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I am powering a 12V 6W lightbulb with two 9V zinc-carbon batteries in series. The voltage should be high enough to power the lightbulb, but the light is only visible for around 40 seconds before it quickly dims out. Can you please help me understand this?

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  • \$\begingroup\$ Supplying 0.5A, those little 9V batteries go flat fast. \$\endgroup\$ Jun 5, 2021 at 17:06

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A 12V/6W bulb has a nominal resistance of V^2/W = 144/6 = 24 ohms.

At 18V a 24 ohm load draws 18/24 A = 750 mA.

A 9V battery typically only has a mAH capacity of around 200 mAH or less -- and note this varies depending on the current draw.

So that's why your 9V batteries don't last very long when powering this light blub.

Here's some tests performed on 9V batteries which give you an idea of how the current load affects battery life:

enter image description here

Source: https://www.powerstream.com/9V-Alkaline-tests.htm

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  • \$\begingroup\$ Ok, thanks. Does the mAh capacity double if I connect two batteries in series? Also, is there a way to predict/calculate the new mAh capacity if I discharge at a different current than what it was tested on? \$\endgroup\$
    – figbooxy
    Jun 5, 2021 at 14:24
  • \$\begingroup\$ No - the voltage doubles in series - capacity stays the same. Placed in parallel the capacity should double... but that assume each battery discharges at the same rate. \$\endgroup\$
    – ErikR
    Jun 5, 2021 at 14:30
  • \$\begingroup\$ Btw - your Zinc-Carbon battery probably is very small -- like only 50 mAH. \$\endgroup\$
    – ErikR
    Jun 5, 2021 at 14:30
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9V zinc-carbon battery has low capacity (200-400 mAh). Two batteries in series will have the same capacity. More important, it has very high internal resistance, about 25 ohms, and that increases as it discharges, which means it is not capable of providing high current.

9V batteries are intended for very low power appliances, like smoke detectors.

The solution to your problem is:

  1. Use LEDs, which provide the same light for 10x less current (at least)

  2. Use batteries with a voltage that suits the LEDs and the chosen driver, to maximize its efficiency.

Even for LEDs, a 9V battery isn't a good choice, as 2-3 AAs cost less, deliver more capacity and higher current, and the voltage matches one LED better.

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  • \$\begingroup\$ If my objective is to drain the batteries as quickly as possible (in a safe manner), is it an effective strategy though? \$\endgroup\$
    – figbooxy
    Jun 5, 2021 at 14:36
  • \$\begingroup\$ @figbooxy No. Always charge/discharge batteries at a safe rate, usually C/10, where C is the AH or mAH rating. So 20mA for a 200mAH battery. Otherwise there is risk of explosion depending on the battery, if the charge/discharge circuit does not have thermal protection. You were lucky that zinc chloride cells have a high internal resistance which reduced the output current. \$\endgroup\$
    – Indraneel
    Jun 5, 2021 at 14:51
  • \$\begingroup\$ If the battery overheats enough, the electrolyte will boil and it will vent, spewing noxious corrosive chemicals. It's not really a good idea. Besides, if you drain it quickly and disconnect the load, it will regain some charge after resting. \$\endgroup\$
    – bobflux
    Jun 5, 2021 at 15:05
  • \$\begingroup\$ @bobflux Is the phenomenon your describing in the last sentence the recovery effect? I used a multimeter to test the voltage of the circuit and after about ~15 minutes it read .5V, but when I disconnected the load the battery read 5 V and slowly increased until now it reads 8.5 V (30 minutes later). \$\endgroup\$
    – figbooxy
    Jun 5, 2021 at 15:28

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