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(A) The circuit consists of a $9\, V$ battery with an internal resistance $r = 1\, \Omega$ connected in series to a parallel combination of two resist

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$2$

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(A) The circuit consists of a $9\, V$ battery with an internal resistance $r = 1\, \Omega$ connected in series to a parallel combination of two resistors ($6\, \Omega$ and $3\, \Omega$).First, calculate the equivalent resistance of the parallel part:$R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\, \Omega$.The total resistance of the circuit is $R_{	ext{total}} = R_p + r = 2\, \Omega + 1\, \Omega = 3\, \Omega$.The total current flowing from the battery is $I = \frac{E}{R_{	ext{total}}} = \frac{9\, V}{3\, \Omega} = 3\, A$.The terminal potential difference $(V)$ across the parallel combination is $V = E - Ir = 9\, V - (3\, A 	imes 1\, \Omega) = 6\, V$.The ammeter is in series with the $3\, \Omega$ resistor. The current through the ammeter is $I_A = \frac{V}{R} = \frac{6\, V}{3\, \Omega} = 2\, A$.

In the figure shown, if the internal resistance of the battery is $1\, \Omega$, the reading of the ammeter will be ................. $A$.

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