Anybody know how many amps a 9v battery puts out?

Question

im trying to build an electrolysis thingy to clean some stuff up, the plans i have say i need an ac/dc converter and suggests 6-12 Volts and 250-500 mili-Amps, i dont know much about electronics and was wondering if a 9v battery would work as a substitute power supply since theyre cheaper and id feel safer not cutting up stuff plugged into the wall and putting it in water


an ac/dc converter. 6-12 Volts and 250-500 mili-Amps would work

someone said its about 100 amps... are amps additive? so could i use 2-3 9v batteries?

Answer 1

Sorry folks but you're all way off. A 9 volt battery is only going to supply about 50 milli-amps of current. Remember the size of the battery and how it is constructed. It is literally a pile of 1.5 volt cells stacked together to make a 9 volt package. Those cells are just so big, and have limited current as a result
For your question, get the 6 to 12 volt AC operated power pack. You'll be dealing with the low voltage side, and barring a failure of the device, your load will never see the 110 volt supply side. To get the kind of current that you need from 9 volt batteries, figure of 50 mA (milli-amps) each, divided into the current the project needs. If it is 500 mA, then you need 10, 9 volt batteries. They are not going to last very long either. So, for your needs and desires, go get the AC operated power supply, just fuse it at a bit less than what you need it to produce in terms of current draw.

Answer 2

if you short circuit a new 9v battery it will start out at about 1.5 amps but it drops steadily down to about 1 amp in less than 10 seconds then it starts getting hot and voltage and current drop at an accelerating rate until the battery is toast at somewhere near the 1 minute mark. Best to not try to get more than about 50 to 100 ma and the less the better if you want it to last longer than it took to read this :-)

Answer 3

It would depend on the load, according to Ohm's law, I = V/R

Where: I is the current in amps;

V is the voltage; and,

R is the resistance of the load in Ohms.

--------------------edit----------------

If you hooked the batteries in series (not parallel), then the voltages will add.

According to Ohm's law, doubling the voltage should double the current, if the resistance of the load remains the same. Likewise, tripling the voltage will triple the current for the same load (and so on).

Why would you want to do this? If the load is designed to take only a 9 volt drop, then you could burn out the circuit that you are supplying.

Hooking up batteries in parallel, by contrast, will make the individual batteries last longer, since the power drain is divided evenly among all of them, but the voltage and the current output should be the same.

Answer 4

Depends on the battery, and how fresh it is...

You have to have more info then that.
What kind of load are you putting on it.
The formula is I=E/R
I=current in amps
E=voltage (9 in your case)
so you have to know "R"

Technically on paper the battery will attempt
to produce whatever amps the formula calls for.
But in reality it doesn't have the stored power
to give lets say 20amps.

After your update- 9v battery would be pretty puny for electrolytic cleaning. A deep cycle 12 battery (like for a boat's trolling motor) would be better suited.

Answer 5

9v = ?Amps x ? Omhs (whatever the resistance of the object
you are powering divided by 9v = Amps = Omhs/Volts
(probably be in mA)

Answer 6

About a hundred milliamperes. In theory as the resistance is lowered the amps will rise but this is the limit.