Explanation

(b) [image:3_image154.emf]
Current though arm $CAD,\ I = \frac{V}{8}amp$
Potential difference between $C$ and $A = V_{C} - V_{A}$
$= \frac{V}{8} \times 4 = \frac{V}{2}volt$
Current through $CBD,\ I^{''} = \frac{V}{4}amp$
Potential difference between $C$ and $B = V_{C} - V_{B}$
$= \frac{V}{4} \times 1 = \frac{V}{4}volt$
Potential between $A$ and $B = V_{A} - V_{B}$
$\therefore V_{A} - V_{B} = V_{C} - V_{B} - \left( V_{C} - V_{A} \right) = \frac{V}{4} - \frac{V}{2} = - \frac{V}{4}$
$\Rightarrow V_{A} - V_{B} < 0$ or, $V_{A} < V_{B}$
as $V_{A} < V_{B},$ so direction of current will be $B$ to $A$