A 6-ohm resistance wire is doubled by folding. What is the new resistance?


Given that

R = 6 Ohms

\(R = \frac{\rho l}{A}\)————-(1)

Wire is doubled so, Area (A) = 2A

length (l) = l/2

New Resistance will be:

\(R_{new} = \frac{\rho (\frac{l}{2})}{2A}\)


Rnew = ρl/4A

By using eqaution (1) we can simplify the above equation as

Rnew = R/4

Rnew = 6/4

Rnew = 1.5 Ohms

Hence, the new Resistance of wire is 1.5 Ohms.

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