# ohm's law

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

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 SmithGuest

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

JS

5. ### Greg NeillGuest

for composing posts.

6. ### Michael A. TerrellGuest

--
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 SmithGuest

Thanks, I missed that ...

Regards,
JS

8. ### Tom BruhnsGuest

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. ### ehsjrGuest

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