An answer from the post "Battery life with hot and cold wires" talks about Peukert's law related to the battery capacity. Now, within the scope of undergrad electromagnetism courses, the following are two common scenarios:
Case 1 : A circuit with only a battery and a resistor.
Case 2 : A circuit with a battery, a resistor and a capacitor.

However, it seems like in none of those cases Peukert's law is taken into account. Furthermore, books always say that voltage is constant and that if you know the resistance, then you can apply (the macroscopic version of) Ohm's law $V=IR$.
For case 1, the power $P=VI$ , so a beginner might believe that a battery keeps giving energy away forever.
For Case 2, one is supposed to use $I(t)=\frac{V_0}{R}e^{\frac{-t}{RC}}$ and assume that $V_0$ is constant.


  1. Peukert's law talks about current. But what about voltage? Does voltage in a battery also decrease in the presence of a resistor?
  2. Is everybody supposed to ignore Peukert's law when doing undergrad electromagnetism ?

1 Answer 1


For an ideal battery, the terminal voltage is constant and independent of the load resistor, $R_L$. But real batteries have internal resistance, $R_b$, which must be considered in series with the load resistor. If $R_{b}<<<R_L$ then the battery terminal voltage will remain constant. If not, the battery terminal voltage will drop.

See the figure below. Note that there will be a voltage drop across $R_b$ when current flows, so that the terminal voltage will be the internal battery voltage minus the voltage drop. The voltage drop will be very small if $R_{b}<<<R_L$.

This also applies if the load resistor is in series with a capacitor.

Insofar as Peukert’s law is concerned, when I was an undergraduate it was not introduced. It is intended to give a better estimate of battery life than simply using the manufacturer’s ampere-hour rating.

Hope this helps.

enter image description here

  • $\begingroup$ Thanks for the answer. If, in an ideal battery, the voltage is constant, why do you call it terminal voltage? Also, if the voltage is constant, then according to $P=V^{2}/R$, the battery outputs energy infinitely nonstop? $\endgroup$
    – evaristegd
    Commented Oct 29, 2018 at 16:59
  • 1
    $\begingroup$ I will update my answer to show the equivalent circuit. $\endgroup$
    – Bob D
    Commented Oct 29, 2018 at 17:01
  • 1
    $\begingroup$ The diagram only addresses the output voltage of the battery as a function of load current and internal battery resistance, not the amount of energy that is available from the battery. The energy delivered by the battery comes from the conversion of the battery"s chemical energy to electrical energy. The battery stops delivering energy when its chemical energy is spent. That's where the ampere-hour rating and Peukert's law come into play. $\endgroup$
    – Bob D
    Commented Oct 29, 2018 at 19:47
  • 1
    $\begingroup$ @evaristegd Where did I say "if Rb is not 'too small', the battery terminal will drop"? I said the voltage drop will be very small if Rb (its internal resistance) is much much less than RL (the load resistor). The undergrad text books I had did not get into the life of batteries. They either treat them as "ideal" which means zero internal resistance and the terminal voltage is independent of the load, or "real" but only with respect to the fact that there is internal resistance which effects terminal voltage, not with respect to how long they last. $\endgroup$
    – Bob D
    Commented Oct 30, 2018 at 13:59
  • 1
    $\begingroup$ @evaristegd I'm not sure if I can go any further with this. Let's just say that as the chemical energy diminishes the ability of the battery to supply current to a load get less and less. In other words, it can only deliver current to a load resistor that that gets smaller and smaller until it reaches some point that no matter how low the load resistor is, the battery will not be able to deliver any current. At that point, it's all over. I can't add anything more. $\endgroup$
    – Bob D
    Commented Oct 31, 2018 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.