A battery has an $\mathrm{e}\mathrm{m}\mathrm{f}$ of $15\mathrm{ }\mathrm{V}$ and internal resistance of $1 \mathrm{\Omega }$ . The terminal to terminal potential difference is less than, equal to or greater than $15\mathrm{ }\mathrm{V}$, if the current in the battery is

(1) from negative to positive terminal,

(2) from positive to negative terminal,

(3) zero current?

Two batteries of emf ${\epsilon }_1$ and ${\epsilon }_2\left({\epsilon }_2>{\epsilon }_1\right)$ and internal resistances $r_1$ and $r_2$, respectively are connected in parallel, as shown in the figure.

The internal resistances of two cells shown are $0.1 \mathrm{\Omega }$ and $0.3 \mathrm{\Omega }$. If $R=0.2 \mathrm{\Omega }$, the potential difference across the cell

Two nonideal batteries are connected in parallel. Consider the following statements.

$\left(\mathrm{I}\right)$ The equivalent emf is smaller than either of the two EMFs.

$\left(\mathrm{II}\right)$ The equivalent internal resistance is smaller than either of the two internal resistance.

Consider the meter bridge circuit without neglecting end corrections. If we used $100 \mathrm{\Omega }$ and $200 \mathrm{\Omega }$ resistances in place of $R$ and $S$, respectively, we get null deflection at $l_1=33.0 \mathrm{cm}$. If we interchange the resistances, the null deflection is found to be at $l_2=67.0 \mathrm{cm}$. If $\alpha$ and $\beta$ are the end correction, then the value of $\alpha +\beta$ should be,