I recently joined a 9 volt cell and two 1.5 volt cells to get an output voltage of nearly 6 volt. For 15 minutes or so, the battery worked perfectly. It did not even show any sign of heating and in the next 2 minutes the 2 1.5 volt batteries hiss out and leave a lot of smoke. I am not even sure if its called an explosion. All the batteries were general purpose walmart batteries.

The batteries were connected in the following way:


simulate this circuit – Schematic created using CircuitLab

My guess is that these particular cells were faulty/defective from the very beginning. Can there be any more possible and realistic reason for this kind of explosion?

PS: All the values in the circuit have been measured using a DMM and rounded of to nearest integer or 10th of an integer.

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    \$\begingroup\$ Be glad no one was hurt, and never do something like that again! Effectively you charged the lower voltage cells in a way they were not intended to tolerate, and are fortunate that they merely vented. Rather obviously your alleged "180 ohm" resistor is much smaller than that if you got over an amp of current through it. If you want a 6v battery, don't try to "subtract" voltages, rather get a collection of cells which will total to that range, for example 4 1.5v-nominal cells, or even an old style lantern battery (which is just four large cells in a common overcase). \$\endgroup\$ Jun 4, 2016 at 18:54
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    \$\begingroup\$ "Without the load", as in, short circuited? Don't do that. \$\endgroup\$ Jun 4, 2016 at 19:06
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    \$\begingroup\$ Your edit reads as though you shorted out the 180 \$\Omega\$ resistor and ran 1.2 A backwards through the two 1.5 V batteries. What did you expect to happen? \$\endgroup\$
    – Transistor
    Jun 4, 2016 at 19:07
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    \$\begingroup\$ If the 1.5V cells are not rechargeable, then it is not surprising that they exploded when charged at 300mA for 15 minutes. The "DO NOT CHARGE" warnings on the labels are there for good reason. \$\endgroup\$ Jun 4, 2016 at 19:51
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    \$\begingroup\$ You got lucky. If you'd have done this with more dangerous batteries, like Lithium ones, you could have seriously damaged yourself and your surroundings. \$\endgroup\$
    – Mast
    Jun 5, 2016 at 14:44

3 Answers 3


What you are trying to do here is conceptually flawed, and as you were fortunate to discover without injury, potentially dangerous.

Do not try to subtract battery voltages

If you want a 6v battery, don't try to subtract voltages, rather get a collection of matching cells which will total to that range, for example 4x 1.5v-nominal cells, or even an old style lantern battery (which is typically just four large cells in a common overcase).

"subtraction" amounts to charging, and charging is something that must only be attempted with care to confine it to what is appropriate for the chemistry and charge state of the cells in question, which for non-rechargeable types (or rechargeable types that are already "full") amounts to "don't do that"

Do not place an ammeter across a battery or power supply

An ammeter is intended to be inserted into a break in a legitimate circuit. If you apply one directly across a source (or for that matter a load resistor) you short-circuit it and end up essentially measuring the internal impedance of the battery in comparison to that of the meter, which is both something of little relevance, and something that can lead to potentially dangerous levels of current flow.

  • \$\begingroup\$ And if you have to "subtract" some voltage, use a resistor instead. \$\endgroup\$ Jun 4, 2016 at 20:55
  • \$\begingroup\$ @ToddWilcox - the voltage dropped by a resistor varies widely with load. To avoid that, one can use a feedback controlled resistor (sold as a "linear voltage regulator") but this is of course wasteful, and it is necessary to design to accommodate the fact that the waste from a power resistor or linear regulator occurs in the form of heat. \$\endgroup\$ Jun 4, 2016 at 20:59
  • \$\begingroup\$ I did not want a specific battery, I just wanted to see ow feasible it is to subtract voltages and get a specific output. That is the reason I did not use a linear regulator or something like that. It was deliberately made like this, to find out what would happen. \$\endgroup\$ Jun 5, 2016 at 9:16
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    \$\begingroup\$ @AadityaSahay "Let's see what happens" is a surprisingly common cause of death among adolescents. \$\endgroup\$ Jun 5, 2016 at 14:03
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    \$\begingroup\$ @JanDvorak And also a surprisingly common cause of adolescents, I'm sure. \$\endgroup\$
    – user253751
    Jun 6, 2016 at 4:32

If your circuit is drawn correctly, then the current is flowing backwards through the 1.5V cells. This is like trying to (re)charge them beyond their design voltage. It is very bad for them, and yes, they may catch fire. An actual explosion is possible but very unlikely, as the case is designed to vent off the pressure and stop the explosion - and it sounds like this is what happened here.

  • \$\begingroup\$ Sorry JackB i wrote down the wrong current ratings, The 1.2 Ampere is without the load and 0.03 Ampere with the load, does that cause any problems? \$\endgroup\$ Jun 4, 2016 at 18:59
  • \$\begingroup\$ Not really. It is the direction, not the size, of the current that is the problem. A smaller current means it will take longer to start a fire, and you might get a smaller fire, but it will still do it. If you had actually put an amp backwards through the batteries, they would probably have exploded in only a few seconds. \$\endgroup\$
    – Jack B
    Jun 4, 2016 at 19:25
  • \$\begingroup\$ @AadityaSahay Are you asking if recharging a nonrechargeable battery is okay as long as you do it slowly? \$\endgroup\$
    – user253751
    Jun 4, 2016 at 21:25

As Chris says, you were charging the 1.5V cells in a way they werent designed to be. If those cells were normal alkaline, they are not designed to be charged at all. Never put alkaline cells in a position where current flows into their anode. Also, never charge a cell beyond its nominal voltage. Charging batteries is a dangerous science depending on the type and the current and voltage have to be carefully controlled.

  • \$\begingroup\$ i wrote down the wrong current ratings, The 1.2 Ampere is without the load and 0.03 Ampere with the load, does that cause any problems? \$\endgroup\$ Jun 4, 2016 at 19:00

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