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

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