Question

A nichrome wire has a length of 80 cm and a resistance of 50 ohms. If the wire is folded four times, find the new resistance of the wire.

Answer 1

Given: L = 80 cm, R = 50 Ω, n = 4

When the wire is folded into four equal parts, the four parts act like four equal resistances connected in parallel. Each part has one-fourth the original length, so its resistance is also one-fourth the original resistance. The equivalent resistance of four identical resistances in parallel is one-fourth the resistance of each part.

Formula:

\[R_{\text{new}}=\frac{R}{n^{2}}\]

\[=\frac{50\ \Omega}{4^{2}}\]

\[=\frac{50}{16}\ \Omega\]

= 3.125 Ω

The new resistance of the wire is 3.125 Ω.