Q.100 cells each of emf 5V and internal resistance 1Ω are to be arranged so as to produce maximum current in a 25Ω resistance. Each row contains equal number of cells. The number of rows should be (A) 2 (B) 4 (C) 5 (D) 100

[!TLDR] For a fixed total number of cells arranged m rows (parallel) by n in series (mn=100), current is maximum when the combination's internal resistance nr/m equals the …

Arrange the 100 identical cells (each emf $\varepsilon=5\,\text{V}$, internal resistance $r=1\,\Omega$) as $m$ rows in PARALLEL, with $n$ cells in SERIES per row, so $mn=100$ (total cell count fixed). Each row has emf $n\varepsilon$ and resistance $nr$; the $m$ identical rows in parallel share this same emf (parallel rows don't add emf) and combine to an equivalent internal resistance $\dfrac{nr}{m}$. The current delivered to the external resistor $R=25\,\Omega$ is $I=\dfrac{n\varepsilon}{R+\frac{nr}{m}}=\dfrac{n\varepsilon m}{mR+nr}$. Since $nm=100$ is fixed, the numerator $n\varepsilon m=100\varepsilon=500$ is a CONSTANT, so maximising $I$ means MINIMISING the denominator $D=mR+nr$, subject to $mn=100$ (i.e. $n=100/m$): $D(m)=mR+\dfrac{100r}{m}$. Setting $\dfrac{dD}{dm}=R-\dfrac{100r}{m^2}=0$ gives $m^2=\dfrac{100r}{R}=\dfrac{100\times1}{25}=4\Rightarrow m=2$ rows (each of $n=50$ cells in series). Substituting back, the equivalent internal resistance at this optimum is $\dfrac{nr}{m}=\dfrac{50\times1}{2}=25\,\Omega$, which exactly EQUALS the external resistance $R=25\,\Omega$ -- confirming the well-known …

Arrange n cells in series per row and m rows in parallel (mn = total cells, a fixed constant); write the delivered current I as a function of m alone using n=N/m, then minimise the denominator (mR+nr) using calculus to find the row count m that maximises I -- equivalently, use the kno …

Applying the maximum-current condition (internal resistance of the WHOLE combination equals R) to only the per-cell resistance r instead of the combination's equivalent nr/m, or forgetting that the row count m and cells-p …