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Consider two identical cells each of emf E and internal resistance r connected to a load resistance R.
Column I Column II
i. Maximum power transfer to load if cells are connected in series p. $\frac{E^{2}}{4 r}$
ii. Maximum power transfer to load if cells are connected in parallel q. $\frac{E^{2}}{2 r}$
iii. For series connection of cells r. $E_{eq} = E, r_{eq} = \frac{r}{2}$
iv. For parallel connection of cells s. $E_{eq} = 2 E, r_{eq} = 2r$
Now, match the given columns and select the correct option from the codes given below.

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i-p, ii-q, iii-p, iv-s

i-r, ii-p, iii-q, iv-s

i-q, ii-q, iii-s, iv-r

i-p, ii-p, iii-q, iv-s

answer is C.

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Detailed Solution

If cells are connected in series: $E_{eq} = 2 E, r_{eq} = 2 r .$

Maximum power will be transferred if $R = r_{eq} = 2 r .$

Then current in R: $I = \frac{E_{eq}}{R + r_{eq}} = \frac{2 E}{2 r + 2 r} = \frac{E}{2 r}$

$Power = I^{2} R = {(\frac{E}{2 r})}^{2} 2 r = \frac{E^{2}}{2 r}$

If cells are connected in parallel: $E_{eq} = E, r_{eq} = \frac{r}{2}$

For maximum power: $R = r_{eq} = \frac{r}{2},$ Then current in R:

$\begin{array}{l} I = \frac{E_{eq}}{R + r_{eq}} = \frac{E}{\frac{r}{2} + \frac{r}{2}} = \frac{E}{r} \\ Power : I^{2} R = {(\frac{E}{r})}^{2} \frac{r}{2} = \frac{E^{2}}{2 r} \end{array}$

Hence [i - q] [ii - q] [iii - s] [iv -r]

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