Answer Power Electronic Questions Research Assignment - 550 Words

Answer Power Electronic Questions Research Assignment (Term Paper Sample) Content: Chapter 3 #1: Determine the equation for the waveform in the figure. The point at which the sine is turned on is . Hint: the answer should have 3 parts.02540Part 1Since gradient is 0 for the line y=0,0dÃ' ² = Part 2Sin Ã' ² dÃ' ²= [-Cos Ã' ²+Cos ]=1+Cos Part 32dÃ' ² = Therefore, the equation for the area covered by the curve is+ (1+Cos )+ 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.0101600Equation of a straight line is given by: y = mx + cWhere m is the gradient and c the y interceptFor positive gradient;M1 = 2IPTTherefore, Y1 = 2IPTXFor negative gradient;M2 = - 2IPT and Y intercept = 2IPTherefore, Y2 = - 2IPTX + 2IPArea under the triangular wave is given by;0TY1+Y2.dt0T2IPTX - 2IPTX + 2IP.dtSubstituting the values of y1 and y2 in the above equation and integrating by parts you get;=2IPX0TChapter 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 limits0 V average = 2 VpSin Ã' ². dÃ' ²Vp is the peak voltageV average = 2VP - CosÃ' ² for the limits 0 Substituting the limits;V average = 2VP - CosÃ' ²-Cos But Cos 180o = -1Thus, V average = 2Vp 1-Cos * Given an amplitude of 170 Vp and a firing angle, =60, calculate the average value.Given Vp = 170 V and =60 DegreesCos 60 = - 0.5Substituting this in the above equation we got;V average = (2 170) 1+0.5V average =162.3VChapter 3 #4: * Calculate the equation for the average value of the triangle wave from Figure 3-34.Average value of triangular waveConsidering the first quarter of the wave, we haveV average = 2T0T2 2T Ip Ã' ². dÃ' ²V average = 4/ 2 Ip 0T2Ã' ² . dÃ' ²V average = 2/ T2 Ip [ Ã' ²22] for the limits 0 T/2Substituting the limits, we get