Question

To get maximum current in a resistance of $3\Omega $, one can use m rows of n cells (connected in series) connected in parallel. If the total number of cells is 24 and the internal resistance of the cells is $0.5\Omega $ then
(A). $m=12,\,n=2$
(B). $m=8,\,n=3$
(C). $m=2,\,n=12$
(D). $m=6,\,n=4$

Answer 1

Hint: The batteries can be connected in series or parallel combination. Given, that cells are connected in series as well as parallel in rows. Using the formula for equivalent voltage in series and parallel, we can calculate the equivalent voltage and resistance. According to Ohm’s law, the current will be maximum when the resistance is minimum.

Formulas used:
$V={{E}_{1}}{{r}_{1}}+{{E}_{2}}{{r}_{2}}+{{E}_{3}}{{r}_{3}}+.....+{{E}_{n}}{{r}_{n}}$
$V=\dfrac{{{E}_{1}}}{{{r}_{1}}}+\dfrac{{{E}_{2}}}{{{r}_{2}}}+\dfrac{{{E}_{3}}}{{{r}_{3}}}+......+\dfrac{{{E}_{m}}}{{{r}_{m}}}$

Complete answer:
Just like other appliances, the batteries can also be connected in series and parallel combinations. When batteries are connected such that they are connected one after the other, they are in series. The equivalent potential in a series combination of batteries is calculated as
$V={{E}_{1}}{{r}_{1}}+{{E}_{2}}{{r}_{2}}+{{E}_{3}}{{r}_{3}}+.....+{{E}_{n}}{{r}_{n}}$
Here, ${{E}_{n}}$ is the potential of each battery
${{r}_{n}}$ is the internal resistance of each battery
$V$ is the equivalent potential
When the potentials of all the batteries in a series combination are equal to E and the internal resistance for all batteries is r. then substituting in the above equation we get,
$V=nER$ - (1)
When all batteries are connected between two points, then the batteries are connected in parallel. The equivalent potential in a parallel combination is calculated as
$V=\dfrac{{{E}_{1}}}{{{r}_{1}}}+\dfrac{{{E}_{2}}}{{{r}_{2}}}+\dfrac{{{E}_{3}}}{{{r}_{3}}}+......+\dfrac{{{E}_{m}}}{{{r}_{m}}}$
When the potential, of all batteries is E and the internal resistance of all batteries is r, then substituting in above equation we get,
$V=m\dfrac{E}{r}$ - (2)
Given, n cells are connected in series in each of the m rows, then applying ohm’s law, we get,
$\begin{align}
  & R=\dfrac{V}{I} \\
 & \Rightarrow \dfrac{nr}{m}+R=\dfrac{nE}{I} \\
 & \Rightarrow I=\dfrac{nE}{\dfrac{nr}{m}+R} \\
 & \Rightarrow I=\dfrac{mnE}{nr+mR} \\
\end{align}$
Substituting given values in the above equation wherein $R=3\Omega ,\,r=0.5\Omega $, we get,
$I=\dfrac{mnE}{n0.5+3m}$
In order for current to be maximum, $n0.5+3m$ should be minimum. Therefore, it will be minimum for
$R=\dfrac{n0.5}{m}$
$\Rightarrow 3=\dfrac{n0.5}{m}$ - (3)
Also, $mn=24$ - (4)
Substituting eq (4) in eq (3), the values of m and n will be,
$\begin{align}
  & 3=\dfrac{n0.5}{\dfrac{24}{n}} \\
 & \Rightarrow \dfrac{3\times 24}{0.5}={{n}^{2}} \\
 & \Rightarrow {{n}^{2}}=144 \\
 & n=12 \\
\end{align}$
Substituting n in eq (4), we get
$\begin{align}
  & 12m=24 \\
 & \Rightarrow m=\dfrac{24}{12} \\
 & \therefore m=2 \\
\end{align}$
Therefore, the value of m is 2 and the value of n is 12.

Hence, the correct option is (C).

Note:
When batteries are connected in series, the current is the same whereas when they are connected in parallel, the potential drop is the same. The ohm’s law gives the relation between resistance, voltage and current. The emf of a cell is the maximum potential whereas the voltage is the potential drop in the circuit.