Question

You are provided with $48$ cells each of emf $2\text{volts}$ and internal resistance $4\text{ohms}$. What maximum current can flow in the circuit having an external resistance of $12\Omega$?

• A. $1\text{A}$
• B. $1.2\text{A}$
• C. $0.96\text{A}$
• D. $1.08\text{A}$

Answer 1

The correct option is A $1\text{A}$

In a circuit containing battery, internal resistances and an external resistance, power output (and hence current) is maximum when $R_{ext}=(r_{int}{)}_{\text{equvalent}}$

Here let's assume to maximise current, we put $n$ rows and each row has $m$ cells.

So, $m\times n=48.......\left(1\right)$

For each row, $r_{eq}=mr$ (Series combination)

for $n$ such rows total, $r_{eq}=\frac{mr}{n}$ (Parallel combination)

for maximum current, $R_{ext}=\frac{mr}{n}$

$\Rightarrow 12=\frac{m\times 4}{n}$

$\Rightarrow m=3n.....\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$ we have,

$\Rightarrow 3n^2=48$
$\Rightarrow n^2=16$
$\Rightarrow n=4$ and $m=12$

So $4$ rows with each row having $12$ cells.
Each row ${\epsilon }_{eq}=12\times 2=24\text{V}$

So circuit can be redrawn as:

$\Rightarrow ({\epsilon }_{eq}{)}_{Total}=24\text{V}$

$(r_{eq}{)}_{Total}=\frac{48}{4}=12$ $\Omega$ and $R_{ext}=12$ $\Omega$

So, $i=\frac{({\epsilon }_{eq}{)}_{Total}}{(r_{eq}{)}_{Total}+R_{ext}}=\frac{24}{12+12}$

$\therefore i=1\text{A}$