Question: 6: The equivalent resistance of the infinite network given below is :
(1) 2 \Omega
(2) (1+\sqrt{2}) \Omega
(3) (1+\sqrt{3}) \Omega
(4) (1+\sqrt{5}) \Omega
Answer: Option (3)
Explanation:
Let the equivalent resistance of the infinite network be R. Due to the infinite and repetitive nature of the network, removing one repeating section does not change the overall resistance.
From the left terminals, the network consists of a 1\Omega resistor in series on the top and bottom, followed by a vertical 1\Omega resistor connecting the two paths, after which the same infinite network repeats.
The resistance of the remaining infinite part is again R.
The vertical 1\Omega resistor is in parallel with the equivalent resistance R of the remaining network between the two junctions.
So, the parallel combination gives:
\frac{1 \cdot R}{1 + R}This parallel combination is in series with the two 1\Omega resistors on the top and bottom paths, which together contribute 2\Omega.
Hence, the total equivalent resistance is:
R = 2 + \frac{R}{1 + R}Multiplying both sides by (1+R), we get:
R(1+R) = 2(1+R) + R R^2 + R = 2 + 2R + R R^2 - 2R - 2 = 0Solving the quadratic equation:
R = \frac{2 + \sqrt{4 + 8}}{2} = 1 + \sqrt{3}Thus, the equivalent resistance of the infinite network is (1+\sqrt{3})\Omega.