Combination of Resistors question from JEE Main 2017 (8 Apr, Shift 1)

Answer: (a)

1. Each section of the ladder is $1\ \Omega + 1\ \Omega$ in series with a $4\ \Omega$ shunt. For an infinite ladder, adding one section does not change $R_\infty$.
2. $R_\infty = 2 + \frac{4R_\infty}{4 + R_\infty} \Rightarrow R_\infty^2 - 2R_\infty - 8 = 0 \Rightarrow R_\infty = 4\ \Omega$
3. $I = \frac{9}{4 + 0.5} = 2\text{ A}$, so $A_1$ reads 2 A.
4. Voltmeter reads terminal voltage $V = 9 - 2 \times 0.5 = 8\text{ V}$, so (c) and (d) are wrong.
5. So the answer is (a) Reading of $A_1$ is 2 A.