0
$\begingroup$

"If we connect two batteries in parallel, the battery with higher voltage will discharge into the other one." What is the intuition behind this?

$\endgroup$
  • $\begingroup$ why would you connect two batteries in parallel? :P I would expect an explosion too! $\endgroup$ – shortstheory Jan 2 '14 at 18:30
1
$\begingroup$

The voltage across the ends of a battery is estimated by $$ V = E - I r$$ where $E$ is the volt rating, $I$ is the current and $r$ is the (small) internal resistance of the battery.

So with two batteries $E_1$ and $E_2$ connected in parallel to some load $R$ they must share a common voltage

$$ V = E_1 - I_1 r_1 = E_2 - I_2 r_2 $$

and the total current through the load is the sum of the battery currents

$$ I = \frac{V}{R} = I_1 + I_2 $$

with solution

$$\begin{aligned} I_1 & = \frac{E_1 (R+r_2)-E_2 R}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_1 -E_2}{r_1+r_2} \\ I_2 & = \frac{E_2 (R+r_1)-E_1 R}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_2 -E_1}{r_1+r_2}\\ V & = \frac{R (E_1 r_2 + E_2 r_2)}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_1 r_2 + E_2 r_1}{r_1+r_2} \end{aligned} $$

given that $r_1 \ll R$ and $r_2 \ll R$.

So $I_1$ and $I_2$ are equal and opposite to each other. If $E_1-E_2$ is postive (first battery has more voltage) then $I_1$ is positve (supplying power to system) and $I_2$ negative draining power.

$\endgroup$
0
$\begingroup$

A battery is powered by a redox reaction. In the reaction the anode is oxidised and produces electrons while the cathode is reduced and consumes electrons. This is what produces the flow of electrons through your external circuit. As the reaction procedes and the current flows the battery is discharged.

However if you apply an external potential larger than the battery voltage you can drive electrons back into the anode and reduce it. This in turn pulls electrons out of the cathode and reduces it. The net effect is to recharge the battery, and indeed this is exactly what happens every time you recharge your phone.

In your example of two batteries the higher voltage one acts like the wall charger of your phone and it charges the smaller battery. In the process it is itself discharged.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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