A 9, V battery with an internal resistance of | vedclass

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(A) Let the equivalent resistance of the infinite network be $x\,\Omega$. Since the network is infinite,adding one more section does not change the eq

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Reading of $A_1$ is $2\, A$

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(A) Let the equivalent resistance of the infinite network be $x\,\Omega$. Since the network is infinite,adding one more section does not change the equivalent resistance. The circuit can be simplified by replacing the infinite part with $x\,\Omega$. The $4\,\Omega$ resistor is in parallel with $x\,\Omega$,and this combination is in series with two $1\,\Omega$ resistors (one in the top branch and one in the bottom branch). Thus,$x = 1 + 1 + \frac{4x}{4+x} = 2 + \frac{4x}{4+x}$. Solving for $x$: $x = \frac{2(4+x) + 4x}{4+x} = \frac{8 + 2x + 4x}{4+x} = \frac{8 + 6x}{4+x}$. $x(4+x) = 8 + 6x \implies x^2 + 4x = 8 + 6x \implies x^2 - 2x - 8 = 0$. Factoring the quadratic: $(x-4)(x+2) = 0$. Since resistance cannot be negative,$x = 4\,\Omega$. The total resistance of the circuit including the internal resistance $r = 0.5\,\Omega$ is $R_{eq} = x + r = 4 + 0.5 = 4.5\,\Omega$. The reading of ammeter $A_1$ is $I = \frac{V_{battery}}{R_{eq}} = \frac{9}{4.5} = 2\, A$. The reading of the voltmeter $V$ is the terminal voltage of the battery: $V_{terminal} = V_{battery} - I \cdot r = 9 - (2 	imes 0.5) = 9 - 1 = 8\, V$. Comparing with the options,the correct statement is that the reading of $A_1$ is $2\, A$.

$A$ $9\, V$ battery with an internal resistance of $0.5\,\Omega$ is connected across an infinite network as shown in the figure. All ammeters $A_1, A_2, A_3$ and the voltmeter $V$ are ideal. Choose the correct statement.

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