When an ammeter of negligible internal resistance is inserted in series with circuit it reads $1A$. When the voltmeter of very large resistance is connected across $X$ it reads $1V$. When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10\, V$ across the battery. The internal resistance of the battery is equal to .............. $\Omega$
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Initially ammeter measures current $1 A$ through the circuit, when voltmeter measures voltage $1 \mathrm{V}$ the resistance is,
$R=\frac{V}{I}=1 \Omega$
When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10 \mathrm{V}$ across the battery.
Hence the current is
$I=\frac{10}{1}=10 A$
Now, the voltage across the terminals of the battery is
$V=E-I r$
$10=12-10 r$
$10 r=2$
$r=0.2 \Omega$
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