question_answer

The n rows each containing m cells in series are joined in parallel. Maximum current is taker, from this combination across an extreme resistance of 3 Q resistance. If the total number of cells used are 24 and internal resistance c: each cell is 0.50 then :

A)  m = 8, n = 3                       

B)  m = 6, n = 4

C)  m = 12, n =  2                    

D)  m = 2, n = 12

Correct Answer: C

Solution :

                 In mixed grouping the current in the external circuit will be maximum when the internal resistance of the battery is equal to the external resistance,                 \[R=\frac{mr}{n}\] Given,   \[R=3\,\Omega ,\text{ }r=0.5\,\Omega \] \[\therefore \]  \[3=\frac{m}{n}\times 0.5\] \[\Rightarrow \]               \[\frac{m}{n}=6\] \[\Rightarrow \]               \[m=6n\]               ... (i) Total number of cells \[=m\times n=24\]         ... (ii) From Eqs. (i) and (ii), we get \[6n\times n=24\] \[\Rightarrow \]               \[6{{n}^{2}}=24\] \[\Rightarrow \]               \[{{n}^{2}}=4\] \[\Rightarrow \]               \[n=2,\text{ }m=12\]