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UEENEEG102ADr. Kenneth Meyer

Electrical AC Physics Tutorial Exercises for Students

Marking Guide

1Developed and Owned by Kenneth Meyer : Copyright 2020 ©

Solve problems in low voltage A.C. circuitsUEENEEG102A

Dr. Kenneth Meyer

Electrical AC Physics Exercises for Teachers

1. AC Fundamentals

2. Phasor Maths

3. R,L & C in AC circuits

4. RLC in series AC circuits

5. RLC in parallel AC circuits

6. Single phase Power

7. Power Factor Correction & Harmonics

8. Three phase introduction

9. Three phase Star (wye)

10. Three phase Delta (Mesh)

11. Three phase power

12. Earth Fault Loop Impedance

13. Dr. Ken’s Assessment techniques

Students Name: Date:


Electrical Trad Principles ( Phillips 3rd ED) Textbook links

1. AC Fundamentals 15.1 -15.4 (page 332)

2. Phasor Maths 15.5 -15.6 (page 346)

3. R,L & C in AC circuits 16.1 -16.7 (page 359)

4. RLC in series AC circuits 17.1 -17.4 (page 378)

5. RLC in parallel AC circuits 18.1 -18.6 (page 401)

6. Single phase Power 19.1 -19.3 (page 419)

7. Power Factor Correction & Harmonics 19.4 -19.5 (page 419)

8. Three phase introduction 20.1 -20.2 (page 436)

9. Three phase Star (wye) 20.3 & 20.5 (page 440)

10. Three phase Delta (Mesh) 20.4 (page 444)

11. Three phase power 20.6-20.7 (page 444)


13. Dr. Ken’s Assessment techniques


1. AC Fundamentals(1.1) Complete the table.

Angle Sin Cos Tan 12o 0.2079 0.978 0.21324o 0.407 0.914 0.44522.50 0.383 0.924 0.414232.00 0.5299 0.848 0.62547.00 0.7314 0.682 1.07267o 0.921 0.391 2.25662.990 0.891 0.454 1.962669.00 0.934 0.3584 2.61111o 0.934 -0.359 -2.61340 0.5592 0.829 0.6751540 0.438 -0.8988 -0.488157.3o 0.386 -0.923 -0.418

(1.2) A sine wave has a peak to peak voltage of 680 Vpk-pk. Determine the instantons value of voltage at the following angles; Vmax = Vpk=pk/2 = 680/2 = 340v

( a ) 460 Sin 46 * 340 = 245v

( b ) 1220 Sin 122 * 340 = 288v

( c ) 2720 Sin 272 * 340 = -338v


(1.3) A sine wave has a maximum current of 50 amps at a frequency of 50 Hz. Determine the instantaneous value of current after the following times; 1/50 = 20 ms period

( a ) 7ms 7/20 * 360 = 1260 Sin 126 * 50 = 40.5 Amps

( b ) 10ms 10/20 * 360 = 1800 Sin 180 * 50 = 0 Amps

( c ) 17ms 17/20 * 360 = 3060 Sin 306 * 50 = -40.5 Amps

(1.4) For the wave form below determine the following

( a ) peak-to-peak voltage 400 V Pk-Pk

( b ) period 10 ms or 0.01 s

( c ) frequency 1/0.01 = 100 Hz


(1.5) Complete the table below;

RMS Value Vmax * 0.707 Maximum Value rms/0.707 Peak-to-Peak Value Vmax *210A 14.14A 28.29A10.6V 15V 30V6A 8.5A 17A230V 325V 651V15.55A 22A 44A2.12kV 3kV 6kV5A 7.07A 14.14A8.48V 12V 24V1.061A 1.5A 3A

(1.6) For a digital oscilloscope displaying the wave form below determine the following;

( a ) pk-pk voltage 8V pk-pk

( b ) peak voltage 4V pk or max

( c ) Periodic time 6 Div * 2ms/Div = 12ms

( d ) frequency 1/12ms = 83.3 Hz

Channel A = 2 v / Div

Channel B not used

Time base = 2 ms / Div


4V

2V

-4V

Period

2. Phasor Maths

(2.1) Draw a phasor diagram to show a voltage of 220v lagging the reference by 300 and a voltage of 255v lagging the reference by 150. (Indicate the DOR).


