btec hnc - science - application of dc and ac theory

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Application of DC & AC Theory Engineering Science By Brendan Burr

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Page 1: BTEC HNC - Science - Application of DC and AC Theory

Application of DC & AC TheoryEngineering Science

By Brendan Burr

Page 2: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Table of Contents

TABLE OF CONTENTS - 2 -

TASK 1 - 5 -

1.1 Two storage batteries, A and B, are connected in parallel for charging from a DC source having an open circuit voltage of 14V and an internal resistance of 0.15 Ohms. The open-circuit voltage of A is 11V and that of B is 11.5V; the internal resistances are 0.06 Ohms and 0.05 Ohms respectively. Calculate the initial charging currents. - 5 -

Solution:- - 5 -Check:- - 7 -

1.2 In the diagram below find the current through the 8 Ohm resistor. - 8 -Solution (Circuit 1):- - 8 -Check (Circuit 1):- - 10 -Solution (Circuit 2):- - 11 -Check (Circuit 2):- - 13 -Solution (Overall Circuit):- - 14 -Check (Overall Circuit):- - 14 -

TASK 2 - 15 -

2.1 Explain how complex waveforms are produced from sinusoidal waveforms. - 15 -

Solution:- - 15 -

2.2 Synthesise the following complex waveform graphically using a spreadsheet: . - 15 -

Solution:- - 15 -

2.3 Plot a minimum of 100 points at intervals of (Periodic Time – T)/100 Seconds.- 15 -

Solution:- - 15 -

2.4 Produce a print out of the graph and data. - 15 -Solution:- - 15 -

2.5 Describe how electrical and electronic devices produce complex waveforms. - 16 -

Solution:- - 16 -

2.6 Describe the effects of complex waveforms on electrical and electronic systems. - 16 -

Solution:- - 16 -

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Page 3: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

TASK 3 - 17 -

3.1 A circuit consists of three branches in parallel. Branch A is a 10 Ohm resistor, Branch B is a coil of resistance 4 Ohms and inductance of 0.02H, and Branch C is an 8 Ohm resistor in series with a 200uF capacitor. The combination is connected to a 100V, 50Hz supply. Find the various branch currents and then, by resolving into in-phase and quadrature components, determine the total current taken from the supply. - 17 -

Solution:- - 17 -Check:- - 20 -

3.2 A 31.8uF capacitor, a 127.5mH inductor of resistance 30 Ohms and a 100 Ohms resistor are all connected in parallel to a 200V, 50 Hz supply. Find the various branch currents and the, by resolving into in-phase and quadrature components, determine the total current taken from the supply. - 22 -

Solution:- - 22 -Check:- - 25 -

TASK 4 - 27 -

4.1 A 1kOhm resistor is connected across the secondary windings of an ideal transformer whose secondary voltage is 100 Volts. The current in the primary windings is 10mA. - 27 -

(a) Draw a circuit diagram - 27 -Solution:- - 27 -

(b) Determine the secondary current - 27 -Solution:- - 27 -

(c) Determine the primary voltage - 28 -Solution:- - 28 -

(d) Determine the transformer turns ratio - 28 -Solution:- - 28 -Check (Overall Circuit):- - 29 -

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Page 4: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

4.2 An ideal transformer has 1000 primary turns and 100 secondary turns. If the primary winding is connected to a 230V AC supply the secondary is connected to a 100 Ohms resistive load: - 30 -

(a) Draw a circuit diagram - 30 -Solution: - 30 -

(b) Determine the secondary voltage - 30 -Solution: - 30 -

(c) Determine the secondary current - 31 -Solution: - 31 -

(d) Determine the power supplied in the primary circuit - 31 -Solution: - 31 -Check:- - 31 -Check (Overall Circuit):- - 32 -

APPENDIX 1 - 33 -

EVALUATION - 34 -

CONCLUSION - 35 -

Books - 36 -

Catalogues - 36 -

Websites - 36 -

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Page 5: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Task 1

1.1 Two storage batteries, A and B, are connected in parallel for charging from a DC source having an open circuit voltage of 14V and an internal resistance of 0.15 Ohms. The open-circuit voltage of A is 11V and that of B is 11.5V; the internal resistances are 0.06 Ohms and 0.05 Ohms respectively. Calculate the initial charging currents.

Solution:-

Loop A:

Loop B:

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Page 6: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

shows up in Equations 2 and 3, therefore substituting Equation 1 into Equations 2 and 3, as follows, gives:

Simplifying Equations 4 and 5 leaves:

Multiply Equation 4 by :

Then add Equation 5:

Amps

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Page 7: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Substituting into Equation 5, gives:

Amps

Substituting and into Equation 1, gives:

Amps

Check:-

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Page 8: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

1.2 In the diagram below find the current through the 8 Ohm resistor.

Solution (Circuit 1):-

Redraw the circuit with one Voltage Source:

Alternative Circuit:

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Page 9: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Voltage drop across R1 is represented by:

Voltage drop across R2 is represented by:

Voltage drop across R3 is represented by:

