Question

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	$\frac{E^2}{4r}$
ii	Maximum power transfer to load if cells are connected in parallel	q	$\frac{E^2}{2r}$
iii	For series combination of cells	r	$E_{eq}=E,r_{eq}=\frac{r}{2}$
iv	For parallel connection of cells	s	$E_{eq}=2E,r_{cq}=2r$

Now, match the given columns and select the correct option from the codes given below.

• A. i - (p) , ii - (q) , iii - (p) , iv - (s)
• B. i - (r) , ii - (p) , iii - (q), iv - (s)
• C. i - (q) , ii - (q), iii - (s), iv - (r)
• D. i - (p), ii - (p), iii - (q) iv - (s)

Answer 1

Answer: (C)

If cells are connected in series: $E_{eq}=2E,r_{eq}=2r$
Maximum power will be transferred if $R=r_{\text{eq }}=2r$.
Then current in $R:I=\frac{E_{eq}}{R+r_{eq}}=\frac{2E}{2r+2r}=\frac{E}{2r}$
Power $=I^{2}R={\left(\frac{E}{2r}\right)}^{2}2r=\frac{E^2}{2r}$
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 $R$ :
$I=\frac{E_{eq}}{R+r_{eq}}=\frac{E}{\frac{r}{2}+\frac{r}{2}}=\frac{E}{r}$
Power: $I^{2}R={\left(\frac{E}{r}\right)}^2\frac{r}{2}=\frac{E^2}{2r}$
Hence $[i-q][ii-q][iii-s][iv-r]$