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Welcome to blog help for PHY 122! Here's how it works. For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
12 comments:
Problem 3: Derive an answer for part(g), but keep it on your paper. Don't submit it to WebAssign because this requires use of the "physPad beta" tool. (The max score for Problem 3 will be 7/8 of a point based on submittal of 7 of the eight parts.) The phyPad beta tool has syntax that I find to be confusing. For part (h), the hint can be misleading. In SV8, look at Active Figure 21.7, which corresponds to the case for this problem. The orange voltage waveform is a sine wave. The purple current waveform is a minus cosine wave. You need the expression you worked out (but did not enter into WebAssign) for part (g) to answer part (h). If your expression is in terms of a minus cosine wave, you will use the inverse cosine function, not the inverse sine function. However, a minus cosine wave is the same as a sinewave with a minus pi over two phase shift inside the argument. That is, the argument of the sinewave would be (omega times t minus pi over two), and the factor (peak current) multiplying the sinewave would be your answer from part (e). In this case, the hint is correct; you use the inverse sine function. Futher comment: I worked it out for the phase-shifted sinewave and entered that result into part (g), which WA graded as correct. When I entered the result into (g) as a minus cosine wave, WA didn't grade it correctly. That's why I'm telling not to answer an result for part (g).
Problem 4: Since you're given an rms voltage for the source, it's only implicit that the the voltage drops in part (a) and part (b) should also be rms. This comment makes that explicit. To get your problem graded correctly by WA, the answers for part (a) and part (b) should be rms voltages.
hi, i am having trouble with part g of #1. i am using the formula i= imax*sin2*pi*f*t. i plug in .005 for t and i plug in everything else but it doesn't seem to work.
That formula should work...
Make sure your calculator is in "radian" mode.
You can also solve this question by using the equation for the output voltage given in the problem.
ok i got it thanks. my calculator was in degrees.
For proble number 3 part H: at what time after t=0 does the instantaneous first reach 1.00 A? I undertand that you have to use the formula for instaneous current, which is i=I(max)*sin2pi*f*t=
1.00=0.8839*sin2pi*f*t=
1.13=sin2pi*f*t
i can't take the inverse functionof this
I have the same problem as Saira. My Imax is .4922, and so when I divide 1.0 A by that, I get a number close to 2. Inverse sine of 2 does not work. Is there a different way to do this?
I received this from a student by email and asked him (and all of you) please to use the course blog. I transferred the question here for answering below:
I am in your 122 class. In my homework 21, question 3, my max current get is 0.836 which is shown correct, but when I was doing part (h). the confusing part for me is that
I=I0*sin(wt),I/I0=1/0.836=1.196, but for sin I can never get any value over 1, how could this be happening? Please take a look. Thanks.
Reply: I wrote about this on 10/15 and spoke briefly about ths problem at yesterday's (10/19) lecture when I emphasized phase shifts in AC circuits. (Look at the material on phase shifts in the posted lecture notes on Blackboard if you weren't there yesterday.) I gave the mnemonic "ELI the ICE man": E (voltage) leads I (current) by 90 degrees = pi/2 in and inductor L; I (current) leads E (voltage) by 90 degrees = pi/2 in a capacitor C. So for the L case in this problem, the reference waveform is the sinewave given in the problem, which corresponds to the source AC voltage; by Kirchoff, this must be the voltage drop across the inductor. Look at SV8, Fig. 21.7. There you see the reference waveform -- the one with the sinewave with zero phase, the orange curve -- is the voltage across the inductor. The purple one is the current waveform, and, as I wrote before, it is a minus cosine wave, which is the same as a sinewave with a minus pi/2 phase shift. So the formula for the instantaneous current is of the form I = Imax times sin(omegat - pi/2). Since I can never be bigger than Imax, the ratio I/Imax is less than or equal to one, so there is no problem taking the arcsine (inverse sine) of the ratio I/Imax. This will allow you to solve for the time for part (h).
Thanks for the reply, but even if the formula is I = Imax times sin(omegat - pi/2), I is still 1 and I (max)=0.8839, and i still can't take the inverse function of sine. I was in lecture yesterday, but i did not understand that part.
Saira,
I had the same problem as you, because my Imax was less than 1, so I/Imax was a number greater than 1. After reading Professor Koch's response, I just assumed I/Imax to be 1 (since he said this was its maximal value). Then, I solved for t.
Hopefully this helps.
thanks Kiran!
Prof,
Even with Eli...The question is bad unless I have a very poor understanding of the material.
Instantaneous current cannot be greater than Maximum current, Correct?
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