Current Electricity • NEET Revision

Example 3.43

Mixed Grouping of Cells

Question

12 cells, each of emf 1.5 V and internal resistance 0.5 Ω, are arranged in m rows each containing n cells connected in series. Calculate the values of n and m for which this combination would send maximum current through an external resistance of 1.5 Ω.

Given

\(N = mn = 12\) cells
Each cell: \(E = 1.5\ \text{V},\ r = 0.5\ \Omega\)
External resistance \(R = 1.5\ \Omega\)

Asked

Values of \(n\) and \(m\) for maximum current (and the current).

Formula

\(I = \dfrac{mnE}{mR + nr}\)

Maximum current: \(mR = nr\) (external resistance = total internal resistance of battery box)

Solution

Condition: \(mR = nr \Rightarrow 1.5\,m = 0.5\,n \Rightarrow n = 3m\).
With \(mn = 12\): \(3m^2 = 12 \Rightarrow m = 2,\ n = 6\).
Maximum current: \(I = \dfrac{mnE}{mR + nr} = \dfrac{12 \times 1.5}{2 \times 1.5 + 6 \times 0.5} = \dfrac{18}{6} = 3\ \text{A}\).

\(n = 6,\ m = 2;\quad I_{max} = 3\ \text{A}\)

Short Cut

Think of it as impedance matching: maximum current (and power) is delivered when the battery box's internal resistance \(\dfrac{nr}{m}\) equals \(R\). Set \(\dfrac{nr}{m} = R\), pair with \(mn = N\), solve the two equations — done in 30 seconds.