EIN02 Answers 18 June 2011

7/27/2019 EIN02 Answers 18 June 2011

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FINAL - EIN02 - ANSWERS – 18 JUNE 2011 1

FINAL - EIN02 – Electrical Installer Theory Examination Marking Schedule

Notes:1. (1 mark) means that the preceding statement/answer earns 1 mark.

2. This schedule sets out the accepted answers to the examination questions. A markercan exercise their discretion and decide on the overall accuracy of any answer that ispresented in the candidate’s own words.

3. Symbols and terms - alternativesPower W or PVoltage V or E or UPhase Active

Question 1 Reference

Marks

Marking notes

(a) Any TWO of:

● The electrical installer did not testbetween each phase and earth.

● One phase was still live.

● Testing phase to phase is not effectivein these circumstances

(2 marks)

(b) Any FOUR of:

● Short circuit

● Under voltage

● No voltage

● Loss of one phase

(2 marks)

(c) (i) This is the time it takes the fuse tointerrupt the current flow

(1 mark)

(ii) This is the time it takes to interrupt theflow of current and extinguish theflame.

(1 mark)

(d) Any ONE of:

● The final subcircuit neutral conductorprovides the return path from the loadsback to the distribution transformer forthe resultant “out-of-balance” currentfrom the three phases

● The neutral conductor is required toensure that the phase voltage ismaintained at its nominal value withrespect to earth.

(2 marks)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(e) ● Current (load)

● Length of run, or similar

(2 marks)

(f) Any ONE of:● Star/delta starter

● Any type of reduced voltage starter

(2 marks)

(g) Any ONE of:

● An imbalance in current between phaseand neutral

● An earth leakage current exceeding 30mA

(2 marks)

(h) (i) The copper shading ring (1 mark)

(ii) Turn the motor or rotor around (1 mark)

(i) I = V

R

(½ mark)

= 240

24

(½ mark)

= 10A (1 mark)

(j) Any ONE of:

● It provides a low resistance returncircuit to the distribution transformer,primarily though the main neutral

● Under fault conditions a large currentwill flow and circuit protection devicewill operate and disconnect the circuit.

(2 marks)

7/27/2019 EIN02 Answers 18 June 2011



Question 2 Marks Reference Marking notes

(a) (i) Ifault = 230

(7 + 0.5)

(½ mark)

= 30.66A (½ mark)

Itotal = Ifault + Iload = 30.66 + 25 (½ mark)= 55.66A (1 mark)

(ii) The 30A fuses have a fusing factor(gG Utilisation Category) of 1.5

Fusing current = 1.5 x 30 = 45 A

(½ mark)

Total current is 55.66A, which ismore than the fusing current of 45A

(½ mark)

The fuse will operate. (1 mark)

(b) (i) Ifault = 230

(7 + 12)

(½ mark)

= 12.1A (½ mark)

Itotal = Ifault + Iload = 12.1 + 25 (½ mark)

= 37.1A (1 mark)

(ii) Total current is 37.1A, which is lessthan the fusing current of 45A

(½ mark)

The fuse will not operateor

The fuse may take a long time tooperate.

(½ mark)

(iii) Vd across Earth conductor equalsshock voltage

VdE = I x R

= 12.1 x 12

(½ mark)

= 145.2 V (½ mark)Electric shock hazard of 145.2Vframe to earth exists

(1 mark)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(a) (i) Any TWO of:

● Easy to reset

● Provides thermal and magneticprotection

● Current rating cannot beinterfered with

● Multi-pole device, fault on onephase will disconnect all three

● Can be operated many times

● Can be locked

(1 mark)

(ii) Any TWO of:

● Slower operating speeds

● Must wait for thermal to reset

● Fixed fusing factor

● Mechanical mechanism could fail

● Can be reset under faultconditions

(1 mark)

(iii) Any TWO of:

● Good discrimination● Good back-up protection

● Fast operation

● Range of utilisation categories(fusing factor)

● Higher rupturing capacity

(1 mark)

(iv) Any TWO of:

● Not easy to see if it has operated

● Many different physical sizesavailable

● Difficult to discern rating,particularly if been in service forsome time.

● Can be easily bypassed

(1 mark)

(b) Any TWO of:

• If the fuse blows again an arc may beestablished between the fuse terminalscausing damage or injury

• Fuse wire does not have the same

(2 marks)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

Utilisation category (fusing factor) asthe HRC fuse

• Cannot safely interrupt short circuitcurrents of much higher values.

• Fuse wire may protrude past the holderwhich creates an exposure to shock.

• Suitable fixing for the fuse wire is notgenerally available.

• Fuse holder is not fire proof.

● Cannot provide "close" protection.

• Cannot ensure reliable operation withinprescribed limits.

• Poor discrimination.

• Fusing characteristics not constant.• Slower operation/acting.

• Deteriorates over time.

(c) (i) The overload:

● causes a bi-metal to heat up andbend,

(1 mark)

● operating a trip mechanism thatdisconnects the circuit

(1 mark)

(ii) The short circuit

● causes a strong magnetic field (1 mark)

● that attracts a trip mechanismthat disconnects the circuit rapidly

(1 mark)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(a) (i) The isolation method has to show:

● Identification of the correct fuses. (½ mark)

● Re-inserting empty carriers intofuse bases. (½ mark)

● Prove-test-prove to ensure circuitis isolated.

