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How to calculate the running time of a battery?

  1. Oct 27, 2016 #1
    Hello
    I have a circuit. When I power the circuit with a power supply, the power supply reads 5V and 0.025A. This gives me the P=IV=0.125W. Is this the same as 0.125W per second?
    I want to power this circuit with a Li-Ion battery that is rated 5V and 3200mAh. How long will my battery run this circuit?
    Thank you
     
  2. jcsd
  3. Oct 28, 2016 #2
    0.125 W is a measure of power - in other words, energy per unit time (joules/second). "0.125 W per second" doesn't make a lot of sense in this situation - or in any that I can think of right off hand. The 3200 mAh specification on the Li-Ion battery is basically telling you its current capacity. Allegedly, it can deliver 3200 mA for 1 hour, or deliver 1600 mA for 2 hours, or deliver 1 mA for 3200 hours. So if your circuit draws 0.025 A (or 25 mA), how long will the Li-Ion battery last? Of course, all of those numbers are approximations. Your performance may vary.
     
  4. Oct 28, 2016 #3
    I don't think anyone would measure Watts per second unless you wanted to account for some change in power consumption over time.

    The Amp-hour rating or mAh is a unit of charge. Since current is coulombs per second, Amp-hours scaled to seconds is Coulombs per second times seconds which is just Coulombs. Watt-hours is a unit of energy in the same sense. They could just say a battery stores so many coulombs and so many Joules, but Amp-hours and Watt-hours are more practical so that's what is used.

    As stated a battery with 3200 mAh can deliver 3.2A for an hour. That would be 3.2A for 3600s which is 11,520 Coulombs, the charge capacity of the battery. Simply divide Ah rating by current demand and you get run time in hours.

    Watt-hours is the integral of charge capacity across voltage. Nominal voltage for a standard Li-Ion battery is 3.7V though voltage varies from 4.2V fully charged down to around 3V fully discharged. 3.7V is the nominal value used to simplify calculation of energy capacity down to charge capacity times voltage.

    The caveat in calculating run time is that energy and charge capacity are not constant across all loads. Higher loads result in more energy wasted across battery internal resistance reducing the effective use of that energy. Typically Watt-hour and Amp-hour ratings are provided by manufacturers under a light load such as 1/5C. "C" stands for charge so for that particular battery it's 1/5 times 3200mAh for 640mA. Higher loads result in lower effective energy and charge capacities which need to be taken into consideration when calculating run times.
     
  5. Oct 31, 2016 #4
    Thank you.
     
  6. Nov 28, 2016 #5
    ##1 \text{mAh}=\frac{18 \text{Ampere} \text{Second}}{5}=3.6 \text{Ampere} \text{Second}##

    ##1 \text{mAh} \text{Volt}=3.6 \text{Joule}##

    ##3200 \text{mAh} \times 3.7 \text{Volt}=\frac{42624. \text{Kilogram} \text{Meter}^2}{\text{Second}^2}=42624. \text{Joule}##

    ##3.2 \text{Ampere} \times 1 \text{Hour}=11520. \text{Coulomb}##

    ##1 \text{Ampere}=\frac{1. \text{Coulomb}}{\text{Second}}##
     
    Last edited: Nov 28, 2016
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