Question

The resistance of three wires BC, CA and AB of same uniform cross-section and material are a, b and c respectively in ohms. Another wire from A of resistance $d=5\Omega$ can make a sliding contact with BC. A battery of constant $emf\epsilon =4V$ is connected between A and point of contact with BC. Further it is known that value of d is equal to sum of the three resistance a, b and c. the minimum current drawn from the battery is $i_0$ in amperes. Find the value of $i_0$

Answer 1

$i=\frac{\epsilon }{R}$
$\frac{1}{R}=\frac{1}{d}+\frac{1}{b+a-x}+\frac{1}{c+x}$
For minimum i, $\frac{1}{R}$ will be minimum
i.e. $\frac{d}{dx}\frac{1}{R}=0$
Gives, $x=\frac{(a+b)-c}{2}$
Put $i_{min}=\frac{\epsilon (a+b+c+4d)}{(a+b+c)d}$
$=\frac{\epsilon \left(5d\right)}{d\times d}=\frac{5\epsilon }{d}$.