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Understanding Electronics Basics #1

Discussion in 'General Electronics Chat' started by cowboybob, Feb 16, 2012.

  1. Muttley600

    Muttley600 New Member

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    The frequency of 6hrs daylight, you are trying to make me see units over numbers, I'm showing you numbers that relate to your units, so we end up with a happy balance, like everything in life, it is always a nicer place when balanced :)
     
  2. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Try this on for size. The period is 24 hours or one day. It doesn't matter if there 1 one hr of daylight or 23 hours of daylight, the period is 24 hours. 24 wasn't in the problem for a reason. I wanted to see if you could relate what a period is. So what's the frequency?
     
  3. Muttley600

    Muttley600 New Member

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    11.5uHz but your not after the number really are you, I feel like your trying to get me to relate to period but I'm just not seeing it am I

    Just found if you don't cancel memory on calculator it keeps adding it up.lol
    So my answer for 6hrs should be different
     
    Last edited: Mar 24, 2012
  4. dave

    Dave New Member

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  5. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Chortle...

    Good one, Graham.

    A few quickie sim quizzes (take your time). And for 1. below, what is the equivalent resistance of network made up of R1, R2 and R3? You may NOT use TINA to derive the answer.

    1. View attachment 62621 2. View attachment 62622 3. View attachment 62623 4. View attachment 62624 5. View attachment 62625

    I should say up front (this one time), there may be information displayed that has NO bearing on the the answer.

    Enjoy.
     
    Last edited: Mar 24, 2012
  6. Muttley600

    Muttley600 New Member

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    Will do in morning CBB, just off to bed
    Night both :)
     
  7. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Yep, that's almost the number I'm after. Since the number is 11.57407407 uHz to many decimal places and there are some error bars that are hard to calculate especially with leap years, leap seconds and their isn't exactly 365 days in a year.

    The answer has to be 11.6 uHz, not 11.5.

    In reality the problem is 1/24 {hr/day} * 60 {min/hr] * 60 [s/min} The units should cancel to s/day. No Latex, so it's hard to demonstrate. The same units on the top cancel the same units on the bottom and that my friend is a VERY powerful concept. You don't have to know that their is 3600 sec in a hr, but I've used the conversion enough that I can use it directly.

    I know that there are 60 sec in a minute and 60 min in an hr so you end up with 60 s/m and 60 m/hr. You can use these fractions to convert whatever you need until you end up with the required units.

    The reason why 11.5 is wrong is because of rounding. when you round a number "up", you look at the next decimal place which is a 7. 7 is bigger than 5, so you have add 1 to the previous number, so the closest number to 11.574 to 1 decimal place is 11.6, not 11.5.

    Obviously, if you were cutting lumber, you would like it to always be larger than the final size.

    Congratulations of applying the suffix of uHz. My boss once said that people are comfortable with numbers between 1 and 10

    Now, do you think you understand frequency and period?

    Your supposed to be able to take a "concept" and apply it.
     
    Last edited: Mar 24, 2012
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  8. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Last edited: Mar 24, 2012
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  9. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    KISS,

    Very interesting.

    I used to program in HyperCard on a Mac (back in the early '90s). At that time, there were precious few drivers for the "serial" interfacing Apple was using on the Mac back then (really an RS-485 variant) to use with HyperCard. Made it very tough to control the RF modems I was using.

    Curious how they sorted the "Index to HyperPhysics" - did not seem to be alphabetical or any other scheme. Do you have a clue?

    Anyway, very interesting. Thanks for the link.

    BTW. Graham did a pretty good job, eh?!?

    CBB
     
    Last edited: Mar 24, 2012
  10. Muttley600

    Muttley600 New Member

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    So you were after the number, fair enough, I'd have given you the full number but I thought I was getting sidetracked & distracted by numbers & you were driving at something else

    Do you know, for the first time (maybe fate intervened to help me get the concept) its a shame latex isn't working, I'd have never got that with a formula, it would have taken many more questions, but now I understand what your showing me, can you remember to show me the formula when it is so I can relate math to it :)

    Good point on rounding up, it's what I do when cutting steel or wire, you can always take a bit off but can't add it back on

    I think I'm on top of frequency & period now, your boss was right about numbers, but that's probably because people never normally have a need, much like me, to use numbers between 1-10 for everyday stuff :)
     
    Last edited: Mar 25, 2012
  11. Muttley600

    Muttley600 New Member

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    Don't panic CBB, I can't use Tina as I'm at work.lol
    But that's cheating anyhow & the only person I'd be cheating is myself if I don't understand :)

    Will go through sims on ciggy breaks

    Did you read that off KISS though, do you think he's ok, I swear I read 'congratulations' eek!
    I wasn't expecting that............

