A uniform metallic wire having resistance $4\Omega$ is bent to form a square loop ( $ABCD$ ) (see fi

Question

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:

Accepted Answer

Sol. [Key Concept: Symmetry of a resistive square loop] → Each side will have resistance of $1\Omega$- It is balanced wheat stone bridge→ No current in the resistance of $2\Omega$$${ightarrow \ \left\lbrack R_{	ext{effective}	ext{~}} ightbrack_{AC} = 1\Omega }{ightarrow \ I = \frac{E}{R_{	ext{eff.}	ext{~}}} = \frac{2}{1} }{ightarrow \ I = 2	ext{ }A}$$

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