Let me clarify. I know how series and parallel setups work. I know that in series the voltage adds but the current stays the same. In parallel current adds but voltage doesn't.

But that doesn't really answer my question about the energy consumption of a given setup.

Let's use an inverter setup as an example.

Say we have a light that uses 10 watts (an LED bulb)

And we have two different inverter setups:

  • an inverter on a 12V battery with 20Ah
  • an inverter on two 12V batteries with 20Ah (wired in series for 24V)

The second battery setup doesn't have more current, just more voltage.

If the LED bulb is consuming the same amount of power in both setups, what happens in the second example?

Does the current draw per battery get divided in half?

I think my misunderstanding comes from the fact that I don't know how electricity storage works with batteries. Elsewhere I've read that I shouldn't think of it in terms of watts but in terms of Ah storage. However, that goes against my way of thinking that \$volts \times amps = watts\$.

Say we have the same set up as above, will the available watts in the second example be more than then first?

$$ 2 \times 12V \times 20Ah = 480W $$

compared to:

$$ 1 \times 12V \times 20Ah = 240W $$

  • \$\begingroup\$ The key point to all this is that you are presupposing Ohm's Law. However Ohm's Law only applies to Ohmic devices, which is a bit circular. Anyway, an Inverter by its very nature is a non-Ohmic device. \$\endgroup\$
    – Aron
    Aug 16, 2015 at 18:03

2 Answers 2


I'll give you an answer covering a bit of general principles, since the specific outcome will have some dependence on the specific type of inverter.

An inverter seen from the terminals of the battery pack (however it is arranged, 12V, 24V, etc.) will look as a constant power load, i.e. as a two-terminal device that always absorbs the same amount of power, as long as its load (the bulb) draws the same amount of power. That's because the internal circuitry will try to deliver a constant voltage to the load and as long as this latter (the bulb) draws the same amount of current at constant voltage, the product \$V \cdot I\$ at the bulb, i.e. the power output of the inverter, will be constant.

Of course inverters don't have 100% efficiency, so the constant power drawn from the batteries will be higher than the constant power delivered to the load to account for losses inside the inverter.

That means that if the same inverter is capable of working with different values of input voltage, it will draw more current when it is connected to a lower voltage battery pack. The value of the capacity of the battery (ampere-hours) here is not relevant. The only thing that matters is that the same current will go through both batteries when they are connected in series.

Let's do some calculations. Let \$P_L=10W\$ be the power drawn by the bulb and \$P_i\$ that drawn by the inverter from the batteries. Assuming a ballpark efficiency \$\eta=0.9=90\%\$ you have \$P_i = P_L / \eta \approx 11 W \$.

This means that the current drawn from the battery pack will be \$I_b = P_i / V_b\$, where \$V_b\$ is the battery pack voltage, i.e. the voltage the inverter input sees.

So we have two cases:

\begin{align*} (1)&& V_b &= 12V &&\text{(single 20Ah battery)} \\[2 em] (2)&& V_b &= 24V &&\text{(two 20Ah batteries in series)} \\[2 em] \end{align*}

The current in the two cases will be:

\begin{align*} (1)&& I_b &= P_i / V_b = 11W / 12V \approx 900mA \\[2 em] (2)&& I_b &= P_i / V_b = 11W / 24V \approx 450mA \\[2 em] \end{align*}

So the capacity of the battery doesn't come into play here. It comes into play, of course, to calculate how many hours of service those batteries would deliver. In fact the capacity \$C\$ of a battery is a rough estimation of how much time the battery can deliver a given current (don't mistake C for the capacitance of a capacitor, though! They're not the same thing at all, despite the confusingly identical symbol!). Mathematically:

$$ C = I_b \cdot t \qquad\Rightarrow\qquad t = \dfrac {C} {I_b} $$

Therefore if \$C = 20 Ah\$ the estimate time of service will be:

\begin{align*} (1)&& t &= \dfrac {C} {I_b} = \dfrac {20Ah}{900mA} = \dfrac {20Ah}{0.9A} \approx 22h \\[2 em] (2)&& t &= \dfrac {C} {I_b} = \dfrac {20Ah}{450mA} = \dfrac {20Ah}{0.45A} \approx 44h \\[2 em] \end{align*}

