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Those same 0.1 \( \Omega \) leads on a one ampere secondary would contribute a burden of

  • A.0.1 VA
  • B.2.5 VA
  • C.0.5 VA
  • D.0.02 VA

Answer: A

Correct answer: 0.1 VA

\[ 0.1 \times 1^{2} = 0.1\ \mathrm{VA}, \] against 2.5 VA for the same leads on a 5 A secondary. The burden goes as the square of the current, so a 1 A secondary imposes a twenty-fifth of the lead burden of a 5 A one. That is the whole reason 1 A current transformers are specified where the cable run between switchgear and panel is long.

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