$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

  • A
    $4$
  • B
    $4.5$
  • C
    $8$
  • D
    $2$

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