Q. A battery of internal resistance $4\ \Omega\ $is connected to the network of resistance as shown. In order to given the maximum power to the network, the value of $R$ (in $\Omega)$ should be
A) 4/9
B) 8/9
C) 2
D) 18
Explanation
(c) The equivalent circuit is as shown in figure (a) and (b)
Since, the network of resistances is a balanced Wheat stone bridge, so resistance between points $A$ and $B$ of network figure (b) is given by
$\frac{1}{R'} = \frac{1}{3R} + \frac{1}{6R} = \frac{2 + 1}{6R} = \frac{1}{2R}\ or\ R’ = 2R$
For maximum power to the network, $R’$ should be equal to internal resistance of the battery. So
$R^{'} = 2R = 4\ or\ R = 4/2 = 2\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.