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(C) The heat generated in a circuit is given by $H = \frac{V^2}{R} 	imes t$. Since the voltage $V$ of the battery is constant,the heat generated is i

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When the wire is divided into four parts and all four are connected to the battery in parallel.

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(C) The heat generated in a circuit is given by $H = \frac{V^2}{R} 	imes t$. Since the voltage $V$ of the battery is constant,the heat generated is inversely proportional to the resistance $R$ $(H \propto \frac{1}{R})$.To maximize heat,we must minimize the total resistance $R$.Let the resistance of the full wire be $R_0$.$(a)$ $R = R_0$$(b)$ Two parts in parallel: $R = \frac{(R_0/2)}{2} = \frac{R_0}{4}$$(c)$ Four parts in parallel: $R = \frac{(R_0/4)}{4} = \frac{R_0}{16}$$(d)$ Half wire: $R = \frac{R_0}{2}$Comparing the resistances,the minimum resistance is $\frac{R_0}{16}$ in case $(c)$. Therefore,the largest amount of heat is generated in case $(c)$.

You are given a resistance wire of length $50\, cm$ and a battery of negligible resistance. In which of the following cases is the largest amount of heat generated?

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