An infinite ladder is constructed with 1Ω and 2Ω resistors as shown in figure.

(a) Find the effective resistance between the points A and B.


(b) Find the current that passes through the 2Ω resistor nearest to the battery.



Concepts/Formula used:
Resistors in Series:


Resistors in parallel:


Ohm’s Law:


Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:



(a)


Let the equivalent resistance between A and B be req Ω .


This means that we can rewrite the circuit as:



Where req has replaced the following infinite combination:



We can redraw the infinite circuit as




Note that de and cb are in parallel. The equivalent resistance is given by:





Now, the 1 Ω resistor and r’eq are in series.




Rearranging and dropping the subscript, we get a quadratic equation:



The roots of this equation are 2 and -1. As resistance can’t the negative the equivalent resistance between A and B is 2 Ω.


(b)


Let the net current be 2I. This current passes through the 1 Ω resistor. Than splits up equally due to symmetry as there are two 2Ω resistors.



Now,




Hence, the current passing through the nearest 2 Ω resistor is 1.5A.


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