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Reading of A₁ is 2 A

Reading of A₁ is 18 A

Reading of V is 9 V

Reading of V is 7 V

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Step-by-Step Solution Step 1: Identify the Infinite Network From the figure (), we notice an infinitely repeating network of resistors. Let the equivalent resistance of this entire infinite network (from point A to B) be $x$. Step 2: Redraw the Circuit We can simplify the given network by focusing on one section of the repeated pattern. The given circuit can be redrawn as (): A $4\,\Omega$ resistor in parallel with the equivalent resistor $x\,\Omega$. A $2\,\Omega$ resistor in series with the above parallel combination. Step 3: Parallel Combination of $4\,\Omega$ and $x\,\Omega$ When two resistors $R_1$ and $R_2$ are connected in parallel, the equivalent resistance $R_{\text{parallel}}$ is given by: $ \displaystyle \frac{1}{R_{\text{parallel}}} \;=\; \frac{1}{R_1} \;+\; \frac{1}{R_2}. $ Hence, for $4\,\Omega$ in parallel with $x\,\Omega$: $ \displaystyle \frac{1}{x_{\text{eq}}} \;=\; \frac{1}{4} \;+\; \frac{1}{x} \quad \Longrightarrow \quad x_{\text{eq}} = \frac{4x}{4 + x}. $ Step 4: Account for the Series Resistor This parallel combination is in series with a $2\,\Omega$ resistor. Since resistors in series simply add up: $ x \;=\; x_{\text{eq}} + 2 \;=\; \frac{4x}{4 + x} + 2. $ Step 5: Solve the Equation for $x$ Substitute $x_{\text{eq}} = \frac{4x}{4 + x}$ into the equation: $ x \;=\; \frac{4x}{4 + x} + 2. $ Rearrange and simplify: $ x \;=\; \frac{4x}{4 + x} + 2 \quad \Longrightarrow \quad x(4 + x) \;=\; 4x + 2(4 + x). $ Multiply both sides: $ 4x + x^2 \;=\; 4x + 8 + 2x \quad \Longrightarrow \quad x^2 - 2x - 8 = 0. $ Solve the quadratic equation $ x^2 - 2x - 8 = 0 $: $ \displaystyle x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (-8)}}{2} \;=\; \frac{2 \pm \sqrt{4 + 32}}{2} \;=\; \frac{2 \pm 6}{2}. $ Hence: $ x = 4 \,\Omega \quad \text{or} \quad x = -2 \,\Omega. $ The negative resistance is physically not possible, so $ x = 4 \,\Omega. $ Step 6: Calculate the Ammeter Reading We have a $9\,\text{V}$ source, a series combination of the equivalent resistance ($4\,\Omega$) and the internal resistor $r=0.5\,\Omega$. Hence, the total resistance in the circuit is: $ R_{\text{total}} = 4 \,\Omega + 0.5 \,\Omega = 4.5 \,\Omega. $ By Ohm’s Law, the current $I$ measured by ammeter $A_{1}$ is: $ \displaystyle I = \frac{V}{R_{\text{total}}} = \frac{9}{4.5} = 2\,\text{A}. $ Final Answer The correct reading of the ammeter $A_{1}$ is $2\,\text{A}$.

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