A piece of wire is cut into four equal parts | vedclass

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(B) Let the original resistance of the wire be $R$. When the wire is cut into four equal parts,the resistance of each part becomes $R' = \frac{R}{4}$.

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$\frac{1}{16}$ as much

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(B) Let the original resistance of the wire be $R$. When the wire is cut into four equal parts,the resistance of each part becomes $R' = \frac{R}{4}$. These four parts are bundled together side by side,which means they are connected in parallel. The equivalent resistance $R_{net}$ of four resistors,each of resistance $R' = \frac{R}{4}$,connected in parallel is given by: $\frac{1}{R_{net}} = \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} + \frac{1}{R'} = \frac{4}{R'}$ Substituting $R' = \frac{R}{4}$: $\frac{1}{R_{net}} = \frac{4}{R/4} = \frac{16}{R}$ Therefore,$R_{net} = \frac{R}{16}$. Thus,the resistance of the bundle is $\frac{1}{16}$ of the original resistance.

$A$ piece of wire is cut into four equal parts and the pieces are bundled together side by side to form a thicker wire. Compared with that of the original wire,the resistance of the bundle is ..........

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