Qualification - Pearson BTEC Level 5 Higher National Diploma in Engineering (Mechanical Engineering)

Unit Name - Engineering Science

Unit Reference Number - 21003K

Unit Level - Level 5

Assignment Title - DC and AC circuits

Learning Outcome 1: Determine parameters within mechanical engineering systems

Assessment brief

Purpose:

The assignment is designed to assess the student's ability to analyse circuits working on both direct and alternate currents.

Scenario

Imagine yourself to be a senior technician at an electrical company out for a routine test of some basic places. There would be places and components across which you may need to calculate the current through and voltage across, both in dc and ac circuits. As you know, you would need to use circuit laws and theorems to do so. You would also need to do so in circuits which are not simple - but are combined in series and parallel. Occasionally, you'd need to analyse the outcome of two or more ac currents when they mix sinusoidally. You may need to analyse generators and transformers according to the principles of electromagnetic induction. The current assignment assesses your skill in all these domains.

Are you stuck with your assignment? Miracleskills.com is the place which can give you a well-written Unit 3 Engineering Science - Pearson BTEC Level 5 Higher National Diploma in Engineering assignment help quickly.

Part 1

(i) Use the superposition theorem to calculate,
(a) the current i,
(b) and the voltage v in the DC circuit across the E kΩ resistor as shown in Figure 1 below.

Solution:

i. Superposition theorem
a. The circuit diagram is shown in Figure 1 below

When 40 V source is acting alone , the circuit diagram is shown in Figure 2

Total resistance = [7 ||8 ] + 6
= (7×103×8×103)/(7×103+ 8×103 ) + 6 ×103
= 3.73×103 + 6 ×103
= 9.73×103Ω

Total current be I = 40/(9.73×10)3 = 4.1mA

The current flowing through the 8kΩ = by current division rule

I1= 4.1mA ×(7/(6+7))k
= 1.913mA
When 50 V is acting alone, the circuit diagram is shown in Figure 3

Total Resistance = [6 ||8 ] + 7
= ((6×10)3×(8×10)3)/((6×10)3+(8×10)3 ) + 7 ×(10)3
= 3.43×(103 + 7 ×(10)3
= 10.43×103Ω
Total current be I = 50/(10.43×10)3 = 4.74mA

The current flowing through the 8kΩ = by current division rule
I2= 4.74mA ×(6/(6+8))k
= 2.03mA
According to the superposition theorem the circuit diagram is shown in Figure 4

i = I1+ I2
= 1.913 mA+ 2.03mA
= 3.943mA
Voltage flowing through the 8kΩ resistor = i × 8kΩ
= 3.943 ×10-3× 8 ×103
= 31.544 V

(ii) Use the Thevenin'stheorem to calculate,
(a) the current i,
(b) and the voltage v in the DC circuit across the E kΩ resistor as shown in Figure 1 above.

Solution:

Thevenin'stheorem
Thevenin's resistance = RThshown in Figure 5

Figure 5
= 6 || 7
= ((6×103)×(7×103)/((6×103+(7×103 )
= 3.23 kΩ
To find the Thevenin's voltage VTh
We can use the superposition theorem to find the Thevenin's voltage
VTh1 = By voltage division rule the figure is shown in Figure 6

= 40 ×7/(7+6) = 21.53 V
VTb = By voltage division rule the figure is shown in Figure 7

= 50 ×6/(7+6) = 23.08 V
So the Thevenin's voltage VTh= VTh1 + VTb
= 21.53 + 23.08 = 44.61V
The Thevenin's equivalent circuit is shown in the Figure 8

i = 44.61/((3.23+8×103)) = 3.972mA
Voltage flowing through the 8kΩ resistor = i × 8kΩ
= 3.972 ×10-3× 8 ×103
= 31.776 V

Don't know what time value of money homework is all about? Get in touch with our experts who can provide you with the great guidance and best solution with HND assignment help!

Part 2

(i) For the series RLC ac circuit shown in Figure 2 below, calculate the current through the circuit and the voltage across the inductor. Give your answers in both the time and frequency domains.

Solution:

V = 4 cos (5t +600)
In polar form V = 4 ∠ 600
In rectangular form V = 2 + 3.46j
Xc = 1/jwC = 1/(j×5×700×(10)^(-6) F) = -j285.71 Ω
XL= jwL = j × 5 × 8 = j40Ω
The circuit diagram is shown in Figure 9

I = V/Z
= (4 ∠ 60)/(j40-j285.71+90)

= (4 ∠ 60)/(90-j245.71)

= (4 ∠ 60)/(261.674∠-69.883)

= 0.015∠ 129.883 A
= -0.010 +j 0.012A
Voltage across the inductor = IR
= (-0.010 + j 0.012) ×j40Ω
= -0.48- j 0.4 V
= 0.625∠39.806 V
In real time form Voltage across inductor will be 0.625cos (5t +39.806)V

