap acts as a potential divider with the upstream impedances - that of the lightning itself and that of the line. The spark gap will keep the voltage low by diverting the fault current away from the equipment. Keeping the voltage low will prevent an over-current in the equipment.

I want to know that is it true that the input current in the surge arrester is greater than the output current which passes on to mext equipment.

If that's what you want to know then it should have been in your question. The answer depends on several factors but mainly on the strength of the strike. If you are asking is the current through the arrestor > current through the load then it just depends on how the current splits and, again, that depends on the strength of the strike.

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  • \$\begingroup\$ I want to know that is it true that the input current in the surge arrester is greater than the output current which passes on to mext equipment \$\endgroup\$
    – Curious
    Nov 15 '20 at 10:05
  • \$\begingroup\$ As V∝I according to Ohm's Law, the surge arrester reduces the voltage. I want to know if the depreciation in the voltage causes current to decrease or not. \$\endgroup\$
    – Curious
    Nov 15 '20 at 10:07
  • \$\begingroup\$ This is in regard with power distribution station \$\endgroup\$
    – Curious
    Nov 15 '20 at 10:08
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    \$\begingroup\$ You still haven't explained what you don't understand about my answer. \$\endgroup\$
    – Transistor
    Nov 15 '20 at 13:01
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    \$\begingroup\$ @Curious yes, it would change, but I think you are asking the wrong question. Although Ohm's law applies here, Kirchhoff's laws are more relevant. This answer describes what happens in a good way. If you don't understand it then I suggest you try to study how the current will divide between two parallell components, where one has practically zero impedance (arrestor during a flash over), and the other one has a comparatively large impedance (transformer). Also, the voltage doesn't pass through anything, the current goes through the arrestor. \$\endgroup\$
    – CG.
    Dec 2 '20 at 10:44

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