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Hi Cleggsan,

I hope you are genki, and not adversely affected by the weather, wherever you are.

A while ago, you helped me with a question about calculating the battery life for a device I was building. I now need to do the same thing again. I am really sorry, I tried to do the calculations based on your previous answer at that time, but I am getting more and more lost.

I have a device with 3,443,334 solenoids. Each one is powered with 0.00486 milliamps for 0.32 seconds, but the solenoids are powered twice in one operation.

I am trying to calculate how many operations I can perform with a 3.7V 1750mAh battery.

Could you tell me how I can calculate how many operations of the solenoids I can power with a 1750 mAh battery?

By my calculations it is around 11 times. Have I made a mistake here? (I am assuming I have)

Once again, I hope you are safe over there!

Best regards,

Eddie

hello,

Doing just fine. We are no where near to the storm area of the USA. Everything is just perfect for us.

This problem is a little bit different from the previous calculations. So, I am going to let you do the arithematic but I will explain how to do it. First, we are not going to use wattage terms. Since the solonoids are not running continuous we must look at the energy per solonoid and multiply the number of each.

Best, then, to calculate the number of joule required to operate on solonoid. Then multiply by the number of solonoids in the system. The sum total of joules can be compared to the available joules provided by the battery.

A joule is one amp per second. Here is an explanation of joules and conversion factor or formats. It will help you decide the numbers to use:

http://en.wikipedia.org/wiki/Joule#Practical_examples

Then, let me see you calculations and I will critique your methodology and answers.

Domo

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