Scale = 50v/10mm

Reff

250v <150

220v <300

DOR Anticlockwise

(2.2) (a) What is the addition of two currents of 12A < -450 and 17 A < -250.

(b) What is the phase angle between the result and the reference?


REF

Scale = 5A/10mm

22A < -30 0

17A <-250

12A <-450

(2.3) Determine from a phasor diagram the supply voltage and the phase angle relationship to the current.


Ref I

Scale 5V = 10mm

30V <-600

20V < 300

10V < -300

20V < 300 + 10V <-30

42.5V <-25 0

20V < 300 + 10V <-30 + 30V <-600

(2.4) Determine the total value of current given the branch currents in the diagram below.

Blank


Scale 5A = 10mm

Ref V

24A <120

12A <-300

24A <-420

24A <120 + 12A <-30

54A < -18 0

24A <120 + 12A <-30 + 24A <-420

3. R,L & C in AC circuits

(3.1) A purely resistive lamp is connected to a 32v, 50 Hz supply. The lamp draws a current of 2.2 amps. Determine ;

(a) Circuit resistance R= V/I = 32/2.2 = 14.54 Ohms

(b) Circuit impedance Lamp is resistive so R = Z = 14.54 Ohms

(c ) Power consumed by the circuit P = I2 * R = 2.22 * 14.54 = 70.37 W

(3.2) A 2.7 k resistor is connected to a variable frequency source. If the frequency is 2 kHz and the circuit current is 4mA, determine the voltage of the supply.

V= R * I = Z * I = 2.7k * 4mA = 2.7 * 103 * 4 * 10-3 = 10.8 V

(3.3) A heater draws a current of 10 amps and has a resistance of 22; Determine;

( a) The applied voltage V = R * I = 10 * 22 = 220V

( b) The circuit impedance Z = R = 22 Ohms

( c) Power supplied by the heater I2 * R = 102 * 22 = 2.2 kW


(3.4) An ideal inductor draws a current of 12 Amps when connected to 220v, 60 Hz. Determine the current if the same circuit is connected to 120v, 100 Hz.

XL = Z = V/I = 220/12 = 18.33 Ohms (60 Hz)

XL = 2* * F * L L = XL/ 2 * * F = 18.33 / 2 * * 60 = 48.62 mH

XL (100Hz) = 2* * F * L = 2* * 100 * 48.62 * 10-3 = 30.55 Ohms

I = V/ XL = 120 / 30.55 = 3.93A

(3.5) Calculate the capacitance of a motor start capacitor if its capacitive reactance is 110 Ohms at 50 Hz.

Xc = 1/ 2* * F * C

110 = 1/ 2* * 50 * C

C = 1/ 2* * 50* 110

C = 28.94 F

(3.6) Determine the current taken by a 110 F capacitor when it is connected to 240v, 60 Hz supply for the purpose of power factor correction.

Xc = 1/ 2* * F * C = 1/ 2* * 60 * 110 * 10 -6 = 24.11 Ohms

Ic = V/ Xc = 240/24.11 = 9.95 A


Blank


4. RLC in series AC circuits

(4.1) The circuit below that consist of an inductor is series with a resistive heater element. The voltage across the heater is 200v and in phase with the current. The voltage drop across the inductor is 120v, leading the current by 750 Using a phasor diagram determine the value of the supply and the phase angle.


I Ref.

