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(B) The reading of the voltmeter when no current is drawn gives the electromotive force $(EMF)$ of the battery,so $E = 5\,V$.When the ammeter is conne

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$2$

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(B) The reading of the voltmeter when no current is drawn gives the electromotive force $(EMF)$ of the battery,so $E = 5\,V$.When the ammeter is connected,it measures the short-circuit current,which is given by $I_{sc} = E/r$,where $r$ is the internal resistance.Thus,$10 = 5/r$,which gives $r = 5/10 = 0.5\,\Omega$.When a resistor $R = 2\,\Omega$ is connected across the battery,the total resistance in the circuit is $R_{total} = R + r = 2 + 0.5 = 2.5\,\Omega$.The current $I$ flowing through the resistor is given by Ohm's law: $I = E / R_{total} = 5 / 2.5 = 2\,A$.

$A$ voltmeter connected across the terminals of a battery reads $5\,V$. When an ammeter is connected across the same terminals,it reads $10\,A$. If a resistor of $2\,\Omega$ is connected across the terminals of this battery,the current flowing through it will be ...........$A$.

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