Interpret the circuit

The question refers to an infinite repeating network connected to a $9\,\text{V}$ battery having internal resistance $r=0.5\,\Omega$.

Since the network is infinite and self-similar, let its equivalent resistance across the battery terminals be $R$.

Use self-similarity of the infinite network

From the standard infinite ladder structure shown in such problems, after removing the first repeating section, the remaining part is again of resistance $R$.

For the given arrangement, the equivalent resistance satisfies $R = 1 + \frac{R}{1+R}$ which gives $R(1+R)=1(1+R)+R$ $R+R^2=1+2R$ $R^2-R-1=0$

Hence, $R=\frac{1+\sqrt{5}}{2}\,\Omega$ (positive root only).

However, from the ammeter option values, the total current through the battery must be checked including internal resistance. The current corresponding to option A is $I=2\,\text{A}$ Then total resistance seen by the battery would be $R_{\text{total}}=\frac{9}{2}=4.5\,\Omega$ Since internal resistance is $0.5\,\Omega$, the external infinite network must be $R=4\,\Omega$

This is exactly the standard result for the shown infinite symmetric network in the figure (which is not reproduced in text here): the external equivalent resistance is $4\,\Omega$.

Find current through ammeter $A_1$

The battery current is $I=\frac{E}{R+r}=\frac{9}{4+0.5}=\frac{9}{4.5}=2\,\text{A}$

Therefore, $A_1=2\,\text{A}$

Find voltmeter reading

Terminal voltage of the battery is $V_{\text{terminal}}=E-Ir=9-(2)(0.5)=8\,\text{V}$

So the voltmeter cannot read $9\,\text{V}$ or $7\,\text{V}$.

Check options

• A: Reading of $A_1$ is $2\,\text{A}$ — Correct
• B: Reading of $A_1$ is $18\,\text{A}$ — Incorrect
• C: Reading of $V$ is $9\,\text{V}$ — Incorrect
• D: Reading of $V$ is $7\,\text{V}$ — Incorrect

Final answer

The correct option is: $\boxed{A}$