Scale 50v = 10mm

200V <00

120V <750

258 <26 0

(4.2) A resistance of 180 Ohms is connected in series with a coil which has an inductance of 0.7H and an internal resistance of 50 Ohms. The circuit is connected to 220v, 50 Hz supply. Calculate the following;

(a) Total circuit resistance = 50 + 180 = 230 Ohms

(b) Inductive reactance = XL = 2 * * F * L = 2 * * 50 * 0.7 = 220 Ohms

(c ) Circuit impedance Z = (2202 + 2302 ) = 320

(d) Circuit current TI = V / Z = 220/320 = 688 mA

(e) Circuit phase angle Cos-1 (230/320) = 44 0


180R

50R

0.7 H

220V

Z

230 R

220 R

320 R

230 R220 R

(f) Inductor impedance ZL = (2202 + 502 ) = 226 Ohms

(g) Inductor phase angle L = Cos-1 (50/226) = 77 0

(h) Voltage drop across the resistor First find current: I = V/Z 220/320 = 688 mA

V = I * R = 0.688 * 180 = 123.8 V

( i) Voltage drop across the inductor VL = I * ZL = 0.688 * 226 = 155.5 V


50 R

ZL220 R

50 R

226 R220 R

(4.3) A resistance of 80 is connected in series with 67F capacitor to 379V, 60 Hz supply; calculate;

( a) Capacitive reactance

Xc = 1 / 2FC = 1/ 2 * * 60 * 67 * 10-6 = 39.59 Ohms

( b) Circuit impedance Z = (39.592 + 802 ) = 89.26 Ohms

( c) Circuit current I = V/Z = 379 / 89.26 = 4.24A

( d) Circuit phase angle = Cos-1 (80/89.36) = 26.3 0

( e) Voltage drop across the resistor First determine current from above 4.24A

VR = R * I = 80 * 4.24 = 339V

( f) Voltage drop across the capacitor Vc = Xc * I = 39.59 * 4.24 = 168V


80R 67F

379V 60 Hz

80 R

Z39.59 R

80 R

89.2639.59 R

( g) Draw the phasor diagram of the circuit.


Scale 50v = 10mm

Ref. I

376V < -26 0

168V <- 900

336V < 00

(4.4) A series circuit consists of the following, 20 Ohm resistor. 0.3 Henry pure inductor, a 63 micro-farad capacitor and an applied voltage of 200v, 50 Hurts. Determine the following;

( a) Circuit impedance

XL = 2FL = 2 * 50 * 0.3 = 94.24 OhmsXc = 1/ 2FC = 1/ 2 * 50 * 62 *10-6 = 50.53 Ohms

( b) Circuit current IT = V/Z 200/48.07 = 4.16A

( c) Circuit phase angle = Cos-1 (20/48) = 65.41 0

( d) Voltage drop across the resistor VR = I * R = 4.16 * 20 = 83. 2V

( e) Voltage drop across the inductor VL = I * XL = 4.16 * 94.24 = 392V

( f) Voltage drop across the capacitor Vc = I * Xc = 4.16 * 50.53 = 210v


200 50 Hz

20R 0.3H 63 F

20 R

Z= (432 + 202 ) = 48.07R94-51 R

20 R

48.07 R

( g) Draw the phasor diagram


I Ref.

Scale 100 v = 10mm

210v < 900

390v < -900

83v < 00

390v -210v = 180v < -900

200v < +65 0

Blank


5. RLC in parallel AC circuits

(5.1) A parallel circuit consists of a motor and a heater connected in parallel. The motor has a current of 5 amps and lags the supply by 300. The heater draws 3 amps and is in phase with the supply at 220v 50Hz. Use a phasor diagram to determine the total current and the phase angle.


V Ref.

Scale 1A = 10mm

3A <00.

5A <-300. 7.7A < -18 0

(5.2) In a parallel circuit there is a resistive load of 21 Ohms, in one branch, and a capacitive branch with a capacitive reactance of 30 Ohms. If the supply voltage is 240v at 50 Hz, determine the total circuit current and draw the phasor diagram.


Scale 2A = 10mm

V Ref.

