I have some device with ac/dc adapter [9v, 500mA], and I want to replace the adapter with batteries.

If I used [6 * 1.5v] battries, is that acceptable? do I have to worry about the max current of the batteries? could the current of the batteries exceeds 500mA?

whenever I try to measure the current of the device connected to the battries, it gives me "1" "the upperlimit of the Multimeter", why? how?

And if so, what should I do?



You don't have to worry about providing too much current. The current is determined by the voltage and the load. To need to make sure that you can provide enough current.

In your case, it should work out fine. An AA battery (alkaline) can provide at least one amp, so your 500mA is within spec.

However, AA's have around a 2.5 Amp-hour capacity, so you won't get more than 5 hours (or so) of run time. Of course, the circuit likely doesn't take the entire 500mA, so it'll probably last longer than that.

Good luck!

  • \$\begingroup\$ Could the device be made to drain the whole 500mA of its adapter(could have low resistence)? and what if I connected the 2.5A batteries and the device drained all 2.5A? \$\endgroup\$ – Tito Tito Apr 11 '14 at 13:51
  • \$\begingroup\$ @TitoTito The adapter's rating is 500mA, which means that it con continue to provide at least 500mA continuously at it's rated voltage (9V). (oops, I pressed Enter too soon... just a sec, I'm typing more...) \$\endgroup\$ – bitsmack Apr 11 '14 at 15:44
  • \$\begingroup\$ @TitoTito So, you won't ever "drain the adapter," even if your device needs all 500mA. If your device needs, for example, 250mA, then that's all it will take, no matter how much current the power adapter can supply. As far as batteries, your device will tke current from them at the same rate is does from the adapter. Except, in this case, the batteries will discharge over time. As they do, the voltage will fall, and the device finally won't have enough voltage to function. No explosions, though :) \$\endgroup\$ – bitsmack Apr 11 '14 at 15:49
  • \$\begingroup\$ I just re-read your comment. If the device was very-low resistance, then it would try to take more current that the suply can handle. In this case, a few things can happen: 1) The voltage will drop, and the device won't work right. 2) The supply might protect itself by turning off its output voltage. 3) The supply might get hot and/or become damaged. But these will only happen if the device fails. The power supply that came with it will have enough current capability for the device. If you short batteries, its possible that they explode, but I've never heard of it actually happening. \$\endgroup\$ – bitsmack Apr 11 '14 at 15:55

If you wire your batteries up in series to get the voltage you require, the current will be the same as from a single battery, if you do need to change this you need to wire another set of 6 in parallel.

That having been said, a very quick search showed that most AA batteries are rated to give a max of about 2400mA, which is ample. HOWEVER 500mA is a fairly large current draw assuming that the device actually requires this much, (which it won't) this means you will run the batteries down quickly, although you do have 6 of them...

To use your multi meter to measure current you must place it in series with the supply, so, positive end of supply - positive multi meter probe, (goes through multimeter) - negative end of probe to positive input of your device - negative end of device to negative end of supply (as normal).


  • \$\begingroup\$ I connected the multimeter in series, but i'm not sure about which probe to which end, whatever. could the device be made to drain the whole 500mA(could have low resistence)? and what if I connected the 2400mA batteries and the device drained all 2400mA? would the device blow?!! \$\endgroup\$ – Tito Tito Apr 11 '14 at 13:47
  • \$\begingroup\$ As Bitsmask says, don't try and measure the current just across the battery, you might damage/fry something. measure the current draw through the device you want to power, connect the meter as I stated in my answer. \$\endgroup\$ – Tim Mottram Apr 11 '14 at 14:03
  • \$\begingroup\$ Sorry, I meant blup1980. \$\endgroup\$ – Tim Mottram Apr 11 '14 at 14:18

As @Tim Mottram said: Amp meter has to be connected in series.


A multimeter in amp meter mode is a device that measure a voltage accross a very small resistor that is connected internally between the two probes of the meter.

If you connect thoses probes to the power supply you will actuall load your power supply with a very small resistor.

Because I = U / R, with R very small, I will be very big. For sure it will be big enough to blow the fuse included into you multimeter. (or blow your multimeter if it's a very very low cost model). A single cell battery would have an internal resistance that is big enough to limit that current and save your meter. But a power supply would for sure have enough power to blow your meter.


Ohms Law, Ohms Law, Ohms Law.

If the voltage is 9V and the load is 30 ohms (for example), the current taken by the load is:-

I = \$\dfrac{V}{R}\$ = 9/30 = 300mA.

If your power supply (or battery pack) is capable of supplying 1000 amps it still has to obey ohms law and that means I = V/R and if the load is 30 ohms it'll take 300mA from a 9V power source.

Don't connect your meter (measuring current) across your battery because you'll measure the full output current from the battery and this might damage your meter or burn small wires.

You current charger is rated at 500mA and is adequate - this means your load takes 500mA or less (maybe 300mA as per my math above). You don't need to measure this to know this.


Put your meter back in its case and make sure the leads are plugged back into measuring voltage because you might forget next time you use it across the AC in your home and then there will be smoke and it'll be your meter burning possibly.

If you want to know how much current a battery can produce, look at its data sheet and don't try measuring it.


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