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Answer Power Electronic Questions Research Assignment (Term Paper Sample)



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.
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.
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;
0T2IPTX - 2IPTX + 2IP.dt
Substituting the values of y1 and y2 in the above equation and integrating by parts you get;
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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