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Name: Ani
Status: Student
Grade: 6-8
Location: PA
Country: United States
Date: January 2008

I am working on a science fair project involving fruit and vegetable batteries (e.g. lemon battery). I finished my experiments testing various electrolytes and electrode materials. I want to calculate the power (I*V) of each of my fruit and vegetable battery for comparison. I am little confused about the correct way to calculate power. I have used I*V1, where V1 is the closed circuit voltage, as there is no current flowing without external resistance and the closed circuit voltage does reflect internal resistance. In this case Power is ((V1^2)/Rext ), where Rext is external resistance. Is this correct? Or should I have used I*V0, where V0 is open circuit voltage, in this case power comes out to be (V1*Vo/Rext)? Please confirm the right way to calculate.

You should use either I*V1, or (V1^2)/Rext, (and they should be the same) because this is the power that the battery actually delivers. V0 is interesting information by itself, because it tells you the maximum voltage that the battery can apply to a very high-resistance load.

I*V0 is not a useful quantity. It does not tell you what the battery actually does or could do, since I and V0 are measured under different conditions.

Tim Mooney

Hi Ani,

This is a good question! Your original method of calculating power is correct. The power you are interested in here, is the power dissipated (or work done by) the load that the battery is driving this case, the load is your resistor. As you have stated, this is calculated by (using your notation) V1^2/R, where V1 is the closed circuit voltage (or in other words, the voltage that appears across the load resistor as current flows though it).

Open circuit voltage is of no interest, because it only occurs when no power is being produced. Power dissipated by a battery's internal resistance is wasted power, and is of little interest when comparing a battery's ability to deliver power to an external load. You are trying to measure how much useful power your battery can deliver to the load (your resistor), so all measurements must be taken at the load resistor. This can be determined either by your method of the square of the voltage across the resistor, divided by the resistor's value in ohms, or simply by multiplying the voltage across the resistor, by the current in amps flowing through it. The result is the same, but since you already know the resistor's value, your method is easier since you don't have to measure the current flowing though the resistor.


Bob Wilson


I am so glad you have your thinking cap on. Yes, V1, the voltage under load, is the right voltage to use.

It helps to really define what you want to measure: a) the power being delivered to the external resistor, or
b) the total power being released into the world as heat, whether outside at the resistor or inside the battery, so long as it's caused by that chemical reaction.

For (a), V1 is right, because power P(resistor)= V(resistor) * I(resistor). Vo is never seen by the resistor, so it can't be the right one.

For (b), Vo _may_ be about right, but even that's not a sure thing. This is based on the idea that when current flows and V1 shows, somewhere inside the battery Vo still exists, but voltage drops are incurred as the current goes from chemicals into the electrodes, which reduce the output to V1. Sometimes this is quite true, other times, maybe the chemicals present have been changed (depleted) by the current that has recently been used, so even Vo goes down.

As an engineer, I would really like to know how much outside work I can get done using this battery. Can I run my electric watch with it? (0.1 milliAmp at 1.5v) Can I run a nice bright LED light with it ? (~10 mA at 2-3v) Definition (a) is the right question for this, so I think V1 is the right voltage.

Of course, a scientist may want to know the whole power being "created", out of academic curiosity. Vo would be right for that.

Jim Swenson

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