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(C) Initially,the ammeter measures a current of $I = 1 \, A$ through the circuit. When the voltmeter measures a voltage of $V_X = 1 \, V$ across resis

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$0.2$

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(C) Initially,the ammeter measures a current of $I = 1 \, A$ through the circuit. When the voltmeter measures a voltage of $V_X = 1 \, V$ across resistor $X$,the resistance $X$ is given by:$X = \frac{V_X}{I} = \frac{1 \, V}{1 \, A} = 1 \, \Omega$.When points $A$ and $B$ (across $Y$) are shorted,the total resistance of the circuit becomes just $X = 1 \, \Omega$. The voltmeter now measures the terminal voltage of the battery as $V = 10 \, V$. The current in this new circuit is:$I' = \frac{V}{X} = \frac{10 \, V}{1 \, \Omega} = 10 \, A$.Using the terminal voltage equation $V = E - I'r$,where $E = 12 \, V$ is the $EMF$ of the battery and $r$ is the internal resistance:$10 = 12 - 10r$$10r = 12 - 10$$10r = 2$$r = 0.2 \, \Omega$.

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