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Ohm's law

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

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  1. Guest

    Hello all,

    Sorry for the repost but I could not get my schematic to show up

    Hi everyone,

    I have the following set up:

    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?

  2. By "cable", do you mean a single wire, or a pair: is the total
    resistance .03 ohms or .06 ohms?
    That would be 0.12 V.
    Yes, if the voltage is measured at the motor.
  3. Guest

    Great. Many thanks.
  4. Guest

    One last question...

    Why is it that the voltage drop is the same on each wire? One is
    connected to a 5V power supply and one is connected to the ground. How
    come the voltage drop is 0.06V for each of the wire?

  5. Guest

    Because current flows in a loop, you will have +A amps in one wire,
    and -A amps in the other, the sign just tells you the direction.
  6. It's the meaning of "voltage drop" : a change in voltage, which is
    independent of the absolute voltage. Voltage is a measurement between
    two points, one of which is taken as a zero reference. Now consider
    the voltages in your circuit, measured with the negative side of the
    power supply as the reference. Starting at the +5 V, going clockwise
    through the motor, you have +5 V, +4.94 V, +0.06 V, 0 V. Do you see
    how it works now?
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