Ohms Law

Discussion in 'Homework Help' started by LizzyLea, Feb 26, 2016.

  1. LizzyLea

    Thread Starter New Member

    Feb 26, 2016
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    I'm a 31yr old female taking an electricians class to take my Apprentice test in May. I don't understand how I would show any work for this question or if I'm posting in the correct category... My question is about Ohms law. There are 3 different formulas to plug numbers into Ohms law E/I E2/P P/i2. I couldn't figure out how to make a square sign for the E & I. So my question is can someone tell me how I figure out what numbers or where I find the numbers to plug into this formula?? Also, can someone give me an example scenario where I can figure out which formula to use and see if I'm correct?? ​
     
  2. #12

    Expert

    Nov 30, 2010
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    You are mixing up Ohm's Law and Watts Law.
    Every form of the equations uses 2 factors to find the third factor. Just write down all 12 equations, figure out which contains the two factors you know and the answer you want to get.
     
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  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It is all about the power. Electrical power.

    E I assume stands for Electrical Potential.

    I is usually means current.

    R is usually means resistance.

    Ohm's Law states: E=I\times{R}
    Meaning that Electrical Potential is product of resistance and current.

    So.
    \frac{E}{I}=R
    This formula is used when you know any two variables:
    - you know E and I
    - you know E and R
    - you know I and R

    P is usually used to denote power.
    P=E\times{I}
    but, from Ohm's Law, we know that I is E/R. Which means that:
    P=E*I=E*(E/R)=E^2/R
    then
    P*R=E^2
    and
    R=\frac{E^2}{P}

    Same sort of manipulation for the next one.
    P=E\times{I}<br />
E=I\times{R}<br />
P=I\times{R}\times{I}<br />
P=I^2\times{R}<br />
R=\frac{P}{I^2}<br />
    If you know any two parts, you can find the third.
     
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  4. WBahn

    Moderator

    Mar 31, 2012
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    It sounds like you are trying to memorize a bunch of formulas and when to plug some numbers into each of them instead of trying to understand electrical concepts.

    There are two base relationships that are involved (and both can be derived from basic physics concepts in short order, but we can start here).

    First, for certain materials (not ALL materials), the voltage across the device is proportional to the current through the device. We call the proportionality constant the resistance. So

    V = I·R

    This is commonly known as Ohm's Law and materials that obey this law (again, not all materials do) are called ohmic materials.

    Simple algebra will let you manipulate this equation into two additional forms. If we divide both sides by R we get

    I = V/R

    while if we divide both sides by I we get

    R = V/I

    The second base relationship, which holds for all materials, not just ohmic ones, is that the power delivered to something is the produce of the voltage appearing across it and the current flowing through it. So

    P = V·I

    As before, we can divide both sides of this equation by either V or I giving us

    V = P/I

    and

    I = P/V

    So that's six equations. But wait, for ohmic materials we can intermingle them. For instance, we can take P = V·I and substitute the equation for V from Ohm's Law to get

    P = V·I = (I·R)·I = I²R

    Or we can substitute in the equation for V and get

    P = V·I = V·(V/R) = V²/R

    Each of these gives us new equations for each of the variables. For instance, the first equation gives us

    R = P/I²

    and

    I = √(P/R)

    while the second gives us

    R = V²/P

    and

    V = √(P·R)

    But there is no need to memorize all of these formulas. Just understand the concepts of Ohm's Law (V = I·R) and power (P = V·I) and you can derive whichever relationship you need in a matter of moments.
     
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  5. crutschow

    Expert

    Mar 14, 2008
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    If you don't understand how to convert from one from of the equation to the other to solve for the unknown you want, then you need to brush up on some basic algebra.
     
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  6. LizzyLea

    Thread Starter New Member

    Feb 26, 2016
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    Yes I definitely need to brush up on some basic as algebra. I used to love math!! I hope once I have a few more classes it will come easier!!
     
  7. crutschow

    Expert

    Mar 14, 2008
    13,001
    3,229
    Here's a simple example:
    One form of Ohm's law is V = I * R.
    But if you want to solve for I, for example, then you need to isolate I on one side of the equation.
    To do that, just divide both sides of the equation by R.
    This gives V / R = (I * R) / R or
    V / R = I.
    Since both sides of the equation are equal, you can swap sides, giving
    I = V / R, the equation you wanted.
     
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  8. dannyf

    Well-Known Member

    Sep 13, 2015
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    It depends on what's known. Those formula are identical to each other.
     
  9. withoutego

    New Member

    Dec 22, 2015
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    [​IMG]
     
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  10. Tonyr1084

    Active Member

    Sep 24, 2015
    535
    86
    I LOVE that magic wheel.

    Here's something I troubled with for a long time, until someone explained how simple this was: I struggled with E=IR. IR? What's IR? Simple: IR is I x R.

    VA was another thing I had trouble with. VA is V x A, which is equal to P (or Watts).

    Knowing ANY two factors you can calculate all other factors involved. IF you know I & R you can quickly and easily calculate P (watts or power) or E (volts)

    For instance, 220 ohms and 0.7 amps means your voltage is 154 v.
    154 = 220 x 0.7 OR 220 x 0.7 = 154

    OR using the wheel, I^2 (which means I squared) x R equals P.
    I^2 x R = IIR = I x I x R = 0.7 x 0.7 x 220 = 107.8 watts (of Power)
    (Note: The I is not 1, don't confuse multiplying 1 x 1 x R with I x I x R or IIR)

    Notice that in my formula I used ^2 to represent "Square". ^3 would be "Cubed". ^4 represents a power of FOUR. ^N (or ^n) represents any number that may be used, such as R1 + R2 + •••Rn (meaning any number of resistors).