(ii) For the series RLC circuit shown below in Figure 3, determine the voltage across the capacitor in the time domain.

ii. The circuit diagram is shown in Figure 10

Let V1 = 4 cos (5t + 600)
In polar form V = 4 ∠ 600
In rectangular form V = 2 + 3.46j

Let V2 = 7 cos (8t + 900)
In polar form V = 7∠900
In rectangular form V = 0 + 7j

By using the superposition theorem we can find the voltage across the capacitor
When V1 is acting alone
Xc = 1/jwC = 1/(j×5×700×(10)^(-6) F) = -j285.7 Ω
XL= jwL = j × 5 ×6 = j30Ω

I1 = V1/Z = (4 ∠ 60)/(j30-j285.7+170)

= (4 ∠ 60)/(-j255.7+170)

= (4 ∠ 60)/(307.05 ∠-56.38)

= 0.013∠116.38 A
= -0.006 +j0.012 A
Now Voltage across capacitor VAB1= I1 R
= (-0.006 +j0.012) × -j285.7
= 3.43 + 1.71j V
=3.833∠26.498

When V2 is acting alone Xc = 1/jwC = 1/(j×8×700×10-6 F) = -j178.57 Ω
XL= jwL = j ×8 × 6 = j48Ω

I2 = V2/Z = (7 ∠ 90)/(j48-j178.57+170)

= (7 ∠90)/(-j130.57+170)

= (7 ∠ 90)/(214.35∠-37.53)

= 0.033∠127.53 A
= -0.020 +j0.026 A
Now Voltage across capacitor VAB2= I2 R
= (-0.020 +j0.026) × -j178.57
= 3.571j + 4.64 V
Now voltage across the capacitor according to the superposition theorem is
VC= VAB1 +VAB2
3.43 + 1.71j + 3.571j + 4.64 V
= 8.07 + j 5.281 V
= 9.644∠33.201 V
In the real time form 9.644cos (5t +33.201) + 9.644cos (8t+33.201) V

Have you stuck with your Diploma assignment? Get our faultless diploma assignment help and finish your work without any stress!

3. Refer to problem (3) above. The voltage we obtained across the capacitor was of the form vc = Acos(Bt+C)+D cos( Et+F). Use the numbers from A to F as assignned by your instructor andplot the complex waveform and describe how it was produced by combining the two sinosoidal waveforms. Include as part of your description, the following :

(a) the harmonics and fundamental, if any,
(b) the amplitude of the harmonics,
(c) the (angular) frequency of the harmonics,
(d) the time period of the harmonics,
(e) phase of the harmonics,

Solution:

The voltage across the capacitor is
VC = 9.644 cos (5t +33.201) + 9.644 cos (8t+33.201) V
V1 = 4 cos (5t + 600)
V2 = 7 cos (8t + 900)
The combined graph is shown in Figure 11

Fundamental frequency w = 2π f
1/f = 2π/w
= 2π/5
= 72 s
f = 0.014 Hz
Harmonics = 9.644 cos (8t+33.201)
Amplitude of the harmonics = 9.44 V
Angular frequency of the harmonics
1/f= 2π/w
= 2π/8
= 45 s
f = 0.02Hz
Time period of the harmonics
T = 1/f= 2π/w
= 45sec
Phase of the harmonics = ∅
= 33.201 deg

Don't what you should include in your Engineering Assignment? Get in touch with the experts of our Engineering assignment Help for top-notch quality assistance at reasonable prices!

4. Refer to the problem (3) above. A capacitor is addedas shown in Figure 4 below between the branches and a small change is made to the second AC voltage source so that both sources have the same frequency. A newcombined series-parallel RLC circuit is provided.

(i) Calculate the current through the new capacitor using the two techniques of your choice.

Solution:

The circuit diagram

Kirchhoff's voltage law
Let V1 = 4 cos (5t + 600)
In polar form V = 4 ∠ 600
In rectangular form V = 2 + 3.46j
XL= jwL = j × 5 × 6 = j30Ω
Xc = 1/jwC = 1/(j×5×900×10-6 F) = -j222.22 Ω

Let V2 = 7 cos (8t + 900)
In polar form V = 7 ∠ 900
In rectangular form V = 0 + 7j
Xc = 1/jwC = 1/(j×5×700×(10)^(-6) F) = -j285.7 Ω
When KVL is applied to loop 1
-4 ∠ 600 + j30I1 - j222.22I1 + 90I1 -j222.22I2 = 0
-4 ∠ 600 + (j30- j222.22 + 90)I1 -j222.22I2 = 0
-4 ∠ 600 + (j192.22 + 90) I1 -j222.22I2 = 0
(j192.22 + 90) I1 -j222.22I2 = 4 ∠600Equation1
When KVL is applied to loop 2
-7 ∠ 900 - j222.22I2 + 80I2 -j285.7I2 - j222.22I1 = 0
-7 ∠ 900+(- j222.22+ 80-j285.7)I2 - j222.22I1 = 0

