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# 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<...

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