Ic = V/ Xc = 240/30 = 8A <900

IR = V/ R = 240/21 = 11.4A <00

Ic + IR = 14.4A < 35 0

(5.3) A resistor of 6 Ohms, a pure inductive reactance of 7 Ohms and a capacitor with a reactance of 10 Ohms are connected in parallel across a 24v ac supply. Draw the phasor diagram to scale and show the following;

The position of the supply Coil current phasor Capacitor current phasor Resistor current phasor Determine the total current phasor, values of current, phase angle, lead or Lag.


Scale 1A = 10mm

IR= 24/6 = 4A <00

Ic= 24/10 = 2.4A <900

IL= 24/7 = 3.4A <-900

IL-IC= 2A <-900

IT = 4.1A <-14 0

24v Ref

Blank


6. Single phase Power

(6.1) A single phase load require 12kW of power. Determine the rating of the alternator and current carrying capacity of the connecting conductors if the load supply is at 220v and a P.F. of;

(Unity) 12 kW = 12kVA I = VA/V = 12kVA/ 220v = 54.55A

(0.87 Lag) I = VA/V = 13.79kVA/ 220v = 62.7A

(0.59 Lag) I = VA/V = 20.29kVA/ 220v = 92.68A


12kW

VA = W/PF 12kW/0.87 =13.79 KVA

12kW

VA = W/PF 12kW/0.59 =20.29 KVA

(6.2) A single phase 110kW load is supplied from a 230V 50 Hz supply. If the current taken by the load is 500 Amps, determine the power factor.

PF = Cos 110/115 = 0.957

(6.3) The following loads are connected in parallel across a 440V 60 Hz supply. Capacitor bank at 83 kVAR, Inductive load of 330 kVA @ 0.65 P.F. Lag and a resistive load of 400 kW.Draw the power triangle and calculate the installation power factor and the line current.


110kW

VA = 230 * 500 =115 KVA

IR = 400 KW/440 = 909A <00

Ic = 83 kVA /440 = 189A <900

IL = 330 kVA /440 = 750A <-49.50 (<= Cos-1 0.65 = 49.50)

V Ref.

IL-IC.

IT = 1450A < -15 0 (0.966 PF)

Scale 100A = 10mm

(6.4) A single phase motor is connected to a 400v, 50 Hz supply and draws 8Amps at a phase angle oh 410 lag. Determine the apparent, true and reactive power taken from the supply and draw the power triangle to scale.


True Power = VA * PF = 3.2 * 0.755 = 24.16 kW

Apparent PowerVA = 400 * 8 =3.2 KVA

-410

PF = Cos 41= 0.755

Reactive Power = VA * Sin 41 = 3.2 * Sin41 = 2.07 kVAR

Blank


7. Power Factor Correction & Harmonics

(7.1) A single phase installation is connected to a 220V, 60 Hz supply and consists of a 7.08 kW inductive load with a power factor of 0.7 lag in parallel with a2.1 kVAR capacitor bank in parallel with a 2.4 kW heater. Determine (a) power factor of the installation and (b) the supply (line) current.

Three currents in parallel: IR = 2.4kW/220v = 10.9AIC = 2.1kVAR/220v = 9.55A < 900

IL = 10.11kVA/220v = 46A <-45.60


Scale 5A = 10mm

V RefIR

IC

IL

IL - IC

IT= 48A < -28 0

(7.2) A 390 V, 50 Hz single phase installation draws 140 Amps from a supply at a power factor of 0.55 lag. Determine the rating of a capacitor back to be connected in parallel with the load to achieve an installation power factor 0.92 lag.

0.55 Lag = -560 0.92 = 23.070


Scale 25A = 10mm

V ref( 1 ) Blue side = Cos 56 * 140A = 78.28A

( 2 ) Red side = Sin 56 * 140A = 116A

560

230

140A <-56

( 3 ) Black side = Tan 56 * 78.3A = 33A

( 4 ) Green = Ic = 116- 33 = 83A

( 5 ) Ic * V = kVAR required83 * 390A = 32.27 kVAR (Capacitors)

( 3* ) Hyp = 78/Cos 23 = 85A <-23

(7.3) Determine the value of a static VAR compensator (fixed capacitor) and the line current for a 40 Amp 0.45 PF lag load supplied by 230 V AC at 50 Hz. Power factor to be be improved to 0.85 lag. (Draw the circuit !)

