I understand the concept of a wire having a working voltage and a peak voltage. My question is, physically why is this? What is happening?

I was under the impression insulators had a dielectric strength and when the charge across the two became greater than this strength it broke through? Does the peak voltage take longer to produce charge across the insulator?

  • \$\begingroup\$ I highly doubt it takes longer, but like everything it does take time. So a MOMENTARY, higher pulse might be tolerable, similar to current pulses producing heat. \$\endgroup\$ – DKNguyen May 14 '19 at 13:43
  • \$\begingroup\$ Isn't Wikipedia's Electrical Breakdown enough explaining this? \$\endgroup\$ – Unknown123 May 14 '19 at 15:03
  • \$\begingroup\$ @Unknown123 It's not obvious if it is there. \$\endgroup\$ – DKNguyen May 14 '19 at 16:02
  • \$\begingroup\$ peak takes much less time. ( us) so much higher level. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 14 '19 at 16:12

High voltage insulators have an impulse rating that can be more than 10x the working voltage due to the rise time. This is normally for lightning protection with a specified rise time and tail time. I'm not sure if that is what you meant by peak voltage.

Each commodity item may have different specs and ratios. It takes longer to ionize air or surface contamination or charge up wire insulation dielectric with a series inductive wire.

There can be spark tests at 60 Hz, 3kHz and Impulse test (peak) which increase in this order.

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