36 cells, each of internal resistance 0.5 Ω and emf 1.5 V each, are used to send current through an external circuit of 2 Ω resistance. Find the best mode of grouping them and the current through the external circuit.
Given
\(N = mn = 36\) cells
Each cell: \(E = 1.5\ \text{V},\ r = 0.5\ \Omega\)
External resistance \(R = 2\ \Omega\)
Asked
Best grouping (\(m\) rows × \(n\) cells) and the maximum current \(I\).
Formula
\(I = \dfrac{mnE}{mR + nr}\)
Maximum current when \(mR = nr \;\Rightarrow\; R = \dfrac{nr}{m}\)
Solution
Condition for maximum current: \(mR = nr \Rightarrow 2m = 0.5n \Rightarrow n = 4m\).
With \(mn = 36\): \(m(4m) = 36 \Rightarrow m^2 = 9 \Rightarrow m = 3,\ n = 12\).
At the matched condition, \(I_{max} = \dfrac{nE}{2R} = \dfrac{12 \times 1.5}{4} = 4.5\ \text{A}\) — denominator is just \(2R\), no need to recompute the full fraction.