In the circuit shown, AB is a uniform wire of resistance  $6\Omega$. With battery connected between C and D, length of AD for which current drawn from the battery is minimum will be

 
Equivalent resistance of the circuit between C and D is given by,
 $\frac{1}{R}=\frac{1}{12}+\frac{1}{2+\text{x}}+\frac{1}{10-\text{x}}$       
For R to be maximum or I to be Minimum  $\frac{\text{d}\left(\frac{1}{R}\right)}{dx}=0$
 $\Rightarrow \text{ 2+x=10-x}$
 $\Rightarrow \text{ x=4}\Omega$
 $\begin{array}{l}\Rightarrow \text{ AD =}\frac{2L}{3}\\ \end{array}$