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Discuss 2330 202 sample test 2 in the Industrial Electrician Talk area at ElectriciansForums.net

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M

Moog

answers 1d 2d 50/75 * 100 = 66.6% 3b R=pl/a so R=pl*(1/2)/2a = pl/a * 4= 1.25ohm 4b 5b 6c 7a 0.00001 = 1E-5 = 10E-6= 10us 8a (conductive, self lubricating, -ve temp coeff) 9d 10d 11b MgO 12c power = 4*1.5 + 2 +3.5 =11.5kW Power = current * voltage (ignoring Power Factor) current= power/ voltage = 11.5E3 / 230 =50 A 13c 14a (best fit) 15a 16b (would be PELV) 17a 18b 19 none of them, RCD does not give short circuit protection the rest are thermal operation 20b 21b 22b 23c 24a ( I am taking the question to refer to exposed conductive parts) 25d or a too thick for cutters 26c 27d 28c 29a ( now mandatory ) 30a
 
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