Voltage drop across R4 is represented by:

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Page 10: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Check (Circuit 1):-

Therefore, :

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Page 11: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Solution (Circuit 2):-

Redraw with second Voltage Source:

Alternative Circuit:

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Page 12: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Voltage drop across R5 is represented by:

Voltage drop across R4 is represented by:

Voltage drop across R3 is represented by:

Voltage drop across R2 is represented by:

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Page 13: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Check (Circuit 2):-

Therefore, :

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Page 14: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Solution (Overall Circuit):-

By merging the two circuits together we get:

Check (Overall Circuit):-

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Page 15: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Task 2

2.1 Explain how complex waveforms are produced from sinusoidal waveforms.

Solution:-

Complex waveforms are comprised of a fundamental component together with a number of harmonic components, each of which have a specific amplitude and phase relative to the fundamental.When a fundamental signal receives a harmonic they are added together. This deforms the shape of the overall waveform which is directly related to the sum of the two waveforms.

2.2 Synthesise the following complex waveform graphically using a spreadsheet: .

Solution:-

See Appendix 1

2.3 Plot a minimum of 100 points at intervals of (Periodic Time – T)/100 Seconds.

Solution:-

See Appendix 1

2.4 Produce a print out of the graph and data.

Solution:-

See Appendix 1

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Page 16: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

2.5 Describe how electrical and electronic devices produce complex waveforms.

Solution:-

Electrical and electronic devices can produce complex waveforms (due to creating Harmonics), in many different ways.The main cause of harmonics being created is due to the mechanical properties of the device, for example “tooth – ripple” in AC Generators, caused by the effect of the slots that accommodate the windings, or because of the non-sinusoidal air-gap flux distribution.For a rectifier to create a half cycle the waveform needs to be removed of its negative cycle, due to the low impedance in one direction and a high impedance in the opposite, diodes create complex waveforms.Harmonics are produced in devices that have a non-linear response to their inputs. Non-linear circuit element (i.e. those in which the current flowing through them is not proportional to the applied voltage) include rectifiers and any large-signal electronic amplifier in which diodes, transistors, valves or iron-cored inductors are used.

2.6 Describe the effects of complex waveforms on electrical and electronic systems.

Solution:-

If a complex waveform is fed into electrical and electronic devices then the device may fail to work correctly. This is because the amplitude of the waveform is altered, so may damage the device if too much voltage/current is fed into it.The input waveform will also have a possible phase shift which will mean that the device may not operate to the required time response and affect output.The output would also be irregular as the components will produce even more of a complex waveform, as explained above, and therefore the overall effect could be something that is extremely difficult to calculate/predict.

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Page 17: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Task 3

3.1 A circuit consists of three branches in parallel. Branch A is a 10 Ohm resistor, Branch B is a coil of resistance 4 Ohms and inductance of 0.02H, and Branch C is an 8 Ohm resistor in series with a 200uF capacitor. The combination is connected to a 100V, 50Hz supply. Find the various branch currents and then, by resolving into in-phase and quadrature components, determine the total current taken from the supply.

Solution:-

BRANCH A =

BRANCH B =

BRANCH C =

Current,

Impedance, Z, for the three branch parallel circuit:

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Page 18: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

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Page 19: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Current for BRANCH A, is given by:

Amps (Rectangular)OR

Amps (Polar)

Current BRANCH B, is given by:

Amps (Rectangular)OR

Amps (Polar)

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Page 20: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Current BRANCH C, is given by:

Amps (Rectangular)OR

Amps (Polar)

Total Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Total Current is given by:

Total Current,

Amps (Rectangular)OR

Amps (Polar)

Check:-

On the following Graph:The Red Waveform represents Total Current ( ) in Polar.The Blue Waveform represents BRANCH A ( ) in Polar.The Magenta Waveform represents BRANCH B ( ) in Polar.The Dark Green Waveform represents BRANCH C ( ) in Polar.

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Page 21: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

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Page 22: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

3.2 A 31.8uF capacitor, a 127.5mH inductor of resistance 30 Ohms and a 100 Ohms resistor are all connected in parallel to a 200V, 50 Hz supply. Find the various branch currents and the, by resolving into in-phase and quadrature components, determine the total current taken from the supply.

Solution:-

BRANCH A =

BRANCH B =

BRANCH C=

Current,

Impedance, Z, for the three branch parallel circuit:

22

Page 23: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

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Page 24: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Current for BRANCH A, is given by:

Amps (Rectangular)OR

Amps (Polar)

Current BRANCH B, is given by:

Amps (Rectangular)OR

Amps (Polar)

Current BRANCH C, is given by:

Amps (Rectangular)OR

Amps (Polar)

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Page 25: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Total Admittance, :

Siemens (Rectangular)OR

Siemens (Polar)

Total Current is given by:

Total Current,

Amps (Rectangular)OR

Amps (Polar)

Check:-

On the following Graph:The Red Waveform represents Total Current ( ) in Polar.The Blue Waveform represents BRANCH A ( ) in Polar.The Magenta Waveform represents BRANCH B ( ) in Polar.The Dark Green Waveform represents BRANCH C ( ) in Polar.

