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Electronics Engineering QUESTION #11736
Question 1
An insulation resistance tester (megohmmeter) applies $500\,\text{V}$ DC to motor winding insulation and reads $50\,\text{M}\Omega$. The leakage current is:
  • $10\,\mu\text{A}$✔️
  • $100\,\text{nA}$
  • $1\,\mu\text{A}$
  • $10\,\text{nA}$
Correct Answer Explanation
$I_{leak} = \dfrac{V}{R} = \dfrac{500\,\text{V}}{50 \times 10^6\,\Omega} = \dfrac{500}{5\times10^7} = 10 \times 10^{-6}\,\text{A} = \mathbf{10\,\mu\text{A}}$. For motors, a minimum insulation resistance of $1\,\text{M}\Omega$ per kV of operating voltage is a common field acceptance criterion (IEEE 43). Readings below this threshold indicate compromised insulation requiring rewinding or drying.