Question

iii) In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and Bis 102 112 192 AM 112 w 292 - to oo 202 292 B -. to oo n! (A) infinite (C) 22 (B) zero (D) 1.5 ​

Answer 1

Answer:

8373828263838363638738484

Answer 2

Answer:

(c) 2Ω

Explanation:

As the whole circuit extends symmetrically to the right, take it as Req leaving the first 1Ω and 2Ω resistor.

This leaves us with a 2Ω resistor in parallel with Req. Thus we get the equivalent resistance as,

R2' = $\frac{1}{R1} + \frac{1}{R2}$

=>R2' = ($\frac{1}{2}$) x ($\frac{1}{Req}$)

=> R2' = $\frac{2 Req}{2 + Req}$

Now this is in series with the 1Ω resistor, thus we get,

Req = R1 + R2'

Req = 1 + $\frac{2 Req}{2 + Req}$

=> $Req^{2}$ - Req - 2 =  0

Upon solving for roots we get, Req = 2Ω ,-1Ω

*We took Req again to solve for in series as we assumed the entire circuit as Req before to solve in parallel, thus a new variable shouldn't be used to simply our calculations*

As resistance cannot be negative, our final answer is 2Ω