    Perhaps you can give us an example that you're having trouble understanding.
     
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  11. LizzyLea

    Thread Starter New Member

    Feb 26, 2016
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    I've tried to upload a picture since my first post. Until I have better service I won't be able to upload... Thanks everyone for the input!!!
     
  12. ISB123

    Well-Known Member

    May 21, 2014
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    You need to understand how Ohm's law works.

    For example If you want to calculate current you'll need to know voltage and resistance.
    Now if voltage increases but resistance stays the same that means that current must increase also, if resistance increases but voltage stays the same current decreases.
    Current is directly proportional to Voltage and inversely proportional to Resistance!
    Your unknown value is current, your known values are voltage(24V) and resistance(350Ω).
    So the formula goes:
    I=V/R
    I=24 / 350
    I=0.068

    Now if voltage was unknown you would have to convert the equation:
    V=IxR
    V=0.068 x 350
    V=24

    If you seek resistance you convert the equation like so:
    R=V/I
    R=24 / 0.068
    R=350

    Once you understand that (Current is directly proportional to Voltage and inversely proportional to Resistance)you will be able to figure out how to get the equations.
     
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  13. WBahn

    Moderator

    Mar 31, 2012
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    You are probably working with an image that is too large given your connection. Use a tool like Paint (or an equivalent in whatever OS you have) and scale the image so that it is about 300 to 500 pixels wide. Then save it as either a PNG file (if it is computer generated line art) or JPG (if it is a photo image). You can always save it both ways and just see which is smaller. For a problem like this, your final image should only be a couple dozen kilobytes and even a slow connection can handle it.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    I don't -- anyone that has to rely on that crutch has such a weak grasp of the concepts and/or such weak algebra skills that they all they are going to be able to do is plug numbers into formulas that someone else gave them and which they don't comprehend (which is why they see that wheel as "magic"). In most cases that means that they are going to regularly use formulas in ways that are not valid and won't have a clue why their solutions are garbage.
     
  15. withoutego

    New Member

    Dec 22, 2015
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    If the student will look at the wheel and see the symmetry involved
    and have some feeling for the interplay of the variables in a circuit
    the wheel is useful. But you need to learn hands on I believe. Otherwise
    there is no intuition, just dry use of the equations.
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    I would still argue that a much better feeling for the interplay of the variables in the circuit comes from being able to start with V=IR and P=IV and having the math skills to easily, preferably in their head but in no more than three to five trivial lines of algebra, arrive at the formula for any one of those variables in terms of any two of the others. In being able to do so, they don't just see some symmetry of some letters, but they understand the application of concepts.

    Example: I in terms of P and R.

    Two equivalent routes:

    Start with the concept that voltage is proportional to current (and that R is defined as the proportionality constant):

    V = IR

    The solve for the desired variable:

    I = V/R

    Now use the concept that power is the product of voltage and current and solve for the variable in the former equation (V) that you want to eliminate:

    P = VI
    V = P/I

    Substitute this back into the prior equation

    I = (P/I)/R = P/(IR)

    And solve for I

    I^2 = P/R
    I = √(P/R)

    Or come at it from the power equation first

    P = VI
    I = P/V

    Now we used Ohm's Law to express the voltage in terms of the current and the resistance.

    V = IR

    I = P/V = P/(IR)

    And solve for I

    I^2 = P/R
    I = √(P/R)

    I've gone into extreme detail here. The actual process should be something akin to

    P = VI
    I = P/V
    I = P/(IR)
    I^2 = P/R
    I = √(P/R)
     
  17. Tonyr1084

    Active Member

    Sep 24, 2015
    535
    86
    Well, everyone has an opinion. Still, if you want to know something about your circuit or what you can expect (or need) it's good to know the numbers. Otherwise, how do you know what resistance you need? How would you know if you need a quarter watt resistor or a 20 watt resistor? Everyone starts somewhere. And learning these formulas have helped me many times before.

    Everybody learns the ways that best suit themselves. If I try to do it your way I may be prone to mistakes and end up leaking smoke. True, the most basic calculations are V, I, R & P. Knowing there are many ways to achieve these numbers just means you can use whatever formula you're most comfortable with. And no, I haven't memorized the wheel. I tend to stick with E=IR & their reciprocals, and P=EI (and their reciprocals).
     
    Last edited: Feb 28, 2016
  18. LizzyLea

    Thread Starter New Member

    Feb 26, 2016
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    Thank you very much!! Your explanation was excellent!! Very explanatory!! I had another class tonight and realized why I was getting so confused... I had thought we (my class) where only focusing on one quarter of the circle, but tonight I realized we were using the whole circle. I've only had about 7 classes. My test is in May and I wasn't too learn as much as I can because it's REALLY, REALLY important to me that I pass this class... THANK YOU AGAIN SO MUCH!!
     
  19. LizzyLea

    Thread Starter New Member

    Feb 26, 2016
    5
    0
    Thank you for the explanation and also for the examples!!! :)
     
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