Question

To draw a maximum current from a combination of cells, how should the cells be grouped?
(A) Parallel
(B) Series
(C) Mixed grouping
(D) Depends upon the relative values of internal and external resistances.

Answer 1

Hint : To answer this question, we have to use the formulae of the series and the parallel equivalent emf. Then we have to see for a given value of the external resistance in a circuit, which combination gives more current.

Formula used: The formulae used for solving this question are given by
 $ \dfrac{E}{{{r_p}}} = \dfrac{{{E_1}}}{{{r_1}}} + \dfrac{{{E_2}}}{{{r_2}}} + \dfrac{{{E_3}}}{{{r_3}}} + ...... $
Here $ E $ is the equivalent emf, $ {r_p} $ is the equivalent internal resistance of the cells of emfs $ {E_1} $ , $ {E_2} $ , $ {E_3} $ ,.. having internal resistances $ {r_1} $ , $ {r_2} $ , $ {r_3} $ ,….. respectively, when these are arranged in parallel combination.

Complete step by step answer:
Consider $ n $ cells each of emf $ e $ and internal resistance of $ r $ . Also let the value of the external resistance be $ R $ .
Let the current in the circuit when the cells are arranged in the serial combination be $ {I_s} $ , and that corresponding to the parallel combination be $ {I_p} $ .
Now, we consider the cells to be connected in series. We know that the equivalent emf for the serial combination is given by
 $ E = {E_1} + {E_2} + {E_3} + ............. $
Since there are $ n $ cells each of emf $ e $ , so the equivalent emf of all the cells in this case is given by
 $ E = ne $ (1)
Since the cells are connected in parallel, each of which have an internal resistance of $ r $ , so all the internal resistances are also connected in serial combination. We know that the resistances in serial combination get added up. So the net internal resistance of all the cells is given by
 $ {r_i} = nr $
Now, the external resistance is in series with this net internal resistance. So the total resistance in the circuit becomes
 $ {R_T} = nr + R $ (2)
Now, from Ohm’s law we have
 $ E = {I_s}{R_T} $
From (1) and (2)
 $ ne = {I_s}\left( {nr + R} \right) $
So the current in the circuit becomes
 $ {I_s} = \dfrac{{ne}}{{\left( {nr + R} \right)}} $ (3)
Now consider the cells to be connected in parallel combination.
Since the internal resistances of all the cells are connected in series with them, so they are connected in parallel combination with each other. So the equivalent internal resistance is given by
 $ \dfrac{1}{{{r_p}}} = \dfrac{1}{r} + \dfrac{1}{r} + \dfrac{1}{r} + ...... $
 $ \Rightarrow \dfrac{1}{{{r_p}}} = \dfrac{n}{r} $
Taking the reciprocal, we get
 $ {r_p} = \dfrac{r}{n} $ (4)
Now, we know that the equivalent emf in the parallel combination is given by
 $ \dfrac{E}{{{r_p}}} = \dfrac{{{E_1}}}{{{r_1}}} + \dfrac{{{E_2}}}{{{r_2}}} + \dfrac{{{E_3}}}{{{r_3}}} + ...... $
Substituting $ {E_1} = {E_2} = ........ = e $ , and $ {r_1} = {r_2} = ........ = r $ , we get
 $ \dfrac{E}{{{r_p}}} = \dfrac{e}{r} + \dfrac{e}{r} + \dfrac{e}{r} + ....... $
 $ \Rightarrow \dfrac{E}{{{r_p}}} = \dfrac{{ne}}{r} $
From (4)
 $ \dfrac{{nE}}{r} = \dfrac{{ne}}{r} $
 $ \Rightarrow E = e $ (5)
The external resistance of $ R $ is in series with this combination of the cells. So the net resistance in the circuit becomes
 $ {R_N} = R + {r_p} $
 $ \Rightarrow {R_N} = R + \dfrac{r}{n} $ (6)
From the Ohm’s law we have
 $ E = {I_p}{R_N} $
From (5) and (6) we have
 $ e = {I_p} \times \left( {R + \dfrac{r}{n}} \right) $
 $ \Rightarrow {I_p} = \dfrac{e}{{R + \dfrac{r}{n}}} $
On simplifying we get
 $ \Rightarrow {I_p} = \dfrac{{ne}}{{nR + r}} $ (7)
Dividing (3) by (4) we get
 $ \dfrac{{{I_s}}}{{{I_p}}} = \dfrac{{ne}}{{\left( {nr + R} \right)}} \times \dfrac{{\left( {nR + r} \right)}}{{ne}} $
 $ \Rightarrow \dfrac{{{I_s}}}{{{I_p}}} = \dfrac{{\left( {nR + r} \right)}}{{\left( {nr + R} \right)}} $ (8)
Now, for deciding whether the current is maximum in the first or the second case, we need to have the value of the ratio $ \dfrac{{{I_s}}}{{{I_p}}} $ . From (8) we need a relation between the internal resistance $ r $ and the external resistance $ R $ . Thus, the grouping of the cells is dependent upon the relative values of internal and external resistance.
Hence, the correct answer is option D.

Note:
The value of the equivalent emf could be calculated without any calculations. As all the $ n $ cells having the same emf are connected in parallel combination, so by symmetry the equivalent emf will be equal to the emf of each individual cell.