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(B) Let the potential at $c$ be $V_c$ and at $d$ be $V_d$. Since the battery is connected across $c$ and $d$,we have $V_c > V_d$.For the branch $c-b-d

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From $b$ to $a$

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(B) Let the potential at $c$ be $V_c$ and at $d$ be $V_d$. Since the battery is connected across $c$ and $d$,we have $V_c > V_d$.For the branch $c-b-d$,the resistances $X$ and $Y$ are in series. Since $X = Y$,the potential at $b$ is the midpoint potential: $V_b = \frac{V_c + V_d}{2}$.For the branch $c-a-d$,the resistances $A$ and $B$ are in series. The potential at $a$ is given by the potential divider formula: $V_a = V_d + (V_c - V_d) \frac{B}{A + B}$.Since $A > B$,we have $\frac{B}{A + B} Therefore,$V_a Comparing the two,we find $V_b > V_a$. Since current flows from higher potential to lower potential,the current flows from $b$ to $a$.

In the Wheatstone's bridge (shown in figure) $X = Y$ and $A > B$. The direction of the current between $a$ and $b$ will be

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