Question

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

Answer 1

Answer:

given that

given thatr = 6 ohms

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2a

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\)

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\)

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4a

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4aby using eqaution (1) we can simplify the above equation as

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4aby using eqaution (1) we can simplify the above equation asrnew = r/4

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4aby using eqaution (1) we can simplify the above equation asrnew = r/4rnew = 6/4

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4aby using eqaution (1) we can simplify the above equation asrnew = r/4rnew = 6/4rnew = 1.5 ohms

given thatr = 6 ohms\(r = \frac{\rho l}{a}\)————-(1)wire is doubled so, area (a) = 2alength (l) = l/2new resistance will be:\(r_{new} = \frac{\rho (\frac{l}{2})}{2a}\) rnew = ρl/4aby using eqaution (1) we can simplify the above equation asrnew = r/4rnew = 6/4rnew = 1.5 ohmshence, the new resistance of wire is 1.5 ohms.

Answer 2

Explanation:

Given :

R=6ohms

R=ρl/A-------------(1)

Wire is doubled so, Area =A=2A

length=l=l/2

New Reistance will be:

Rnew=ρ(l/2)/2A

Rnew=ρl/4A

Rnew=R/4-----(from equ 1)

Rnew=6/4=1.5 ohms

Hence new Reistance of wire is 1.5ohms.