A battery is made by joining $n = 3$ identical cells (each of EMF $2\ \text{V}$ and internal resistance $0.5\ \Omega$) in series to form one row, and $m = 2$ such rows are then connected in parallel with each other (a mixed grouping). Find the value of the external resistance $R$ for which the current delivered to $R$ is maximum, and calculate this maximum current.

A battery is made by joining n = 3 identical cells (each of EMF 2V and internal resistance 0.5Ω) in series to

Numerical · Q25

n = 3

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Row EMF (n=3 cells in series per row): $\varepsilon_{\text{eq}} = n\varepsilon = 3\times2 = 6\ \text{V}$ .

Row internal resistance: $nr = 3\times0.5 = 1.5\ \Omega$ ; with $m=2$ such rows in parallel,

r_{\text{eq}} = \frac{nr}{m} = \frac{1.5}{2} = 0.75\ \Omega

For a mixed grouping, the current delivered to an external resistance $R$ is maximum when $R$ equals the equivalent internal resistance:

R = r_{\text{eq}} = 0.75\ \Omega

At this value, total circuit resistance $=R+r_{\text{eq}}=0.75+0.75=1.5\ \Omega$ , so …