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(D) Let the total resistance of the wire be $R$. In figure $(1)$,the resistance between $M$ and $N$ is $R_1 = R$.In figure $(2)$,the wire is bent into

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Heat produced in the wire and current through the wire.

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(D) Let the total resistance of the wire be $R$. In figure $(1)$,the resistance between $M$ and $N$ is $R_1 = R$.In figure $(2)$,the wire is bent into a square,so each side has resistance $R/4$. The points $M$ and $N$ are at opposite corners of the square. This creates two parallel branches,each with a resistance of $R/2$ (two sides in series). The equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{4}{R}$,so $R_2 = R/4$.Since $R_2 According to Ohm's law,$I = V/R$. Since $R$ decreases,the current $I$ increases.The heat produced is $H = \frac{V^2}{R} t$. Since $R$ decreases,the heat produced $H$ increases.Therefore,both the current through the wire and the heat produced in the wire increase.

$A$ wire is connected to a battery between the points $M$ and $N$ as shown in figure $(1)$. The same wire is bent in the form of a square and then connected to the battery between the points $M$ and $N$ as shown in figure $(2)$. Which of the following quantities increases?

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