Re: Ohm's law
- From: luca.pamparana@xxxxxxxxx
- Date: Wed, 16 Jan 2008 15:23:41 -0800 (PST)
On Jan 16, 8:26 pm, John O'Flaherty <quias...@xxxxxxxxx> wrote:
On Wed, 16 Jan 2008 12:03:50 -0800 (PST), luca.pampar...@xxxxxxxxx
wrote:
Hello all,
Sorry for the repost but I could not get my schematic to show up
correctly.
Hi everyone,
I have the following set up:
http://farm3.static.flickr.com/2098/2197444321_ffd34dd2d4.jpg?v=0
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.
By "cable", do you mean a single wire, or a pair: is the total
resistance .03 ohms or .06 ohms?
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.
That would be 0.12 V.
Is this reasoning correct? Or is there a special way to add them?
Yes, if the voltage is measured at the motor.
--
John
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?
Thanks,
Luca
.
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