Suppose I connect a 9 V DC battery to a DC-to-AC converter. Then connect the output of the converter to a step-up transformer, which increases the voltage to 2000 V. Now, if I put myself between the ends of the terminals of the secondary circuit of the transformer (i.e. I'm the load), will I be electrocuted? (Assuming my body resistance to be 1000 Ohms.)

I think 2000 V and 1000 Ohms can produce about 2 A of current through my body, which could be fatal. This means I'm drawing about 4000 J of energy per second from the battery. A quick google search tells me that a standard 9 V battery may contain 20,000 J of energy. If that's true, 2 A current will go through my body for about 5 seconds before the battery is completely drained. I'll be fried like a potato for sure if that happens.

Are my reasoning and numbers fairly correct? Can a 9 V battery kill you?

Also, what would happen if the 9 V DC is coming from an adapter connected to a 230 V AC power source and not from a DC battery? If I'm the load, would 2 A current flow non-stop through my body in such a scenario (until the AC power source is depleted)?

  • 2
    $\begingroup$ A standard 9 V battery is comprised of 6 series AAAA batteries, more or less: you can carefully take a dead one apart with pliers and see for yourself. These cells are smaller than AAA cells, so this post might help: electronics.stackexchange.com/a/591473/223146. In college, I once built a tickler circuit that stepped up the output of a 9 V battery to around 1200 V. I was going to use it for a prank, but got zapped right after I finished construction. Decided to skip the prank. No real harm, but not funny in any event. Ps. I did not down vote. $\endgroup$
    – Ed V
    Jan 3, 2022 at 20:07
  • $\begingroup$ I do not know what batteries or circuitry is used in stun guns, but maybe someone here can comment on that, if relevant. $\endgroup$
    – Ed V
    Jan 3, 2022 at 20:24
  • 3
    $\begingroup$ The 9 volt battery probably can't deliver enough power to kill you, but if you charged a capacitor and used that, you would be in real danger. $\endgroup$ Jan 3, 2022 at 20:28
  • 4
    $\begingroup$ Even dead shorted, you cannot get 2 amps from a 9 V battery for very long, so I don’t think your numbers are realistic. Basically, the battery will overheat and polarize (be rate limited at the electrodes). It is basically the power issue, as @DavidWhite commented. $\endgroup$
    – Ed V
    Jan 3, 2022 at 20:34
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    $\begingroup$ Your estimate for human resistance is quite low, on average the resistance of a (dry) human is more like $1 \text{ M}\Omega$ if the leads are directly attached to your outer skin layers. $\endgroup$
    – Triatticus
    Jan 4, 2022 at 2:21

1 Answer 1


If you had 2A at 2000V, it would be a minimum of 444 A for the batterie, so the voltage will drop to almost zero since the inner resistance of a 9V batterie is about 0.5 Ohm.

  • $\begingroup$ The cells are connected in series, so it is still 2 A per cell. $\endgroup$
    – dominecf
    Jan 3, 2022 at 21:22
  • $\begingroup$ In the question it was a transformer which made the 2000V not some 220 cells in series, an for ideal transformers I*U=const . so 2000V*2A =9V*444.4..A, so if this is the reason of down voting , please correct it. $\endgroup$
    – trula
    Jan 3, 2022 at 21:31
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    $\begingroup$ Transformer is actually too complicated a device; any inductor massive enough that would keep the 100mA+ current for some milliseconds after disconnecting the battery by bare hands could kill a human. And I just measured the secondary of a MOT to be some 100 Ω. Do not try at home. $\endgroup$
    – dominecf
    Jan 3, 2022 at 21:45

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