This is a rough estimation, since the capacity of a battery is not an exceptionally precise parameter when it comes to evaluate the service time. To be precise you need what are called the discharge curves of the battery (batteries, surprisingly, are not linear devices), as you can see in this datasheet for a lead acid battery:

enter image description here

Here is a zoomed pic of the curves:

enter image description here

Note how the service time changes with the discharge current. For higher discharge currents the battery will last less, of course, but if you do your math you'll see that it is not a proportionality law, i.e. doubling the discharge rate (i.e. the discharge current) reduces the time by more than half.

BTW, the energy in the battery is neither measured in ampere-hours (Ah) nor in watts (W), but the correct unit is the watt-hour (Wh), which is an alternative unit to the joule (J), which is the standard unit for energy. The conversion is simple:

$$ 1 Wh = 1 W \cdot 1h = 1W \cdot (3600 s) = 3600 W \cdot s = 3600 J = 3.6 kJ $$

The watt-hour is a more convenient unit when talking about devices that operate for long times and end up drawing large quantity of energy (not necessarily high power levels).

The watt (W) is the unit of power, instead, i.e. the rate of conversion of energy. \$1W = 1 J/s\$ Something drawing 1W from a source means that the device is converting/transferring/utilizing 1 joule of energy per second, i.e. for each second it is working a joule of energy is drawn from the source of power it is connected to.

The capacity of a battery C is a rough estimation of the energy stored in the battery. I cannot say why the industry introduced this strange unit, since you could well give the energy stored in the battery as a figure in watt-hours (and the manufacturers often do that on their technical datasheets). I can only guess that, since a battery is a roughly constant voltage source you can (very roughly) say that the capacity is proportional to the energy stored inside it, so it is easier to talk about ampere-hours, since amperes are more easily measured with common instruments, such as digital multimeters (which ordinarily cannot measure power directly).

  • \$\begingroup\$ In the second to last paragraph, "100W" should be "1W". \$\endgroup\$
    – Dan D.
    Aug 16, 2015 at 8:00
  • \$\begingroup\$ Wow. This is more than enough information. Way more than I was expecting. It answers my questions and then some. I'll refer to this and keep coming back because digesting this much information in one sitting doesn't work well for me. Thanks a bunch! \$\endgroup\$ Aug 16, 2015 at 9:02
  • \$\begingroup\$ It is amazing how for an already complex abstraction level you can write a whole chapter, and then in the next chapter could go on and say "Now considering the previous chapter, an inverters efficiency is not constant throughout changing supply voltages" and open a new whole can of worms to calculate through... \$\endgroup\$
    – PlasmaHH
    Aug 16, 2015 at 12:06
  • 1
    \$\begingroup\$ @PlasmaHH If you can highlight something wrong I said, I'll be glad to edit my answer. I always welcome positive criticism. If you can provide an answer that you deem better suited for the OP, you are welcome either. You could also gain more votes. What's the point with your comment? \$\endgroup\$ Aug 16, 2015 at 12:17
  • 1
    \$\begingroup\$ @PlasmaHH And, BTW, as it is the common guidelines of SE sites, the answer to a question should strive to be useful not only for the questioner, but also for other people browsing the site. Again, if you feel you could post a better answer, I'll be glad to have a chance of upvoting it. \$\endgroup\$ Aug 16, 2015 at 12:26

If you have an inverter it generally means it converts DC to AC so I'm assuming you means a buck regulator or flyback regulator - these are highly efficient energy convertors and if the load wattage is 10 watts then the power taken by the input of the converter is 10 watts plus maybe 1 or 2 watts in losses.

So, if the input supply rail is 12 volts, the current drawn by the converter to feed 10 watts to the load will be 12 watts / 12 volts - 1A. If the converter were 100% efficient the current taken would be 0.8333 amps.

If the voltage were increased to 24V (assuming the losses were 2 watts and the converter can handle 24 V), the input current to the converter will be: -

12 watts / 24 v = 0.5 amps.

If the 24V battery is made up from 2 12V batteries in series it makes no difference whatsoever.


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