1 all cells are connected in series combination
2 all cells are connected in parallel combination
3 4 cells in each row are connected in series and 6 such rows are connected in parallel
4 6 cells in each row are connected in series and 4 such rows are connected in parallel
Explanation:
D Given,
Number of cells, $\mathrm{n}=24$ cells, $\mathrm{E}=1.5 \mathrm{~V}$
Internal resistance, $\mathrm{R}=1 \Omega$
External resistance, $\mathrm{R}^{\prime}=1.5 \Omega$
According to checked by options,
Option (a),
When all cells are connected in series then
Current (I) $=\frac{\mathrm{nE}}{\mathrm{nR}+\mathrm{R}^{\prime}}=\frac{24 \times 1.5}{24 \times 1+1.5}=\frac{36}{25.5}=1.41 \mathrm{~A}$
Option (b),
When all cells are connected in parallel combination then
$\operatorname{Current}(\mathrm{I})=\frac{\mathrm{E}}{\frac{\mathrm{R}}{\mathrm{n}}+\mathrm{R}^{\prime}}=\frac{1.5}{\frac{1}{24}+1.5}=\frac{1.5 \times 24}{1+36}=0.97 \mathrm{~A}$
|Option (c),
4 cells in each row are connected in series and 6 such rows are connected in parallel.
Then, let $\mathrm{n}=4, \mathrm{~m}=6$
$\text { Current }(\mathrm{I})=\frac{\mathrm{nE}}{\mathrm{R}^{\prime}+\frac{\mathrm{nR}}{\mathrm{m}}}$
$\mathrm{I}=\frac{4 \times 1.5}{1.5+\frac{4 \times 1}{6}}=2.77 \mathrm{~A}$
Option (d),
When 6 cells in each row are connected in series and 4 such rows are connected in parallel.
Then, let $\mathrm{n}=6, \mathrm{~m}=4$
$\text { Current }(I)=\frac{\mathrm{nE}}{\mathrm{R}^{\prime}+\frac{\mathrm{nR}}{\mathrm{m}}}=\frac{6 \times 1.5}{1.5+\frac{6 \times 1}{4}}=3.0 \mathrm{~A}$
Hence, out of these four combinations in $4^{\text {th }}$ combination current is maximum.