I guess people have asked slightly similar questions to mine before, but I still haven’t been able to find a complete answer as to WHY the current doesn’t increase even when the voltage in a circuit does.

When batteries are connected in series, their voltages add up. Now let’s imagine we put these batteries in a circuit with a load, but we do this by first connecting the load to ONE battery and finding the I (current,) then after that, adding the second battery in series BUT keeping the load and everything the same. By V=IR, wouldn’t this mean that since we’ve doubled the voltage in this case that the current would also be doubled, considering we keep the resistance (ie load) the same? I’ve done a sketch of an example, attached below:

enter image description here

Please provide a straight-to-the-point answer if you can. I don’t know why this has been bugging me so much but if I can understand this then I’m confident that I’ll be able to grasp pretty much everything else about electronics.

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    \$\begingroup\$ If you increase the votlage and keep the resistive load that same, then yes current will increase. What makes you think that is not the case? Can you provide an example where the current hasn't increased? \$\endgroup\$
    – Puffafish
    Commented Mar 9, 2023 at 13:25
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    \$\begingroup\$ I think you are confusing the amp-hours (or milliamp-hours) capacity of a battery with current. If you connect two of 1.5V/500mAh batteries in series the total voltage will be 3V yet the mAh capacity will stay the same. Is that what you mean? \$\endgroup\$ Commented Mar 9, 2023 at 13:28
  • \$\begingroup\$ @RohatKılıç oh I see, yeah I wasn’t thinking about the capacity (mAh) when I was asking the question but now that you mention it, it makes more sense - thanks!! \$\endgroup\$ Commented Mar 9, 2023 at 13:41
  • \$\begingroup\$ "So why is it said that current doesn't increase when batteries are in series". Can you provide an example of one of the times you've seen this claim? Your understanding is fundamentally correct in the simple examples you've shown with a resistive load. The story is more nuanced with a more complex load, which I suspect was the case in the examples you saw this claim, but would be helpful to add examples to the question. \$\endgroup\$
    – effect
    Commented Mar 10, 2023 at 21:37
  • \$\begingroup\$ You should also review Kirchhoff's laws, which quantify how current flows through a circuit and how voltage varies around a loop in a circuit. Those laws explain that current is equal, at all points along a current loop. Regardless of the number of batteries in series, the current flowing through any one battery is the same as the current flowing through the other batteries, because the universe enforces it. \$\endgroup\$
    – doug65536
    Commented Mar 10, 2023 at 22:24

3 Answers 3


They mean the maximum current doesn't increase. If you have a battery labeled as 9V 1A and you put two in series in a box, the label on the box should say 18V 1A, not 18V 2A.

But you don't have to use the full current that the battery is designed for. If you only take 0.1A from your 9V 1A battery, and you put two of them in a box and put the same resistor in series with the battery box, you get 0.2A, but that doesn't mean the batteries can support 2A.

Note that the full current is going through each battery. Battery 1 has to support the full current and so does battery 2.

  • 2
    \$\begingroup\$ Okay, so the maximum current (and NOT the current flowing through the circuit) is what is discussed when assessing how battery arrangement will affect current in the circuit right? That makes more sense, so thank you! \$\endgroup\$ Commented Mar 9, 2023 at 13:31
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    \$\begingroup\$ That's correct. If you connect them in parallel, they'll deliver more maximum current in total, but the voltage won't increase. \$\endgroup\$ Commented Mar 10, 2023 at 14:31

The current through the load certainly increases when you put a second battery in series with the load - you've used Ohm's law to prove it.

What people mean when they say "current doesn't increase when batteries are in series" is that the maximum current you can get from the batteries doesn't increase.

All batteries have a limit to the current they can deliver. If you have two batteries that can each deliver one ampere of current and you put those two batteries in series, then you can still only get one ampere out of the batteries. The voltage adds, but the maximum deliverable current stays the same.

If you put those same two batteries in parallel, then the current from the batteries adds. You can get double the maximum current out of them - but the voltage is only the voltage of one battery.

Look at it like this:

A battery is an ideal voltage source with a resistor in series with it:


simulate this circuit – Schematic created using CircuitLab

When you put two batteries in series, you also get two of those series resistors in series:


simulate this circuit

You've got twice the voltage, but also twice the internal resistance. That keeps the maximum current the same as with just one battery.

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    \$\begingroup\$ Ohhh okay that makes everything clearer! So all batteries have to have some internal resistance, right? Thanks a lot for the explanation!! \$\endgroup\$ Commented Mar 9, 2023 at 13:39
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    \$\begingroup\$ It isn't that they have to have, it is just that they do. An ideal voltage source has no resistance. A battery isn't ideal, though. It has resistance in the materials it is made of as well as limits to the chemical reactions inside it. \$\endgroup\$
    – JRE
    Commented Mar 9, 2023 at 13:43

It completely depends what the load is. There are different kinds of loads.

If you have a 12V battery and 12 ohm load which draws 1 A, adding another battery will double the voltage to 24V, and a 12 ohm load will draw double the current, 2A. No surprises there for purely resistive loads with constant resistance.

If you have a device with switch mode power supply, and it draws 1A from 12V battery for 12 watts, adding another battery to achieve 24V will make the device draw only 0.5 A, because it still needs only 12 watts. So doubling the voltage will halve the current, for a constant power load.

A constant current load would take 1A regardless of if you apply 12V, 24V, or maybe 6V.


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