Page 295 - CITS - Electrician - Trade Practical

ELECTRICIAN - CITS

> SUB CIRCUIT = L1, L2, F1, F2, S1, S2
DISTRIBUTION > SUB CIRCUIT = L3, L4, L5, L6, F3, F5, F6, S3, S4
BOARD
> SUB CIRCUIT = P1,P2
3 Prepare wiring layout diagram for customer approval

i current through subcircuit-1
(2 x 60) +(2 x 80) + (2 x 100)/230
=120+160+200=480
480/230=2.08Amp.
ii Current through subcircuit -2
(4 x 60) + (3 x 80) + (2 x 100)
= 240+240+200=680
680/230=2.95Amp.
iii Current through sub circuit 3
= 2000/230=8.696Amp.
Total current =2.08+2.95+8.696=13.726Amp.

16Amp,250V flush type DP main switch is sufficient

Length of conduit and cable

19 mm conduit can be used up to ABC length and for remaining length,12mm conduit in sufficient

Horizontal Runs
19mm conduit for length ABC = 1 m
19mm conduit for length at C (wall thickness) = 0.4 m
Total = 1.4 m

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CITS : Power - Electrician & Wireman - Exercise 105