Source: High school physics (Chinese)
Problem
Four identical batteries (each with EMF $\varepsilon$ and internal resistance $r$) supply the same load resistor $R$. Three arrangements are considered:
(a) Two parallel branches, each containing $2$ batteries in series. (b) All $4$ batteries connected in series. (c) All $4$ batteries connected in parallel.
Compute the effective EMF $\varepsilon_{\text{eff}}$ and internal resistance $r_{\text{eff}}$ for each, then $P = \dfrac{\varepsilon_{\text{eff}}^2 R}{(r_{\text{eff}} + R)^2}$.
(a) $2$ series $\times$ $2$ parallel: $\varepsilon_a = 2\varepsilon$, $r_a = r$. $P_a = \dfrac{4\varepsilon^2 R}{(r + R)^2}$.
(b) $4$ in series: $\varepsilon_b = 4\varepsilon$, $r_b = 4r$. $P_b = \dfrac{16\varepsilon^2 R}{(4r + R)^2}$.
(c) $4$ in parallel: $\varepsilon_c = \varepsilon$, $r_c = r/4$. $P_c = \dfrac{\varepsilon^2 R}{(r/4 + R)^2} = \dfrac{16\varepsilon^2 R}{(r + 4R)^2}$.
$P_a > P_b$: $\dfrac{4}{(r+R)^2} > \dfrac{16}{(4r+R)^2}$, so $(4r+R)^2 > 4(r+R)^2$, giving $4r+R > 2(r+R)$, i.e., $R < 2r$. $P_a > P_c$: $\dfrac{4}{(r+R)^2} > \dfrac{16}{(r+4R)^2}$, so $(r+4R)^2 > 4(r+R)^2$, giving $r+4R > 2(r+R)$, i.e., $R > r/2$.Intersecting both: $r/2 < R < 2r$.