Question

$N$ cells each of e.m.f. $E$ & internal resistance $r$ are grouped into sets of $K$ cells connected in series. The $\left(\frac{N}{K}\right)$ sets are connected in parallel to a load of resistance $R$, then:

• A. Maximum power is delivered to the load if $K=\sqrt{\frac{NR}{r}}$
• B. Maximum power is delivered to the load if $K=\sqrt{\frac{r}{NR}}$
• C. Maximum power delivered to the load is $\frac{N{E}^2}{4r}$
• D. Maximum power delivered to the load is $\frac{E^2}{4Nr}$

Answer 1

Answer: (A), (C)

Maximum power delivered to the load is $\frac{N{E}^2}{4r}$

Given that N cells are grouped into sets of $K$ cells connected in series, So total $\frac{N}{K}$ rows each of emf $KE$ and internal resistance $Kr,$ are connected in parallel

Net emf of the circuit in a parallel setup is given by
$E_{net}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}+.....}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+....}$
$\Rightarrow E_{net}=\frac{\frac{KE}{Kr}\times \frac{N}{K}}{\frac{1}{Kr}\times \frac{N}{K}}=KE$
Net internal resistance,
$\frac{1}{R_{net}}=\frac{1}{Kr}\times \frac{N}{K}=\frac{N}{K^{2}r}$
$\Rightarrow R_{net}=\frac{K^{2}r}{N}$


From maximum power transfer theorem, we know that internal resistance is equal to external resistance
$\Rightarrow R=\frac{K^{2}r}{N}$
$\Rightarrow K=\sqrt{\frac{NR}{r}}........\left(1\right)$
Now, power delivered is
$P=I^{2}R$
$\Rightarrow P={\left(\frac{KE}{R+R}\right)}^2\times R$
$\Rightarrow P=\frac{K^2{E}^2}{4R^2}\times R=\frac{K^2{E}^2}{4R}.......\left(2\right)$
put the value of $K$ in eq. $\left(2\right)$
$\Rightarrow P=\frac{\frac{NR}{r}\times E^2}{4R}$
$\Rightarrow P=\frac{N{E}^2}{4r}$