Cos-1 0.45 = 62.30 Cos-1 0.85 = 31.80


Scale 10A = 10mm

V Ref

620

320

( 1 ) Blue side = Cos 62 * 40A = 18.59A

( 2 ) Red side = Sin 62 * 40A = 35.4A

( 3 ) Black side = Tan 32 * 18.59 = 11.6A

( 4 ) Green = Ic = 35.4- 11.6 = 23.8A

( 5 ) Ic * V = kVAR required23 * 230A = 5.47 kVAR (Capacitors)

( 6 ) IT = (18.52 + 11.62) = 21.9A <-32 0

IT

Blank


8. Three phase introduction

(8.1) At what speed must a six pole three phase alternator be driven to produce a frequency of 50 Hz?

Page 573 of the textbook

F=np/12 n= f*120/p n= 50 *120/ 6 = 1000 RPM

(8.2) A two pole alternator is driven at 3000 rpm, determine the supply frequency.

F = np/120 = 2 * 3000/120 = 50 Hz

(8.3) A three phase alternator produces a maximum voltage of 440V . Determine the instantaneous voltages (A-B-C) when the A phase is at a rotational position of 900 .

Determine the position of all three phases in relation to A ay 900.B Phase is at 900-1200 = -300

C Phase is at 900-2400 = -1500

A phase: v = Sin 900 * 440 = 440v

B phase: v = Sin -300 * 440 = -220v

C phase v = Sin -1500 * 440 = -220v


9. Three phase Star (wye)

(9.1) A three phase distribution transformer is star connected and the supply line voltage of 440V. What is the transformers phase voltage. (Draw the circuit!).

Vp = VL / 3 = 440 / 3 = 254V

(9.2) A three phase star connected four wire supply has a phase voltage of 110V. What is the line voltage? (Draw the circuit!).

VL = Vp * 3 = 110 * 3 = 191V

(9.3) A start connected induction motor is connected to a 600 v supply. The motor takes a line current of 52 Amps from the supply. Calculate the impedance of each of the phase windings.

Vp= VL/3 =600/3 = 346

Zp = Vp / Ip = 346/52 = 6.65 Ohms

(9.4) A start connected, symmetrical, three phase load takes 12 A from a 220v three phase, four wire supply. Calculate (a) the impedance of each phase of the load. (b) the line voltage.

Zp= Vp/ Ip = 220/ 12 = 18.3

VL = Vp * 3 = 220 * 3 = 381V


440V

110V

600V

52A

12A

220V

(9.4) On the diagram below demonstrate the connection of the motor in ‘star’ to the three phase supply.

(9.5) A three phase star connected heater is connected to a 400v three phase 4 wire supply. If the resistance of each heater element is 14 Ohms, then determine the magnitude of the neutral current. ( Hint draw phasor diagram an add all the currents).

Vp = 400/3 = 231V

Ip = Vp/Zp = 231/14 = 16.6A

IN = IA + IB + IC = 0A


C2 B2A2

A1 B1 C1

U V W

Supply

Motor Terminals

400V14

16A

16A

16A

B+C

A+B+C=0

(9.6) A three phase, four wire supply has the following connected;A – Phase to neutral, 4A resistive load.B - Phase to neutral, 3A Lag at 600 C – Phase to neutral, 3A Lead ay 500 Determine the value of the neutral current and direction.


Scale 1A = 10mm

A

B

C

C 500 Lead

B 600 Lag

IN 2A <-16 0 (A+B+C Rev 180)

B+C

(9.7) Three, single phase 240v loads are connected to different phases of a 415 v three phase 4 wires supply. Determine the value and direction of the neutral current if;A Phase - capacitor-run motor taking 11A at 0.91 lead P.F.B Phase – split-phase motor taking 15 A at 0.66 lag P.F.C Phase – 2.4 Kw radiator.