-7 ∠ 900+ (-j507.92 + 80) I2 -j222.22I1 = 0
(-j507.92 + 80) I2 -j222.22I1 = 7 ∠ 900Equation 2

This can be solved by matrix method
[((j192.22 + 90)&[email protected]&(-j507.92 + 80) )]×[(I_1@I_2 )] = [(4 ∠ 60@7 ∠ 90)]

Δ = |((j192.22 + 90)&[email protected]&(-j507.92 + 80))|
= (j192.22 + 90)×(-j507.92 + 80)-(j222.22×-j222.22 )
= 154214.11- j30335.2
= 157169.39 ∠ -11.13
Δ1 = [(4 ∠ 60&-j222.22@7 ∠ 90&(-j507.92 + 80) )]

= 361.86-j739.04
= 822.88 ∠ -63.91
Δ2 = [((j192.22 + 90)&4 ∠ [email protected]&7 ∠ 90 )]

=-2114.42+j1074.44
= 2371.75 ∠ 153.06
I1 = (822.88 ∠ -63.91)/(157169.39 ∠ -11.13) = 0.005 ∠ -52.78 A
= 0.003- j0.004

I2 = (2371.75 ∠ 153.06)/(157169.39 ∠ -11.13) = 0.015∠164.19 A
= -0.014+ j0.004
Current through the capacitor 900μ F is
I1+ I2= 0.003- j0.004 + -0.014+ j0.004
= -0.011A

Superposition Theorem
When V1 = 4 cos (5t + 600) is acting alone
The circuit diagram is shown in Figure 12

When KVL is applied to loop 1
-4 ∠ 600 + j30I1 - j222.22I1 + 90I1 -j222.22I2 = 0
-4 ∠ 600 + (j30- j222.22 + 90)I1 -j222.22I2 = 0
-4 ∠ 600 + (j192.22 + 90) I1 -j222.22I2 = 0
(j192.22 + 90) I1 -j222.22I2 = 4 ∠ 600Equation 2
When KVL is applied to loop 2
j222.22I2 + 80I2 -j285.7I2 - j222.22I1 = 0
(j222.22 + 80 -j285.7)I2 - j222.22I1 = 0
( 80 -j63.48)I2 - j222.22I1 = 0 Equation 2

This can be solved by matrix method
[((j192.22 + 90)&[email protected]&( 80 -j63.48))]×[(I_1@I_2 )] = [(4 ∠ 60@0)]

Δ = |((j192.22 + 90)&[email protected]&( 80 -j63.48))|
= (j192.22 + 90)×( 80 -j63.48)-(j222.22×-j222.22 )
= 68783.85 + 9664.4
= 69459 ∠ 8
Δ1 = [(4 ∠ 60&-j222.22@0&( 80 -j63.48) )]

=379.64 + j149.84
= 408.14∠21.54
Δ2 = [((j192.22 + 90)&4 ∠ [email protected]&0 )]

=-768.88 + j444.44
= 888.09∠149.97
I1 = (408.14 ∠ 21.54)/(69459 ∠ 8) = 0.005 ∠13.54 A
= j0.001

I2 = (888.09 ∠ 149.97)/(69459 ∠ 8) = 0.013∠141.97 A
= j0.008
Current through the capacitor 900μ F is
I1- I2= j0.001-j0.008
= -j0.007A
When V2 is acting alone
When KVL is applied to loop 1
-7 ∠ 900 - j222.22I1 + 80I1 -j285.7I1 - j222.22I2 = 0
-7 ∠ 900+(- j222.22+ 80-j285.7)I1 - j222.22I2 = 0

-7 ∠ 900+ (-j507.92 + 80) I1 -j222.22I2 = 0
(-j507.92 + 80) I1 -j222.22I2 = 7 ∠ 900Equation 1

When KVL is applied to loop 2
j30I2 - j222.22I1 + 90I2 +j222.22I2 = 0
(j30+ j222.22 + 90)I2 -j222.22I1 = 0
(j252.22 + 90)I2 -j222.22I1 = 0 Equation 2

This can be solved by matrix method
[((-j507.92 + 80) &[email protected]&(j252.22 + 90))]×[(I_1@I_2 )] = [(7 ∠ 90@0)]

Δ = |((-j507.92 + 80) &[email protected]&(j252.22 + 90))|
= (j192.22 + 90)×( 80 -j63.48)-(j222.22×-j222.22 )
= 184689 - j25535.2
= 186446 ∠ -7.87
Δ1 = [(7 ∠ 90&-j222.22@0&(j252.22 + 90) )]

=-1765 + j630
= 1874∠160
Δ2 = [((-j507.92 + 80) &7 ∠ [email protected]&0 )]

=-1555.54
= 1555.54∠180
I1 = (1874 ∠ 160)/(186446 ∠ -7.87) = 0.01∠167.87 A
= j0.002

I2 = (1555.54 ∠ 180)/(186446 ∠ -7.87) = 0.008∠187.87 A
= -j0.001
Current through the capacitor 900μ F is
I1- I2= j0.002+j0.001
= j0.003A

The current flowing through the capacitor is -j0.004 A

(ii) Evaluate the techniques used by commenting briefly as to which of them was easier to solve the given problem with justification.

