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ohm's law

Discussion in 'Electronic Design' started by [email protected], Jan 16, 2008.

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  1. Guest

    Hi everyone,

    I have the following set up:

    5V
    ----------------------------------------------------|----------------------------------------

    | |

    | |

    Motor Pressure sensor

    | |
    0V 1.5m
    | |
    ---------------------------------------------------------------------------------------------|

    The electric motor and a pressure sensor are running of the same 5V
    power supply. The transducer is connected to the power supply with
    1.5m long 1 mm diameter cable with resistance of 0.02 ohms/meter.

    The motor draws a current of approximately 2 amps when operating.

    Now, what I have to do is calculate the drop in supply voltage across
    the pressure sensor when the motor is turned on.

    Now, what I reasoned is that there will be a drop in both the wires:

    So, the resistance in the wires are: 1.5 X 0.02 = 0.03 ohms

    So the voltage drop when the motor is turned on would be:

    2 X 0.03 = 0.06 V

    However, it should be 0.06 X 2 = 1.2 V because there are two wires.

    Is this reasoning correct? Or is there a special way to add them?

    Cheers,
    Luca
     
  2. Guest

    Sorry the diagram did not come very well:

    You can find this on:

    5V

    ----------------------------------------------------|----------------------------------------

    |
    | |

    |
    | |

    |
    Motor Pressure sensor

    |
    | |
    | 1.5m
    |
    | |
    ---------------------------------------------------------------------------------------------|
    0V

    I hope this shows up better!
     
  3. Guest

    Sorry guys, the diagram does not show up0 good!
     
  4. John Smith

    John Smith Guest

    Set your font to fixed font to view it ...

    JS
     
  5. Greg Neill

    Greg Neill Guest

    Set your newsreader to use a fixed width font
    for composing posts.
     

  6. He is using 'Google groups' not a newsreader.


    --
    Service to my country? Been there, Done that, and I've got my DD214 to
    prove it.
    Member of DAV #85.

    Michael A. Terrell
    Central Florida
     
  7. John Smith

    John Smith Guest

    Thanks, I missed that ...

    Regards,
    JS
     
  8. Tom Bruhns

    Tom Bruhns Guest

    Presumably 2* 0.06 is only 1.2 in your part of the world?? ;-)

    I suspect you've drawn the diagram in a slightly strange way; do you
    mean that the motor, the wires, the pressure transducer (pressure
    switch that becomes a very low resistance to turn the motor on??), the
    5 volt power source, and the wires are all one series loop? If so,
    your reasoning is sound -- just correct that multiplication.

    Cheers,
    Tom
     
  9. ehsjr

    ehsjr Guest

    This might be what you are trying to draw:


    +---------------+------------S +5
    | | U
    Motor Pressure P
    | Sensor P
    | | L
    +---------------+------------Y 0v

    |<-- 1.5m-->|

    Your math was correct until the last step (which should
    equal .12V, not 1.2v), but imprecise, because you said
    the current the motor draws was *about* 2 amps. How
    far off it is is anybody's guess, but it could be far
    enough to make the math irrelevant for whatever it is
    you are trying to do.



    I'm going to take a guess at what might be behind
    your question, and make an assumption.

    I think what you really may need is a stable supply
    to your pressure sensor, rather than knowledge of
    how much voltage drop it "sees". Depending on a lot
    of factors, this is what you may need to do:

    +---------------+---------------------+
    | | a |
    | Diode |
    | | |
    | +---------+ S +5
    | | | U
    Motor Pressure Big P
    | Sensor Capacitor P
    | | | L
    +---------------+---------+-----------Y 0v

    That might keep the voltage to the sensor
    stable enough for your (unstated) needs.

    Ed
     
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