    Thanks both, I feel like my understanding is really coming
    on :)
     
    Last edited: Mar 25, 2012
  12. Muttley600

    Muttley600 New Member

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    CBB
    No1:0.9Ω total or 0.3Ω each R
    No2: 1u
    No3: 2u
    No4: 4u (I thought the max of each one in a circuit like this was always less than lowest value?)
    No5: 12u
     
    Last edited: Mar 25, 2012
  13. Muttley600

    Muttley600 New Member

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  14. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Mornin' Graham.

    1. Correct. The actual R values were R1=1Ω, R2=10Ω an R3=100Ω. Take the time to determine their parallel value (which you already have) to confirm the current as displayed by the multimeter. Just a good exercise since they are all different, and that might be useful in a particular circuit. For fun, determine the current through each, individual resistor.

    2. through 5., Incorrect

    Review the math for determining capacitive equivalence in parallel and series networks.
     
    Last edited: Mar 25, 2012
  15. Muttley600

    Muttley600 New Member

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    Hiya CBB
    I don't get what your asking, surely the I would be broken down as follows:
    0.1
    1.0
    10.0 giving us the 11.1A

    Least I know R inside out, shame about caps but all part of the learning process, I'll keep going til I get it right :)

    Doh......I'm a numpty, they are opposite way round
    No2: 2
    No3: 0.5
    No4: 12
    No5: 4

    New game, guess when I don't write things down to refer too, certainly shows how much I do or don't remember when I haven't got stuff to hand
     
    Last edited: Mar 25, 2012
  16. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    Good.

    Just wanted you to think of "why" would you make a circuit like that, beyond wanting to make a 0.9009Ω equivalent resistor.

    For instance, if you wanted to create 3 different currents (0.1A, 1.0A and 10A), all from one current source (rather than from three different current sources). Not exactly a likely scenario, but possible.

    HINT: As KISS pointed out, if you know how to derive equivalent parallel resistance, you know how to do the same for a capacitive one.
     
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  17. Muttley600

    Muttley600 New Member

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    Cool, duty cycles then :)
    This is just the width of charge at a given time or the width of a trigger pulse

    I think it was the other DC thread where you were showing me five steps of charge

    I've just trawled both threads & I can't find what I'm on about

    Ok, now I get caps R, what's the math behind inductor in series or parallel ?


    Then where are we after this?

    We've had some really good links on here, that reassure me I havent gone mad lol
    Found what I was on about in post #180 in this thread in link off CBB & it is going on about time constants, will have a proper look tonight but we may just of covered it.lol

    Also found link off KISS about inductors in series & parrallel

    So although I didn't understand them at the time I can go back & get a better understanding now :)
     
    Last edited: Mar 25, 2012
  18. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Real quick, we have to make sure you know how to solve simple equations. e.g. V=IR and P=VI. Memorizing these two relations can get you anywhere you need to go with V, I, R and P.

    (1) So from V=IR, there is I=V/R and R=V/I; Hopefully you understood how we got there from V=IR

    Now when P=VI, we can substitute any of the equations from (1) to get

    P=VI = [substitute V=IR) you get P = I*I/R or P= I^2R; similarly, start with P=VI and (substitute I=V/R) you get P = V*V/R or P = (V^2)/R
    and of course one more. Again, this could look better with Latex.

    Yep, very simple algebra.
     
  19. cowboybob

    cowboybob Well-Known Member Most Helpful Member

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    As a basic concept, yes. But there's more to it for a complete understanding.

    Motor controllers , dimmers, radars and the like often use pulse width (PW) to control speed, brightness and radar signal strength. And the signal is, generally, a square-wave.

    Recall the 555 Timer circuit where we looked at an RC network controlling the timing (duration) of the pulse (where the timed-switch controlled frequency)?

    The "duty cycle" is considered the "Time-On" and "Time-Off" components combined, PLUS the repetition rate (frequency, or period), PLUS how much of the MAX pulse amplitude the pulse rises to.

    This site gives a good Duty Cycle Description plus the associated math. Although it refers to radar duty cycle, it is applicable in most of the cases I can think of.

    If the math gives you a problem, don't worry for the time being: it's the concepts that you need to nail down first.

    It also mentions dB and it might help you there since dB is a gain ("power" output and input considerations) concept as well.

    I'll look back at #180 to see what I said, since I don't remember at the moment.

    KISS. I mentioned LateX being broken some time ago to ericgibbs in a PM, who responded that he sent something to EM (?) about it. It worked for a bit, but then broke again.
     
    Last edited: Mar 25, 2012
  20. Muttley600

    Muttley600 New Member

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    I memorised first one by V, as in multiply, I or R /
    Second I need to work on

    P
    -------
    V/I(2)

    E=P*T
     
    Last edited: Mar 25, 2012
  21. Muttley600

    Muttley600 New Member

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    That link makes great reading CBB, but of a pain on phone so will have a good read in a bit :)
    Everything I seem to be reading or understanding at the moment I can relate to transmission/bike radio

    I'm trying not to get distracted but we are covering the bones of electric here & it seems to relate to lots of different things, I simply cannot put into words how much I'm enjoying this

    Thanks both
     

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