The KVL method is easier.
In both the cases the current flowing is calculated using KVL.

Question 5. (i) Refer to Figure 2 in Problem (3) above. An inductor is added in the vicinity of the inductor present as shown in Figure 5.

The following questions are in reference to the principles of electromagnetic induction :

(a) Describe with explanation what happens in the second coil when the given ac source is switched on in the first coil.

(b) Describe with explanation what will happen in the second coil when the ac source is replaced by a dc source and it is switched on.

(c) If in question (b), we changed the value of the dc voltage from lower to higher, describe with explanation what will happen in the second coil.

(ii) The following questions refer to the applications of electromagnetic induction :

(a) For the coil of an ac generator, explain the factors that influence the ac current generated by the coil.

(b) Assume the coil of the ac generator starts with its axis perpendicular to the magnetic field (0o) . Describe with explanation the amounts and directions of the induced current as the coil rotates through a complete cycle.

(c) For an ideal transformer, explain how the secondary voltage and current depend on the voltage, current and the number of turns in the primary circuit.

Solution:

Principles of electromagnetic induction

When the first coil is energized , a non-zero current in each of the two coils produces a mutually induced voltage in each coil due to the flow of current in the other.
There won't be any mutual induction.
As the dc voltage is increased the primary coil will burn.
Applications of electromagnetic induction

It depends upon the emf generated and the resistance.
I = E/R
E = 2πNBAf

The emf attains its peak values when the plane of the coil is parallel to the plane of the magnetic field, passes through zero when the plane of the coil is perpendicular to the magnetic field, and reverses sign every half period of revolution of the coil.

An ideal transformer has no losses. That is secondary is open and primary being energized by ac.There is ac current in primary which means magnetizing one.This will produce alternating fluxproportional to the current and will be in phase. This will link with the both windings. This causes self inducedemf in primary causing a mutually induced emf in secondary and is antiphase whose magnitude being proportional to the rate of change of flux and number of secondary turns.

RELATED COURSES & ASSIGNMENT SERVICE!!

 Analyze the sustainable supply chain management Transport Network Design Assignment Help Strategic Management Plan Assignment Help Facilitating Change in Health and Social Care Assignment Help Level 4 Diploma in Office Administration and PA Assignment Help Unit 1 Business and the Business Environment Assignment Help BTEC Level 2 Extended Certificate in Business Assignment Help HND Diploma in Tourism and Hospitality Management Assignment Help Level 7 Award in Strategic Management and Leadership (RQF) Assignment Help Database Management System Assignment Help Design of systems for Adverse Event Reporting Business BTEC Introductory Diploma Level 1 Assignment Help

#### Are You Looking for Determine parameters within mechanical engineering systems?

Getting a complete Unit 3 Engineering Science - Pearson BTEC Level 5 Higher National Diploma in Engineering assignment help is not an easy task but Miracleskills.com is the top writing service that can give you the best solution at the best price.

 Learning Outcomes and Assessment Criteria Pass Merit Distinction LO2 Determine parameters within mechanical engineering systems D4 Evaluate different techniques used to solve problems on a combined series-parallel RLC circuit using A.C. theory. P8 Calculate currents and voltages in D.C. circuits using circuit theorems   P9 Describe how complex waveforms are producedfrom combining two or more sinusoidal waveforms.   P10 Solve problems on series RLC circuits with A.C. theory. M4 Explain the principles and applications of electromagnetic induction

Avail Pearson BTEC assignment help service for below mention course:-

• Unit 12 Industrial Power, Electronics and Storage Assignment Help
• Unit 7 Digital Principles Assignment Help
• Unit 14 Lean Manufacturing Assignment Help
• Unit 4 Managing a Professional Engineering Project Assignment Help
• Unit 10 Professional Engineering Management Assignment Help
• Unit 3 Engineering Science Assignment Help
• Unit 6 Quality and Process Improvement Assignment Help
• Unit 1 Engineering Design Assignment Help
• Unit 5 Electrical and Electronic Principles Assignment Help
• Unit 8 Electronic Circuits and Devices Assignment Help
• Unit 9 Research Project Assignment Help
• Unit 11 Further Mathematics Assignment Help