To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The…

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 ${\mathrm{V}}_{\mathrm{o}}$ is almost zero, identify the correct statement:

1. The emf of the battery is $1.5 \mathrm{V}$ and its internal resistance is $1.5\mathrm{ }\mathrm{\Omega }$
2. The value of the resistance $\mathrm{R}$ is $1.5\mathrm{ }\mathrm{\Omega }$
3. The potential difference across the battery is $1.5 \mathrm{V}$ when it sends a current of $1000 \mathrm{m}\mathrm{A}$
4. The emf of the battery is $1.5\mathrm{ }\mathrm{V}$ and the value of $\mathrm{R}$ is $1.5\mathrm{ }\mathrm{\Omega }$

Solution

$V=E-Ir$
when $V=V_0=0\Rightarrow 0=E-Ir$
$\therefore E=Ir$
when $I=0, V=E=1.5 \mathrm{V}$
$\therefore r=1.5 \mathrm{\Omega }.$

Asked in: JEE Main 2019 (12 Apr Shift 1)

Practice more Current Electricity questions on Aicharya