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Q.
The equivalent resistance of the figure i.e., infinite network of resistors between the terminals $A$ and $B$ is
Current Electricity
Solution:
Let $R$ be the equivalent resistance. Then addition/subtraction of one more set of resistors $R_{1}, R_{2}$ and $R_{3}$ will not affect the total resistance. Thus,
$R=R_{1}+\left(\right.$ parallel combination of $R$ and $\left.R_{2}\right)+R_{3}$
$R=R_{1}+\left(\frac{R R_{3}}{R+R_{3}}\right)+R_{2}$
$\Rightarrow R^{2}+R R_{3}=R R_{1}+R_{1} R_{3}+R R_{3}+R R_{2}+R_{2} R_{3}$
$\Rightarrow R^{2}-R\left(R_{1}+R_{2}\right)=\left(R_{2}+R_{1}\right) R_{3}=0$
$\Rightarrow R=\frac{\left(R_{1}+R_{2}\right) \pm \sqrt{\left(R_{1}+R_{2}\right)^{2}+4\left(R_{1}+R_{2}\right) R_{3}}}{2}$
As $R$ cannot be negative, hence
$R =\frac{1}{2}\left[\left(R_{1}+R_{2}\right)+\sqrt{\left(R_{1}+R_{2}\right)^{2}+4\left(R_{1}+R_{2}\right) R_{3}}\right] $
$=\frac{1}{2}\left[\left(R_{1}+R_{2}\right)+\sqrt{\left(R_{1}+R_{2}\right)\left(R_{1}+R_{2}+4 R_{3}\right)}\right]$
