Question

Twenty four cells each of emf 1.5V and internal resistance 0.5 ohms are to be connected to a 3 ohm resistance. For maximum current through this resistance the number of rows and number of columns that you connect these cells is?

• A. 12 cells in series 2 rows in parallel
• B. 8 cells in series 3 rows in parallel
• C. 4 cells in series 6 rows in parallel
• D. 6 cells in series 4 rows in parallel

Answer 1

The correct option is A 12 cells in series 2 rows in parallel

Let m cells be in series and n such parallel components are there

$[\Rightarrow mn=24cells]$

In one component

effective emf $=$ mE

and internal resistance$=$ mr

If 'n' cells of mE volt and mr $\Omega$ are connected in parallel

Then effective internal resistance

$=\frac{1}{r_{eff}}=\frac{1}{mr}+\frac{1}{mr}......+\frac{1}{mr}$

$\frac{1}{r_{eff}}=\frac{n}{mr}\Rightarrow r_{eff}=\frac{mr}{n}$

and effective emf

$=\frac{E_{eff}}{r_{eff}}=\frac{mE}{mr}\times n\Rightarrow E_{eff}=\frac{nE}{r}\times r_{eff}$

$\therefore formaximumcurrent,ratioofmandnshouldbeminimumandmmustbemaximum$

$I=E_{eff}R+r_{eff}$

$=mER+\frac{m}{n}r=\frac{mE}{R+\frac{m^{2}r}{24}}$

$\frac{dI}{dm}=0$

$[R+\frac{m^{2}r}{24}]-m^2\left[\frac{r}{12}\right]=0$

$\Rightarrow \frac{m^{2}r}{24}=R$

$m^2=16\times 9\Rightarrow m=12$

$=\text{12 in series and 2 such parallel components}$