I have a 12VDC solenoid and I was trying to measure amp draw. Unfortunately I dont have a 12V battery handy. So I connected it to a 9V-6xAA battery pack.
I first measured the ohms across the terminals of the 12VDC solenoid and got 43Ohm, which would suggest its a 285mA valve.
Then I measured the ampreage when powering it from the 6xAA-9V battery pack, which is know is sub-optimal because its 3V under, but its what I have handy. I got 195mA.
I have 2 questions:
Does it seem about right, comparing the 285mA with the 195mA? I know if I used a 12V source Id probably get more mA running through it.
When measuring the amperage, I noticed that it went to 195mA and upon disconnecting it, I saw the meter jump to 700mA, sometimes 2 and even 3Amps. Is this the back current?
What do you mean by "these are ratios"? The 285 mA I got by dividing 12vdc nominal by 42Ohms as I measure using 2 different meters. I know I got less mA because I used lower voltage, but I the relationship wasn't exactly linear. After all 9/12 is 75% but that would be 213mA.
As for the end, that's what I thought. I used a centech p30756.
Did you measure your 9V under load. If not you may well find that cell internal resistance was contributing to your "non-linearity"
Your current on disconnection was a spurious reading and not "true" current. Induced emf is voltage, not current. An inductor endeavours to maintain current flow by increasing voltage so what you saw induced voltage affecting your meter - in effect probably radio frequency noise.
Measure the 9V battery voltage with the solenoid connected and use that voltage in your equation.
You will find you are not working with 9V when you load the battery down with the solenoid.
I did connect the solenoid to the 9v battery, actually not a 9v, it was a 6xAA pack.
So those jumps from .195 to 2.0 and 3.0 was back emf which has no concrete interpretation? OK.
So you say I should connect the solenoid to the battery pack and then measure the V on the battery pack and it won't be 9, it'll be less, maybe 8.5v. Then I use that voltage:
8.5V = I * 43Ohms
To calculate my mA?
You mean I = 1000*V/43 mA
Where did you get 1,000V?
I = V / R Units of Amps, Volts and Ohms
To convert Amps (A) into milliamps (mA) you multiply its value by 1000. So a factor of 1000 appears on the right hand side of the equation. For example 6Amps = 6*1000mA = 6000mA
Nothing to do with 1000volts
Oh yeah I know that. I didn't notice I wrote out the equation in standard units and then wrote "To calculate my mA".
The 6xAA battery pack reads 8.8V unloaded.
Ok so I get 7.5V at the solenoid terminal while loading the 6Xaa:
7.5V = I x 0.043Ohms
7.5/0.043 = 174mA
Which is indeed closer to 195mA.