(1 mark)

● Steps taken to preventreconnection

- Danger tag for personalprotection

(1 mark)

- Locking off the isolator (1 mark)

(ii) The method used to ensure a safe workarea has to show:

● Ensuring no access to live parts, (1 mark)

● Ensure cables are protectedagainst mechanical damage.

(1 mark)

● Danger tag removed and/or Out-of-service tag attached.

(1 mark)

(b) (i) ● Ensure the circuit is still isolated

by using the prove-test-provemethod.

(1 mark)

● Attaching their own danger tag tothe circuit.

(1 mark)

(ii) ● Earth continuity (protectiveearthing conductor) test.

(½ mark)

● Insulation resistance test. (½ mark)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(a) (i) ● Protective earthing conductor test

● Insulation resistance test

(1 mark)

(ii) and (iii)

The following tests relating to theanswers to (i)

Test 1 – Protective earthing conductortest

● Ohms function (½ mark)

● Lowest range (½ mark)

● Test between the PEC conductorof the flexible cord and the frame

of the motor

(1 mark)

● Result of a lot less than 1 ohm (1 mark)

Test 2 – Insulation resistance test

● Any ONE of:

- Insulation resistance testfunction

- 500V d.c

(½ mark)

● Any ONE of:

- The Megohm range

- 500V d.c.

(½ mark)

● Test between each conductor of the flexible cord and the frame of the motor

(1 mark)

● Result of more than 1 Mohm (1 mark)

(b) Any ONE of:

● Ohms function (½ mark)

Lowest range (½ mark)

Test between the frame of the motorand a known earth other than themotor circuit

(1 mark)

The impedance is low enough to ensurethe protective device operates

(1 mark)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(a) 0.82 (½ mark)

(b) Output kW = HP x 746 Accept answers using750W in the calculation

= 5.36 x 746 (½ mark)

= 4kW (1 mark)

(c) Input kW = √3 x V x I x pf (½ mark)

= √3 x 415 x 8.52 x 0.82 (½ mark)

= 5.022 kW (1 mark)

(d) (i) Any ONE of:● International protection

● Ingress protection

(1 mark)

(ii) ● Protection against the ingress of solid objects

(½ mark)

● Protection against the ingress of dust

(½ mark)

(iii) Protection against the ingress of water. (1 mark)

(e) Any ONE of:

● Delta connection

● DOL starter

● Star/delta starter

(1 mark)

(f) ● A current larger that the current settingof the thermal overload.

(1 mark)

● That is sustained long enough tooperate the overload

(1 mark)

7/27/2019 EIN02 Answers 18 June 2011




(a) (i) Where there is a possibility of personal danger through the supplybeing restored.

(1 mark)

(ii) Where equipment is faulty ordamaged and using that equipmentwould cause damage or injury.

(1 mark)

(iii) Any THREE of:

● Make sure the correct isolatingswitch is tagged.

● Make sure the switch is in the “OFF” position before it istagged

● Fasten the Danger Tagsecurely so that it will notcome off.

● Test to ensure isolation hastaken place.

● Appropriate details areentered on the tag

(3 marks)

(b) “Isolated” means that the motor has

been:● deliberately disconnected from the

electricity supply and

(1 mark)

● precautions taken to preventreconnection

(1 mark)

● “Switched off” means that theelectricity ceases to be supplied tothe motor

(1 mark)

(c) ● The test instrument is checked to

be operating correctly on a knownlive source.

● The equipment is tested to confirm(or otherwise) that it is isolated.

● The test instrument is againchecked on a known live source toensure is still operates correctly.

(1½ marks)

(d) Test between each phase and earth. (½ mark)

7/27/2019 EIN02 Answers 18 June 2011




Marks

Marking notes

(a) (i) Vs = Ns x Vp

Np

(½ mark)

= 1 x 1100047.8

(½ mark)

= 230.13V or 230V (1 mark)

(ii) V1 = Vph x √3 (½ mark)

= 230 x √3 (½ mark)

= 398.6V or 400V (1 mark)

(iii) IL = VA

(V1 x √3)

(½ mark)

= 15000

11000 x √3

(½ mark)

= 7.87A (1 mark)

(iv) IL = VA

(V1 x √3)

(½ mark)

= 15000

398.6 x √3

or

15000

400 x √3

(½ mark)

= 217.27A or 216A (1 mark)

(b) VA or kVA (1 mark)

(c) Any TWO of:

● star – star● star – delta

● delta – star

● delta - delta

(1 mark)

7/27/2019 EIN02 Answers 18 June 2011




(a) (i) IPH = VPH

RPH

(½ mark)

= 230

30

(½ mark)

= 7.66A (1 mark)

(ii) IPH = VPH

RPH

= 400

30

(½ mark)

= 13.33A (1 mark)

IL = IPH x √3 (½ mark)

= 13.33 x √3 (½ mark)= 23.1A (1 mark)

(iii) Star is the most suitable connectionarrangement.

(½ mark)

(b) (i) In star

P = VL x IL x √3 x pf (½ mark)

= 400 x 7.66 x √3 x 1 (½ mark)

= 5307W (1 mark)

(ii) In delta

P = VL x IL x √3 x pf

= 400 x 23.1 x √3 x 1

= 16003W (1 mark)

(iii) Difference in power consumed

= 16003 – 5307

(½ mark)

= 10696W (½ mark)

EIN02 Answers 18 June 2011

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