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

ANSWER POWER ELECTRONIC QUESTIONS

source..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.

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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.

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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 averag

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