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Page 26: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

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Page 27: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Task 4

4.1 A 1kOhm resistor is connected across the secondary windings of an ideal transformer whose secondary voltage is 100 Volts. The current in the primary windings is 10mA.

(a) Draw a circuit diagram

Solution:-

(b) Determine the secondary current

Solution:-

Amps

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Page 28: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

(c) Determine the primary voltage

Solution:-

Volts

(d) Determine the transformer turns ratio

Solution:-

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Page 29: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Check (Overall Circuit):-

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Page 30: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

4.2 An ideal transformer has 1000 primary turns and 100 secondary turns. If the primary winding is connected to a 230V AC supply the secondary is connected to a 100 Ohms resistive load:

(a) Draw a circuit diagram

Solution:

(b) Determine the secondary voltage

Solution:

Volts

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Page 31: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

(c) Determine the secondary current

Solution:

Amps

(d) Determine the power supplied in the primary circuit

Solution:

Watts

Check:-

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Page 32: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Check (Overall Circuit):-

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Page 33: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Appendix 1

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Page 34: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

EvaluationThis assignment required me to use existing knowledge of AC and DC Theories, to establish the answers when circuits are at certain conditions.Beginning with Task 1.1 with the two storage batteries, I felt that this Task went reasonably well. The question, for some reason appeared difficult at first as it was difficult to understand what the circuit looked like and what actually needed calculating. After much consideration I realised how the circuit should be put together and that the currents between the two batteries and the open circuit equivalent battery were required. Using Kirchhoff’s Voltage and Current Laws, along with a general understanding of the circuit, I worked out equations for the two loops. I also worked out an equation for the currents, and using both, through superimposing and transposition, worked out the values for the currents.I checked Task 1.1 on Croc Clips and identical values for the currents were presented. This gave me the assurance that my answers were correct.For Task 1.2 I had to work out a specific current flowing through R4 (8 Ohms Resistor). This was done by splitting the original circuit into two and calculating each circuit with only one voltage source. I could then superimpose the two circuits onto one another and come up with an overall circuit diagram with each relevant current.I also checked Task 1.2 using Croc Clips, and the value for the current flowing through the 8 Ohms resistor was 319mA, which is exactly the same as I calculated.Moving onto Complex Waveforms, which we have been covering in other units, I found that Task 2 was relatively straight forward.It is very easy to generate a complex waveform when you know the equation, as you can enter it into Graphmatica, however this Task required me to use Microsoft Excel. This dragged the Task out for longer than was actually required as there was a lot of messing around required to get a waveform that can be generated in seconds elsewhere. The only thing that I can think of that is beneficial by doing it on Microsoft Excel is that you can see in the formula that you are adding the Harmonics to the Fundamental Waveform.I began to face a few problems when completing Task 3. This was mainly because of the AC condition and that the Check was difficult to follow because the waveforms were constantly moving and always out of phase with one another. For Task 3.1, I managed to get the individual Branch Currents and then using this I could calculate Overall Current being drawn from the supply. I done this by calculating the overall admittance and then multiplying it by the voltage, but this could also be done by adding the three currents together.For Task 3.2, this wasn’t as simple. By calculating the overall admittance and then multiplying by the voltage I managed to get the Overall Current, which was represented in my check. However by adding the three currents, I did not receive the same result. I am still not sure why this is, but my guess would be that Branch A has zero resistance, which isn’t realistic, therefore theoretically the overall calculation would be wrong, but through simulation you get the correct answer.I am confident my answers are correct, as my workings tie in extremely closely to the simulated results from Croc Clips.

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Page 35: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Finishing the assignment with Task 4, I felt that this Task went very quickly. I have already done similar problems to this through the EEP Unit and therefore could understand it immediately. I worked out each answer and checked them accordingly by using Croc Clips, this method was also used for my overall check as shown above.

ConclusionTo conclude, I am pleased with the outcome of this assignment. I am confident that I have correctly and appropriately answered the Tasks, due to my extensive checking methods.This assignment has also taken a lot of time to ensure it is accurate and I feel it meets the standard of my other two assignments in this Unit.I have received absolutely no “help” on this assignment, and will therefore be looking to achieve a good mark.I have one assignment left for this unit which is entirely research based and is already underway.

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Page 36: BTEC HNC - Science - Application of DC and AC Theory

Brendan Burr BTEC Higher National Certificate in ElectronicsApplication of DC & AC Theory

Bibliography

Through guidance from my lecturer, the following text books, catalogues and websites I was able to complete this assignment:

Books

BTEC National Engineering (Mike Tooley & Lloyd Dingle) ISBN: 978-0-7506-8521-4Success in Electronics (Tom Duncan & John Murray)ISBN: 0-7195-4015-1Higher Engineering Mathematics (John Bird) ISBN: 0-7506-8152-7Engineering Science (John Bird)ISBN: 0-7506-4991-7

Catalogues

N/A

Websites

N/A

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