Essay Available:

You are here: Home → Term Paper → Engineering

Pages:

2 pages/≈550 words

Sources:

No Sources

Level:

APA

Subject:

Engineering

Type:

Term Paper

Language:

English (U.S.)

Document:

MS Word

Date:

Total cost:

$ 12.96

Topic:

# Answer Power Electronic Questions Research Assignment (Term Paper Sample)

Instructions:

ANSWER POWER ELECTRONIC QUESTIONS

source..Content:

Chapter 3 #1: Determine the equation for the waveform in the figure. The point at which the sine is turned on is φ. Hint: the answer should have 3 parts.

02540

Part 1

Since gradient is 0 for the line y=0,

0φdѲ = φ

Part 2

φπSin Ѳ dѲ= [-Cos Ѳ+Cos φ]=1+Cos φ

Part 3

π2πdѲ = π

Therefore, the equation for the area covered by the curve is

φ+ (1+Cos φ)+ π

Chapter 3 #2: Determine the equation for the waveform in the figure. This is a symmetric triangle wave. The equation has two parts. Each segment is of the form y=mx+b. The second segment has a negative slope and a y intercept of 2Ip.

0101600

Equation of a straight line is given by: y = mx + c

Where m is the gradient and c the y – intercept

For positive gradient;

M1 = 2IPT

Therefore, Y1 = 2IPTX

For negative gradient;

M2 = - 2IPT and Y intercept = 2IP

Therefore, Y2 = - 2IPTX + 2IP

Area under the triangular wave is given by;

0TY1+Y2.dt

0T2IPTX - 2IPTX + 2IP.dt

Substituting the values of y1 and y2 in the above equation and integrating by parts you get;

=2IPX0T

Chapter 3 #3:

* Calculate the equation for the average value of the partial sine wave from Figure 3-33.

Average value of the partial sine wave: from the figure, the quarter sine wave has the limits

0‹ φ ‹π

V average = 2π φπVpSin Ѳ. dѲ

Vp is the peak voltage

V average = 2VPπ - CosѲ for the limits 0 ‹ φ ‹ π

Substituting the limits;

V average = 2VPπ - CosѲ-Cos φ

But Cos 180o = -1

Thus, V average = 2Vpπ 1-Cos φ

* Given an amplitude of 170 Vp and a firing angle, φ=60, calculate the average value.

Given Vp = 170 V and φ=60 Degrees

Cos 60 = - 0.5

Substituting this in the above equation we got;

V average = (2 ×170)π × 1+0.5

V average =162.3V

Chapter 3 #4:

* Calculate the equation for the average value of the triangle wave from Figure 3-34.

Average value of triangular wave

Considering the first quarter of the wave, we have

V average = 2T0T2 2T Ip Ѳ. dѲ

V average = 4/ Π2 Ip 0T2Ѳ . dѲ

V average = 2/ T2 Ip [ Ѳ22] for the limits 0 ‹ T/2

Substituting the limits, we get<...

02540

Part 1

Since gradient is 0 for the line y=0,

0φdѲ = φ

Part 2

φπSin Ѳ dѲ= [-Cos Ѳ+Cos φ]=1+Cos φ

Part 3

π2πdѲ = π

Therefore, the equation for the area covered by the curve is

φ+ (1+Cos φ)+ π

Chapter 3 #2: Determine the equation for the waveform in the figure. This is a symmetric triangle wave. The equation has two parts. Each segment is of the form y=mx+b. The second segment has a negative slope and a y intercept of 2Ip.

0101600

Equation of a straight line is given by: y = mx + c

Where m is the gradient and c the y – intercept

For positive gradient;

M1 = 2IPT

Therefore, Y1 = 2IPTX

For negative gradient;

M2 = - 2IPT and Y intercept = 2IP

Therefore, Y2 = - 2IPTX + 2IP

Area under the triangular wave is given by;

0TY1+Y2.dt

0T2IPTX - 2IPTX + 2IP.dt

Substituting the values of y1 and y2 in the above equation and integrating by parts you get;

=2IPX0T

Chapter 3 #3:

* Calculate the equation for the average value of the partial sine wave from Figure 3-33.

Average value of the partial sine wave: from the figure, the quarter sine wave has the limits

0‹ φ ‹π

V average = 2π φπVpSin Ѳ. dѲ

Vp is the peak voltage

V average = 2VPπ - CosѲ for the limits 0 ‹ φ ‹ π

Substituting the limits;

V average = 2VPπ - CosѲ-Cos φ

But Cos 180o = -1

Thus, V average = 2Vpπ 1-Cos φ

* Given an amplitude of 170 Vp and a firing angle, φ=60, calculate the average value.

Given Vp = 170 V and φ=60 Degrees

Cos 60 = - 0.5

Substituting this in the above equation we got;

V average = (2 ×170)π × 1+0.5

V average =162.3V

Chapter 3 #4:

* Calculate the equation for the average value of the triangle wave from Figure 3-34.

Average value of triangular wave

Considering the first quarter of the wave, we have

V average = 2T0T2 2T Ip Ѳ. dѲ

V average = 4/ Π2 Ip 0T2Ѳ . dѲ

V average = 2/ T2 Ip [ Ѳ22] for the limits 0 ‹ T/2

Substituting the limits, we get<...

Get the Whole Paper!

Not exactly what you need?

Do you need a custom essay? Order right now:

### Other Topics:

- Analytic Dynamic Mechanic: Bending And Torsion ForcesDescription: The moment which results in bending of objects are called the bending moment. Bending can lead to forces of compression and tensile inside the metallic beam and this makes the beam to start stretching....1 page/≈550 words| No Sources | APA | Engineering | Term Paper |
- Renewable Sources of Energy: Wind Energy TechnologyDescription: The aim of the paper was to identify the alternative sources of renewable energy for homes in Uk, especially the homes located close to the shore line...11 pages/≈3025 words| 10 Sources | APA | Engineering | Term Paper |
- Answer Power Electronic Questions Research AssignmentDescription: Determine the equation for the waveform in the figure. This is a symmetric triangle wave. The equation has two parts. Each segment is of the form y=mx+b. The second segment has a negative slope and a y intercept of 2Ip....2 pages/≈550 words| No Sources | APA | Engineering | Term Paper |