Electronics beginner

Discussion in 'Homework Help' started by b.man58, Aug 6, 2016.

  1. b.man58

    Thread Starter New Member

    Dec 30, 2015
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    Hi folks...
    I am just starting out in electronics and am taking a home study course. The subject material provided is questionable and confusing by times. I tried calling a course advisor, who obviously was reading out of a prescripted manual and by the time I finished, I was worse off.
    Anyway, could someone please help me with the following including a step by step solving process.
    Q: what is the maximum current recommendation for a 1200-ohm, 1/4-watt resistor that is designed to work at 100% of its rating...using the formula I = (square root of) P/R?
     
  2. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Write the formula that was given to you: I = radical (P/R)
    Place the given number for power (1/4 watt) in the place where, "P" is.
    Place the given number for resistance (1200 ohms) in the place where "R" is.
    Use your calculator to solve for P/R
    Use your calculator's square root button to find the square root.
    Write the answer.
     
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  3. WBahn

    Moderator

    Mar 31, 2012
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    4,788
    Feel free to play around with the concepts a bit. You don't have to just plug things into a formula that's been handed to you.

    First, get a feel for the numbers involved.

    Let's say that the current is 1 A. Can you figure out how much power that would dissipate in that resistor? If you need to, feel free to use the formula

    P = I²·R

    If that turns out to be more than the 1/4 W limit, then does the allowed current need to be more or less than 1 A. Repeat this for some other currents, such as 1 mA and then 10 mA. See if you can make successively better estimates of what current will result in 1/4 W being dissipated in the resistor.

    Now you might do something similar, but instead of considering the current through the resistor, play with the voltage across the resistor. Let's say the voltage across it is 1 V. Is that enough to dump 1/4 W into the resistor? Or is it too much. Feel free to use the formula

    P = V²/R

    See if you can walk into the needed voltage in just a few more iterations.

    Now, using the current from the first round, what voltage across the resistor does that result in?

    Then, using the voltage from the second round, what current through the resistor will result?

    Hopefully these results are in close agreement with each other.
     
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  4. b.man58

    Thread Starter New Member

    Dec 30, 2015
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    Thank you very much, my friend! You make perfect sense. Ive got beginner jitters, especially after talking to a "purported" advisor on the topic. The big fear I suppose, is getting off base and not realizing it or not getting a clear and concise response to set me straight. I wish you were my teacher! I thank you very much!
     
  5. b.man58

    Thread Starter New Member

    Dec 30, 2015
    4
    0
    Thanks WBahn...
    You too have provided a reasonable and encouraging response! I really got messed up after talking to the "advisor" in my course. Big time. And that made me gun-shy and almost petrified of making mistakes that ww all make, but having no trustworthy source to get me back on track. You have, as well as the first responder to my question! Now I do know where to come for help! Again, thank you!
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,415
    3,354
    Don't take the simplest of formulas for granted. Make sure you fully understand the meaning behind the formula.

    For electronics, Ohm's Law is the most fundamental and this must become second nature in every aspect of electronics engineering.

    Ohm's Law is not V = IR

    Ohm's Law is I = V/R

    Fully understand what this means.

    Ohm's Law states that the current through a resistance is proportional to the voltage applied across the resistance and inversely proportional to the resistance.

    Putting it in simpler words, as voltage goes up, current goes up. As resistance goes up, current goes down.

    V = IR and R = V/I are simply corollaries of Ohm's Law, i.e. solving for V or R given the other two parameters.

    Power is defined as P = I²·R

    Given this formula for P in terms of I and R, see if you can apply the above formulas to state P in terms of V and R, and V and I.

    Now we have four variables, I, V, R and P.
    Write out every formula for each of these given any two of the other variables.

    There should be twelve such formulas (or formulae for the pedantic).
     
  7. b.man58

    Thread Starter New Member

    Dec 30, 2015
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    And thank you Mr chips! Much appreciated....and very succinct!
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    If you are going to draw a big distinction between V = IR and I = V/R as to which is Ohm's Law, then you shouldn't go and claim that power is defined as P = I²·R. Instead, it is defined as energy transformed per unit time or, mechanically, work through a distance per unit time, which for electrically charged particles is P = V·I. It is only because, for an ohmic material, that V = I·R, that you come up with P=I²·R. This is a far more important distinction than whether Ohm's Law is V = IR or I = V/R, or even R = V/I (which probably has the best claim for being the "proper" description of Ohm's Law) since all three of these are equally valid for an ohmic material and equally invalid for non-ohmic materials. Where as P=I²·R is only valid for ohmic materials wheras P = VI is valid for ANY current passing through ANY voltage differential (though things do get trickier once you start talking about non-conservative electric fields).
     
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