Need help with some circuit problems

[duragezic]

Hmm... I did this homework assignment surprisingly quick but during class, the prof was fielding some questions and I think I did 3/5 of the problems wrong so I'm doing them over. It is pretty simple circuits stuff but I'm not sure about the answers I got.

So here is two problems I'm not sure about.

1) The prof said something about having the parallel impedance of the L and C go to infinity? I didn't understand what he said.

All I got out of 'purely resistive' was that the equivalent impedance would have no imaginary term. Thus, the magnitude of the imaginary terms of L and C should be equal but opposite so they cancel out. So I calculate the impedance of L & C and set them equal to each other, decide I don't like the 'j' and '-j' and decide to just throw them out (aka duragezic math), then I can solve for the radian freq omega which I can then use to get the frequency f = 71.18 Hz.

After more consideration, I understood it as asking to find the resonant frequency, which is given as omega = (sqrt(LC))^-1. That also gives me f = 71.17 Hz. But the explanation from the prof was different. I don't see how the f I got isn't right, because if I plug that back in, there is no imaginary term in the equivalent impedance of the circuit, thus it is purely resistive, right?

2) The he prof said something about finding the RMS value, squaring this current value & multiply by R, then integrating over a period and multiplying by 1/period. The period is 8 seconds. I suck at integrating and I'm not sure what the RMS value has to do with anything (thus using duragezic logic it is ignored), but I ended up with 125 W. Can anyone confirm or deny this?

 

[darthsidious]

for 1), your method gives you the right answer, but you might want to be more careful applying it. when you equate the two magnitudes, 1/wC = wL, which gives you w = 1/sqrt(LC). so the f you got is right...I don't know why you think your answer is wrong.? 71.18 Hz and 71.17 Hz are essentially the same - probably just a rounding difference.

for 2), I think your prof's method works, but is un-necessarily complicated. There are some redundant steps in your explanation. for example, the RMS value is given by integrate i(t)^2 from 0 to 8sec and then take square root. So you might as well not take the square root and square it again later- just integrate i(t)^2 from 0 to 8 seconds. The second integral over a period and the multiplying by 1/period is essentially averaging the power over the whole cycle. But as the RMS calculation already averages the current, and resistance is constant, it amounts to multiplying and dividing by 8 seconds,which just cancels out. So the final power answer should just be (RMS value)^2*R. The RMS value is an representation of the average magnitude of a signal, and RMS^2 is the average value of i(t)^2, which is what you want to calculate power loss here.

In this particular problem, you can calculate the average value of i(t)^2 without integration. Ignoring the signs of the two different pulses(for power dissipation direction doesn't matter), each pulse has magnitude 5A, and the rest of the time (50%) is 0A. So the average|i(t)| is 2.5 A, so the |i(t)|^2 is 2.5^2, or 6.25. So the average power dissipated is 6.25*10, or 62.5 W (Check my numbers), which is half of what you get.

Edit: I was wrong. We need to average I(t)^2, not |i(t)|, which would account for the factor of 2. Your answer was right.

 

[Mark R]

Originally posted by: duragezic
2) The he prof said something about finding the RMS value, squaring this current value & multiply by R, then integrating over a period and multiplying by 1/period. The period is 8 seconds. I suck at integrating and I'm not sure what the RMS value has to do with anything (thus using duragezic logic it is ignored), but I ended up with 125 W. Can anyone confirm or deny this?

This is correct. The reason to use the RMS value is to take account of the duty cycle. Irms = Sqrt (Ion^2 x D). D = 0.5 so Irms = 5 / Sqrt(2) = 3.54 A

Then taking i=RMS current, you calculate the average power i^2.R


There alterantive method to approach this problem is to take account of the duty cycle and leave it until the end.

Calculate the power during the 'on' period: |Ion| = 5A. Pon = 5^2 x 10 = 250 W
Duty cycle = 0.5
Average power = Pon x D = 125 W

The advantage of using the RMS value is that you don't have to worry about the duty cycle when calculating power, as the duty cycle is absorbed into the RMS value. It makes it much simpler when performing subsequent calculations, as I^2 is makes a frequent appearance in electrical calculations.

 

[duragezic]

1) Yeah what I meant was they were both the same (the 71.17 was a typo and meant 71.18 Hz) just I didn't seem to do it the way the prof explained, and he even mentioned something about the resonant frequency, that it could be right depending what resonant frequency you calculated (?). But I might've not understood correctly when he was talking about whatever goes to infinity... it just seemed like a more complicated explanation that what I understood the problem as.

2) Ok, I see now. Again, I didn't understood the prof's explanation. What I now think he said was "You need to find the RMS value... which can be done by squaring the current, integrating..." as opposed to "You need to find the RMS value, then square the current, integrate...". I was talking to someone in class about this problem and he had got 62.5 W doing it one way and 125 W the other and neither of us understood it enough to say what was correct.

Thank you both.