question_answer

The n rows each containing m cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of 30 resistance. If the total number of cells used are 24 and internal resistance of each cell is \[0.5\text{ }Q\]then

A)  \[m=8,\text{ }n=3\] 

B)  \[m=6,\text{ }n=4\]           

C)   \[m=12,\text{ }n=2\]

D)  \[m=2,\text{ }n=12\]       

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,   \[R=3\Omega ,\,\,\,\,\,\,r=0.5\Omega \] \[\therefore \] \[3=\frac{m}{n}\times 0.5\] \[\Rightarrow \] \[\frac{m}{n}=6\] \[\Rightarrow \] \[m=6n\] ?..(i) Total number of cells \[=m\times n=24\]      ... (ii) From Eqs. (i) and (ii), we get \[6n\times n=24\] \[\Rightarrow \] \[6{{n}^{2}}=24\] \[\Rightarrow \] \[{{n}^{2}}=4\] \[\Rightarrow \] \[n=2,\,\,\,m=12\]