A battery consists of N identical cells, each of e.m.f. ε. Is it true that the energy wasted when using the battery to charge a capacitor through a resistor can be reduced by charging it in N stages? That is by connecting it first across a single cell, and then across two cells, and so on, rather

Question

A battery consists of [latex] N [/latex] identical cells, each of e.m.f. [latex] ε.[/latex] Is it true that the energy wasted when using the battery to charge a capacitor through a resistor can be reduced by charging it in [latex] N [/latex] stages? That is by connecting it first across a single cell, and then across two cells, and so on, rather than across the whole battery in a single step.

Accepted Answer

The energy ultimately stored in the capacitor is [latex] \frac{1}{2}CN^2ε^2 [/latex] in both cases. Direct connection results in an equal amount of wasted energy. For charging by stages, the total work done by the battery is

[latex] \Delta Q \sum{(\Delta V)} =Cε (ε +2ε +...+Nε)=\frac{1}{2}Cε^2N(N+1). [/latex]

The wasted energy is thus

[latex] \frac{1}{2} Cε^2[N(N+1)-N^2]= \frac{1}{2} Cε^2N, [/latex]

i.e. only [latex] 1 / N [/latex] of the original loss.

Note. You can obtain the same result by considering the following graph. The triangle [latex] OAB [/latex] represents the stored energy of the capacitor, and the triangle [latex] OAC [/latex] corresponds to the energy wasted by charging it in a single step. The small shaded triangles show the energy wasted when using multiple (e.g. [latex] N = 4 [/latex]) steps. The total area of the shaded triangles is [latex] 1/ N [/latex] of that of [latex] OAC. [/latex]

Question P.155: A battery consists of N identical cells, each of e.m.f. ε. I...

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