Find the minimum number of cells required to produce an electric current of 1.5 A through a resistance of 30 Ω. Given that the emf of each cell is 1.5 V and internal resistance 1.0 Ω.
Given
Required current \(I = 1.5\ \text{A}\) through \(R = 30\ \Omega\)
Each cell: \(E = 1.5\ \text{V},\ r = 1.0\ \Omega\)
Asked
Minimum total number of cells \(N = mn\) and their arrangement.
Number of cells is minimum when current is maximum for given \(N\), i.e. when \(mR = nr\)
Then \(I_{max} = \dfrac{nE}{2R} = \dfrac{mE}{2r}\)
Solution
Series alone fails: \(I = \dfrac{nE}{R + nr} = \dfrac{1.5n}{30 + n} = 1.5 \Rightarrow 1.5n = 45 + 1.5n\) — impossible. So mixed grouping is required.
For minimum cells use the maximum-current condition \(mR = nr\), i.e. \(30m = n\).
Then \(I = \dfrac{nE}{2R}\): \(\;1.5 = \dfrac{1.5\,n}{2 \times 30} \Rightarrow n = 60\).
From \(30m = n = 60\): \(m = 2\).
Minimum number of cells \(N = mn = 2 \times 60 = 120\).
\(N = 120\) cells — 2 rows in parallel, 60 cells per row
Short Cut
At the matched condition each half of the emf drives \(R\): so \(I_{max} = \dfrac{nE}{2R}\). Solve for \(n\) in one line, then get \(m\) from \(mR = nr\). Also remember: at match, \(I_{max} = \sqrt{\dfrac{N E^2}{4Rr}}\) — handy for direct \(N\) questions.