Elec1230 - Waveform, Phasors, & Calculations

Discussion in 'Homework Help' started by dennis.muller, Jun 8, 2008.

Feb 5, 2008
10
0
I'v posted here a copy of my work in the hopes someone could pin point where i'm going wrong or give me some understanding on how to solve the problem correctly. The problems I'm having are with questions # 41 to 51.

These questions are from the following book chapter 15 pg 501,

Circuit Analysis Theory and Practice
by Allan H. Robbins
ISBN-10: 141803861x
ISBN-13: 978-1418038618
Hardcover, 2006
Edition: 4

PG1 Q7-11
PG2 Q13-19
PG3 Q21-23
PG4 Q23-25
PG5 Q27
PG6 Q29
PG7 Q31-33
PG8 Q33-35
PG9 Q35-37
PG10 Q39-41
PG11 Q41
PG12 Q41-43
PG13 Q43
PG14 Q45-51
PG15 Q51
PG16 Q55-57
PG17 Q57-59

2. jorgemotalmeida Member

Oct 23, 2007
23
2
http://home.cogeco.ca/~dennis-muller/ELEC1230-01/ASS2/Image (3).jpg the first part is correct. With TI calculator it is easy. Check the mode to RADIAN. And then put the value desired. Example: press 50, then press the ANGLE (2nd key + ANGLE key) choose the degree symbol and then press twice in ENTER. you will have the radian value quickly.

Feb 5, 2008
10
0
Thanks for the prompt reply to my problem. The problem questions you refured to I have no problems with, my problems start with question 41 and end with question 51. With these questions, yes I have tryied calculating both in rads and degs with no succsess.

Feb 5, 2008
10
0
I requested some help from my professor who had the following below to say. I posted his reply as a resolution to this posting.

Hi Dennis:

I don't have enough time to go over each question, but hopefully the
following remarks will help:

29c: You made a calculation error. You used an f of 120 microseconds
when it was already calculated to be 8.33 kHz. Otherwise method is good.

39) You are given that the waveform reaches its peak at 54 degrees. This
means that it crossed the horizontal axis at -=36 degrees. That tells
you that the phase angle theta is + 36 degrees. You are not given the
period but are told that the portion from crossing the vertical axis to
the end of the cycle is 1620 microseconds. This represents 360 - 36
degrees = 324 degrees. Therefore to get the period, mutilply 1620 by
360/324. Then you can get the angular velocity and be off to the races.

41) You made a mistake I noticed in the drawing. Period is 50 ms and you
are correct in that, but the rising edge of the waveform is not at t=0
but later, so the end of the complete cycle is not at 50 ms but
correspondingly later.
v = Vm sin(wt + theta). Here theta is 45 degrees, as you know. In order
for the instantaneous voltage to be zero, the argument of the sine
function must be zero, and you got this correct in part a). For 23
degrees, you have to say:
v = Vm sin (wt + theta) where you know v is 23 v and you want to find t.
You have to figure out at what value of t the sine function sin (wt -45
deg) equals 23/30. This is done by using the inverse sine function. Say
sin (-1) (wt-45 deg) = 23/30. (excuse my notation -- by sin (-1) I mean
inverse sine.
Same thing for -23 v.
43) You are making a general mistake which makes all your drawings
wrong. I suggest you review the Angles and Phasors handout and also the
solution to 43c which is in the Problem Solutions I posted. Basically if
theta is zero, the waveform crosses the horizontal axis at t=0. Let us
class this waveform 1. If we have another waveform, waveform 2, that
leads waveform 1, then its phase angle is going to be positive and it is
going to cross the horizontal axis BEFORE waveform 1, which means before
t= 0, i.e. at some negative value of t. If your horizontal axis is
scaled in angle (wt) as we usually do, this means that waveform 2
crosses the horizontal axis at a negative angle. Similarly if waveform 3
lags waveform 1, then it crosses the horizontal axis LATER than waveform
1, which means to the right of the vertical axis, at a positive value of
t, at a positive angle.

You need to be clear on this for the test. I will take up such questions
in the Monday class.

45) Probably similar mistake. See number 46 in the problem solutions.

47) Average value is the area under the curve. Areas above the x axis
are positive, areas under the x axis are negative. Since all of these
curves are symetrical above and below the x axis, adding areas gives
zero.

51) This is a case of superiimposed ac and dc. The ac portion is e = 25
v sin(wt). and the DC portion is constant 15 v. Since they are in
series, they add. The effect is that the AC waveform is shifted upwards
by 15 v. The equation is e = 15 + 25 v sin(wt), not what you had. As far
as average is concerned, the formula you used does not apply to this
case. The formula Vavg = .637 Vm is applicable only to a pure sine wave.
Here y ou need to think in terms of the area under the curve. The AC
portion cancels itself out, leaving the Dc value as the average. This is
all explained on page 491 of the text.

55) Your answer for a) and b) are correct. See solution to c) in the
Problem Solutions posted. We did not deal with cosines in class but if
you do, you have to use a trigonometric identity (see inside cover of
book) to convert from a cosine to a sine.

57) a) This is a case of superimposed ac and dc and you need to use