Scale 4A = 10mm


B

A

C

Ic = 2400/240 = 10A

11A <260 (PF 0.91)

15A <-490 (PF 0.66)

B+C

IN = 12A -50 0 (A+B+C Rev 180)

(9.8) Determine the load star point to neutral voltage of a 400v three phase, four wire system if the following voltages are measured when the neutral is disconnected. A Phase to star point 210v. B Phase to star point is 500V. C Phase to the star point is 490v.

Vp = 400/ 3 = 230V


Scale 50V = 10mm

V star point 70V

A=210V

A Phase 230V

C Phase 230V

B Phase 230V

B=490V

B=500V

Neutral 0V

(9.9) Three 32 heating elements are connected in star to a 415v three phase supply. If the fuse to the line C blows, determine the load phase voltages and the line current for; (a) four-wire system and (b) a three-wire system. ( Draw the circuits!).


3 Wire no phase relationship so Ohms Law

32

32 32

415V

O/C

415V

4 Wire neutral present so Phase voltages present

I= V/R = 415V A-B / 32+32 = 6.48A

32

32

32

415V

415V

Neutral

Vp = VL / 3 = 415/3 = 240V

240V

240V

Ip = Vp/Zp = 240/32 = 7.5A

Blank


10. Three phase Delta (Mesh)

(10. 1) A delta connected resistive load takes a current of 28.66 A from the supply. What is the current in each phase?

IL = 28.6A Ip = IL/3 = 28.6 / 3 = 16.5A

(10.2) The phase current in each phase of a delta connected load is 150 Amps. What is the current in the supply lines?

IL = Ip * 3 = 150 * 3 = 260A

(10.3) A delta connected load has phase impedances of 25 Ohms each. The line currents IA = IB and IC = 100A each. ( Draw the circuit).Calculate; (a) the phase currents and (b) the line voltages.

( a ) Ip = IL/3 = 100/3 = 57.7A

( b ) Vp=VL Vp = Zp * Ip = 25 * 57.7 = 1.44kV

(10.4) A balanced delta connected load consists of a three phase motor with a stator winding impedance of 28 Ohms. Determine the motor line current if the supply is a three phase 440v, 50 Hz.

Ip = Vp/Zp = 440/28 = 15.7A

IL = Ip * 3 = 15.9 * 3 = 27.28A


150A

25R25R

25R

100A

100A

28R

440V

440V

440V

(10.5) A delta connected heater with a 500 W element in each phase is supplied from a three phase, three wire 415 V 50 Hz supply. Determine the current in the heating elements if the fuse in line B has blown. ( Draw it!)

No phase relationships, use Ohms Law

Ip = P/Vp = 500/415 = 1.2A Element A-C

I A-B-C = 1.2 / 2 = 600mA

(10.6) An industrial load connected in star has an impedance of 7 Ohms per phase. It is connected to a four wire, there phase 400v supply. (a) calculated the current in each phase of the load. (b) same load now configured in delta. Calculate the new phase current.

(a ) Vp = VL/ 3 = 400/3 = 231V Ip= Vp/Zp = 231/7= 33A

( b ) Ip = Vp/Zp = 400/7 = 57.1A


500W500W

500W

415 V

415 V

O/C

7R

7R7R

400v

400v

400v

400v

400v

400v

7R7R

7R

11. Three phase power

(11.1) A three phase 66kV (line voltage) transformer supplies 800kW balanced three phase load operating at 0.8 P.F. lagging. Determine the line current supplied by the transformer.

True power = Root 3 * V line * I line cos Ɵ 800kW

800kW = Root 3 * 66*103 * I * 0.8

I = 800 *103 / Root 3 * 66*103 * 0.8

I = 8.75A

(11.2) Calculate the power factor of a system using two wattmeter method if W1= 12 Kw and W2 = 15kW. Also determine the total power consumed by the system.

