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(A) The circuit consists of a $9\, V$ battery with an internal resistance of $1\, \Omega$ connected in parallel to two branches. One branch has a $6\,

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(A) The circuit consists of a $9\, V$ battery with an internal resistance of $1\, \Omega$ connected in parallel to two branches. One branch has a $6\, \Omega$ resistor, and the other branch has an ammeter in series with a $3\, \Omega$ resistor.First, calculate the equivalent resistance of the two parallel branches:$R_p = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2\, \Omega$.Now, include the internal resistance of the battery $(r = 1\, \Omega)$ which is in series with the parallel combination:$R_{	ext{total}} = R_p + r = 2\, \Omega + 1\, \Omega = 3\, \Omega$.The total current drawn from the battery is:$I_{	ext{total}} = \frac{E}{R_{	ext{total}}} = \frac{9\, V}{3\, \Omega} = 3\, A$.The terminal potential difference $(V)$ across the parallel branches is:$V = E - I_{	ext{total}}r = 9\, V - (3\, A 	imes 1\, \Omega) = 9 - 3 = 6\, V$.The reading of the ammeter, which is in series with the $3\, \Omega$ resistor, is the current flowing through that branch:$I_{	ext{ammeter}} = \frac{V}{3\, \Omega} = \frac{6\, V}{3\, \Omega} = 2\, A$.

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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