SSE JE Electrical Engineering 24 March 2021 Shift 2 Paper

Question : 163 of 200

Marks: +1, -0

A single-phase 240-V AC supply is used in a house consisting of 4 fans of 100 watts each and 20 lamp points of 60 watts each. What will be maximum permissible leakage current for the house wiring?

2.66 mA
1.33 mA
0.133 A
1.33 A

Validate

Solution:

Total connected load = 4 × 100 + 20 × 60 = 1600 W

\I_\FL = 1600 / 240 = 6.67 \A

According to IE Rule 48:

\I _{{\lek }}={ \I _{ \FL }}/{5000}

={6.67}/{5000}

=0.001334 \A =1.33 \mA

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