Question

To get maximum current through a resistance of $ 2.5\Omega $ , one can use $ m $ rows of cells, each row having $ n $ cells. The internal resistance of each cell is $ 0.5\Omega $ . What are the values of $ n $ and $ m $ , if the total number of cells is $ 45 $ ?
  $
  A)m = 3,n = 15 \\
  B)m = 5,n = 9 \\
  C)m = 9,n = 5 \\
  D)m = 15,n = 3 \\
  $

Answer 1

Hint: When both terminals of a resistor are connected to each and every terminal of one or more resistors, then they are considered to be connected in parallel. The total amount of current flowing from the source is equal to the sum of the currents flowing through each path. The given formula can be used to calculate total resistance in a parallel circuit, $ \dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .... + \dfrac{1}{{{R_n}}} $

Complete answer:
From the question, we have resistance, $ R = 2.5\Omega $ and Internal resistance, $ r = 0.5\Omega $ .
We also know that there are $ m $ rows of cells and each row has $ n $ cells, so the total number of cells, $ mn = 45 $ .
Let us consider the number of rows in parallel to be $ m $ and the number of cells in one row in series be $ n $ . Consider all the cells to be identical to each other. Let every cell have an emf $ E $ and have an internal resistance is $ r $ .
When the cells of each row are in parallel, the total internal resistance $ ({r_p}) $ of a cell is given by,
 $ \dfrac{1}{{{r_p}}} = \dfrac{1}{{{n_r}}} + \dfrac{1}{{{n_r}}} + ... + n $ terms
Since we have $ m $ rows of cells, the total resistance becomes,
 $
  \dfrac{1}{{{r_p}}} = \dfrac{m}{{{n_r}}} \\
   \Rightarrow {r_p} = \dfrac{{{n_r}}}{m} \\
  $
The total resistance in the circuit is the total of external resistance and the internal resistance which is equal to
 $ R + \dfrac{{{n_r}}}{m} $
The effective emf of the cell is $ nE $
The current will be,
 $
  I = \dfrac{{nE}}{{R + \dfrac{{{n_r}}}{m}}} \\
   = \dfrac{{mnE}}{{mR + {n_r}}} \\
  $
The maximum current in the circuit is possible if,
 $ R = \dfrac{{nr}}{m} $
Since we know that internal resistance, $ r = 0.5\Omega $
On substituting the value of $ r $ , we get,
 $
  2.5 = \dfrac{n}{m} \times 0.5 \\
  \dfrac{n}{m} = 5 \\
  $
 $ n = 5m \to $ equation $ (1) $
given, $ mn = 45 \to $ equation $ (2) $
on substituting equation $ 1 $ in $ 2 $ , we get,
 $
  m(5m) = 45 \\
  5{m^2} = 45 \\
  {m^2} = 9 \\
  m = 3 \\
  $
Now we substitute the value of $ m $ in equation $ 1 $ , we get,
 $
  \dfrac{n}{3} = 5 \\
  n = 15 \\
  $
Therefore, the correct option is $ A)m = 3,n = 15 $

Note:
The current in each resistor is the same in a series connection because the first resistor's output current flows into the second resistor's input. All of the resistor leads on one side of the resistors are connected together in a parallel circuit, as are all of the resistor leads on the other side.