Question

The $n$ rows each consisting in cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of $3\Omega $. If the total number of cells used in $24$ and internal resistance of each is $0.5\Omega $, then
(A) $m = 8,n = 3$
(B) $m = 6,n = 4$
(C) $m = 12,n = 2$
(D) $m = 2,n = 12$

Answer 1

Hint We are provided with the internal resistance and the external resistance of a circuit. The cells in the circuit in series are joined in parallel. So it is a series and parallel combination of cells. Use the relation of mixed combinations of cells to solve the problem.

Complete step by step answer
In a circuit consisting of a cell, key, resistance, when the current flows in the external circuit, it also passes inside the cell of the circuit. When current flows inside the cell, the electrolyte of the cell offers some resistance to the current which is known as internal resistance of the cell.
Combination of a cell is known as a battery.
Given,
The external resistance is $3\Omega $
The total internal resistance is $n \times 0.5 = 24 \times .5 = 12\Omega $
We can see that the external resistance is smaller than the internal resistance
When external resistance is very small in comparison to the internal resistance, then the cells are connected in parallel to get maximum current.
Let us consider n cells connected in series and m cells connected in parallel. It is a mixed combination of cells.
These cells are sending current in the external circuit which has external resistance R.
Let the emf of each cell connected in series and parallel be E
Let the internal resistance offered to the current flow by the cells be r.
Then current in the external circuit be I
According to the derivation of current in mixed combination of cells,
$I = \dfrac{{nE}}{{R + \dfrac{{nr}}{m}}}$
Here,
R is the external resistance of the circuit
E is the emf of the cell
$\dfrac{{nr}}{m}$ is the total internal resistance of the battery
r is internal resistance
n is the number of number of cells connected in series
m is the number of cells connected in parallel
Given,
$n$ rows each consisting in cells in series are joined in parallel.
Internal resistance of each cell, $r = 0.5\Omega $
External resistance of the circuit is $R = 3\Omega $
The total number of cells used in the circuit,
 $ \Rightarrow n + m = 24$
$ \Rightarrow n = 24 - m$
It is given that maximum current is taken from this combination across an external resistance of $3\Omega $
In the combination of cells connected in series and parallel, maximum current can be drawn only if the external resistance of the circuit is equal to the total internal resistance of the battery
In the above discussion we have seen that
$\dfrac{{nr}}{m}$ is the internal resistance of the battery
R is the external resistance of the circuit
By the condition,
$ \Rightarrow R = \dfrac{{nr}}{m}$
Where,
R is the external resistance of the circuit
n is the number of number of cells connected in series
m is the number of cells connected in parallel
r is internal resistance
Substitute the given values in the above relation
$ \Rightarrow R = \dfrac{{nr}}{m}$
$ \Rightarrow 3 = \dfrac{{n \times 0.5}}{m}$
\[ \Rightarrow \dfrac{n}{m} = \dfrac{3}{{0.5}}\]
\[M = 0.5{\text{ }}and{\text{ }}n = 3\]
Since, it is a ratio it is in reduced form,
Multiplying by 4 we get
\[ \Rightarrow \dfrac{n}{m} = \dfrac{{12}}{2}\]
Now, \[m = 2,{\text{ }}n = 3\]

From the given options, the option (D) \[m = 2,{\text{ }}n = 3\] is the correct answer.

Note We got the answer in reduced form, since it is a ratio it can be in reduced form. But we don’t have the answer we got in the given options so, relating the answer we got with the given options we get a similar answer \[m = 2,{\text{ }}n = 3\]