A battery of emf $E$ and internal resistance $r$ is connected across a resistance $R$. Resistance $R$ can be adjusted to any value greater than or equal to zero. A graph is plotted between the current $\left(\mathrm{i}\right)$ passing through the resistance and potential difference ($V$) across it. Select the correct alternative (s).

In the given figure, $E= 12 \mathrm{V}, R_1 = 3 \mathrm{\Omega },\mathit{ }{R}_2 = 2 \mathrm{\Omega }$ and $r\mathit{ }=1 \mathrm{\Omega }.$ Then choose the correct option/s.

By mistake, a voltmeter is placed in series and an ammeter in parallel with resistance in an electric circuit, with a cell in series. 

Two cells of unequal emfs ${\epsilon }_1$ and ${\epsilon }_2$ and internal resistances $r_1$ and $r_2$are joined as shown. $V_A$ and $V_B$ are the potentials at $A$ and $B$, respectively. 

In the potentiometer arrangement shown, the driving cell $D$ has emf $\epsilon$ and internal resistance $r$. The cell $C$, whose emf is to be measured, has emf $\frac{\epsilon }{2}$ and internal resistance $2r$. The potentiometer wire is the potentiometer wire is $100-\mathrm{cm}$ long. If balance is obtained at the length $AJ=l.$