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(C) The terminal potential difference $V$ across a battery is given by $V = E - Ir$,where $E$ is the $emf$ and $r$ is the internal resistance.From the

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The $emf$ of the battery is $1.5 \, V$ and its internal resistance is $1.5 \, \Omega$.

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(C) The terminal potential difference $V$ across a battery is given by $V = E - Ir$,where $E$ is the $emf$ and $r$ is the internal resistance.From the graph,when the current $I = 0$,the voltage $V = E = 1.5 \, V$.When the current $I = 1000 \, mA = 1 \, A$,the voltage $V = V_0 \approx 0 \, V$.Substituting these values into the equation $V = E - Ir$:$0 = 1.5 - (1 \, A) 	imes r$$r = 1.5 \, \Omega$.Thus,the $emf$ of the battery is $1.5 \, V$ and its internal resistance is $1.5 \, \Omega$.

To verify Ohm's law,a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current,and the following graph is obtained. If $V_0$ is almost zero,identify the correct statement.

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