The equivalent resistance of resistors connected in parallel is
[tex] \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... [/tex]
In our problem, we have 5 identical resistor of resistance [tex]R[/tex], so their equivalent resistance is
[tex] \frac{1}{R_{eq}} = 5 (\frac{1}{R} ) [/tex]
The problem also says that the equivalent resistance is [tex]R_{eq}= 17 \Omega[/tex], so we can find the resistance R of each piece of wire:
[tex]R=5 R_{eq} = 5 \cdot 17 \Omega = 85 \Omega[/tex]
In the initial wire, it's like the 5 pieces are connected in series, so the equivalent resistance of the wire is just the sum of the resistances of the 5 pieces:
[tex]R_{wire} = 5 R = 5 \cdot 85 \Omega = 425 \Omega[/tex]
