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ohm's law

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

Reply to
luca.pamparana
Loading thread data ...

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!

Reply to
luca.pamparana

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

sensor

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

Sorry guys, the diagram does not show up0 good!

Reply to
luca.pamparana

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

Pressure sensor

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

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

JS

Reply to
John Smith

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

Reply to
Greg Neill

He is using 'Google groups' not a newsreader.

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Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

Thanks, I missed that ...

Regards, JS

Reply to
John Smith

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

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

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

Reply to
Tom Bruhns

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

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

This might be what you are trying to draw:

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

||

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

Reply to
ehsjr

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