While doing an experiment with a potentiometer as shown in the figure,it was found that the deflection is one-sided and $(i)$ the deflection decreased while moving the jockey from one end $A$ of the wire to the end $B$; $(ii)$ the deflection increased while the jockey was moved towards the end $B$.
$(i)$ Which terminal ($+$ or $-ve$) of the cell $E_1$ is connected at $X$ in case $(i)$ and how is $E_1$ related to $E$?
$(ii)$ Which terminal of the cell $E_1$ is connected at $X$ in case $(ii)$?

(N/A) $(i)$ When the jockey is moved from $A$ to $B$,if the galvanometer deflection decreases,it implies that the potential difference across the galvanometer circuit is decreasing. This happens because the potential drop across the length of the potentiometer wire $A$-jockey opposes the $EMF$ $E_1$. For the deflection to decrease as the jockey moves towards $B$ (where potential increases),the terminal of $E_1$ connected to $X$ must be the positive terminal,and $E_1 > E$ must hold true so that the net current is in a direction that opposes the initial deflection.
$(ii)$ When the jockey is moved from $A$ to $B$,if the deflection increases,it implies that the potential difference across the galvanometer circuit is increasing. This occurs when the terminal of $E_1$ connected to $X$ is the negative terminal,causing the $EMF$ $E_1$ to add to the potential drop across the wire segment,thereby increasing the current through the galvanometer.