In the circuit shown, if a conducting wire is connected between points $A$ and $B,$ the current in this wire will
AIPMT 2006, Medium
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Current through arm $C A D$, $I=\frac{V}{8}\, amp$
Potential difference between $C$ and $A=V_{C}-V_{A}$
$=\frac{V}{8} \times 4=\frac{V}{2}\, \mathrm{volt}$
Current through $C B D$, $I^{\prime}=\frac{V}{4}\, amp$
Potential difference between $C$ and $B=V_{C}-V_{B}$
$=\frac{V}{4} \times 1=\frac{V}{4} \,\text { volt. }$
Potential between $A$ and $B=V_{A}-V_{B}$
$\therefore \quad V_{A}-V_{B}=V_{C}-V_{B}-\left(V_{C}-V_{A}\right)=\frac{V}{4}-\frac{V}{2}=\frac{V}{4}$
$ \Rightarrow {V_A} - {V_B} < 0\quad or,\quad {V_A} < {V_B}$
As ${V_A} < {V_B}$, so direction of current will be from $B$ to $A$.
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