Power = W1 + W2 = 12 + 15 = 27 kW

Tan 3 * [ (W2-W1)/(W2+W1)]

Tan 3 * [ (15-12)/(15+12)]

Tan 3 * 111 = 0.19245

Tan-1 0.19245 = 10.890

PF = Cos 10.80 = 0.982


(11.3) A three phase balanced load requires 4 MVA of power when operating at 0.55 lag P. F., connected to an 11kV, 50 Hz three phase supply. Determine (a) the three phase kVAR rating of the capacitance per phase of a start connected capacitor bank that would improve the P.F. to 0.9 lag. (b) the line current before and after the improvement of the power factor.


Scale 50A = 10mm

VRef

I = 4MVA/√11kV Cos 56 =<3750 (.55PF)

260 (.9PF)

( 1 ) Cos 56 * 375 = 209A

( 2 ) Sin 56 * 374 = 310A

( 3 ) Tan 25 * 209 = 97A

( 4 ) 310 -97 =213A Reactive required

( 5 ) 213 *11kV 2.32 MVAR

Blank



(12.1) The supply authorities nominated property supply is fault level at 32kA at 230v. Determine the impedance of the supply.

Z = V/I = 230/ 32000 = 7.187 m

(12.2) The internal impedance of an installations has been determined at 0.0445 Ohms. The supply voltage is 220v. Calculate the fault current.

Fault Current = V * 80%/Z = 220 * 0.8 / 0.0445 = 3.95 kA

(12.3) The supply authority has nominate a maximum fault level of 40kVA @230v. The internal impedance is measured at 0.245 Ohms. Determine the total earth fault loop impedance.

Z Ext= V/I = 230 / 40000 = 5.75m

Z Total = Z Ext + Z Int = 5.75 + 0.245 = 0.251

(12.4) Using as 3008 determine the AC resistance at 50Hz of the following conductor.Conductor is; 6mm2 , Cu, multicore TPS V90, at a length of 174m. Answer in Ohms or Milliohms.

Table 35 ( Multicore) Colum 5 (V90) 6mm2 = 3.93 / 1000m (AS3008:2009)

Ohms for 174m = 3.93 / 1000 * 174 = 0.684 Ohms


Blank


13 Dr. Ken’s Assessment techniques. ( Open Textbook )

1. Before an assessment read the textbook, do all the exercises (Practical or Theory) depending on the assessment type.

2. Before the assessment install sticky tabs on the page edges of your textbook to make finding critical aspects easy. Also do the same on AS3000 and AS3008.

3. Take a clean copy of your equation sheet to the assessment. Once assessment starts, make as many notes as you can remember around the formulas. Doing this beforehand is considered cheating.

4. Read through the entire assessment. Note (write down) the easy and hard question numbers. If the equation for a question is obvious, write the question number on the equation sheet next to the relevant equation.

5. Prioritise your time. Three (3) hours and open book may seem as you have plenty of time, but good time management is key. Do easy question first, then more difficult ones.

6. Answer all the questions, particularly check the multi-choice questions.

7. In maths calculation, try to estimate the size of the answer. If you your calculators answer seem too large or too small, check everything.

8. Answering calculation questions. Put you name and the question number on the working-out sheet. Logically lay-out your answer approach.

(a) Read the question and as you go through, underline the ‘data’.(b) Decide which part of electrical physics is involved. Eg. Ohms Law, Inductance, etc?(c) Determine the variable that is to be solved for. (d) Select the appropriate formula and write in on the working-out sheet. (e) Insert the data into the formula.(f) Transpose the equation if required to make the missing variable the subject.(g) Solve equation, taking special care with Milli * 10-3 & Mico *10-6 & Kilo *103 etc.(h) Check your answer. Don’t forget Units and angle. (i) Underline your final answer/s

9. Review your entire assessment if you have time. Check for missing answers. For questions that have two or more answer are your answers in the correct field?

10. Check again ! This is Competency base assessment. All questions must be correct.


Wired for Imagination · Web viewSolve problems in low voltage A.C. circuits UEENEEG102A Dr....

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