Question

A wire of resistance 5 ohms is stretched so that its diameter is halved. Determine its new resistance

Answer 1

Answer:- 20 ohms

Area of conductor a = $\pi r^{2}$

r = D /2

a = $\pi (D/2)^{2}$

Resistance and Diameter is inversely proportional by a factor of 2 since it is square.

So if diameter is halved, the resistance will increase 4 times.

In the given question, earlier resistance was 5 ohms, so the new resistance will be 20 ohms.

Answer 2

Given that,

A wire of resistance 5 ohms is stretched so that its diameter is halved.

To find,

New resistance of the wire.

Solution,

Resistance is given by :

$R=\rho\dfrac{l}{A}$

l is length of wire

A is area of cross section,

$A=\pi r^2$, r is radius of wire

Since, r = d/2, d is diameter

Resistance becomes,

$R=\rho\dfrac{l}{\pi (d/2)^2}\\\\R=\dfrac{\rho l}{(\pi/4)d^2}$

If diameter is halved, d' =d/2

So,

New resistance becomes,

$R=\rho\dfrac{l}{\pi ((d'/2)/2)^2}\\\\R'=\dfrac{\rho l}{(\pi/4) (d^2/4)}\\\\R'=4\times \dfrac{\rho l}{\pi r^2}\\\\R'=4\times R$

So, the new resistance becomes 4 times of the initial resistance.

Learn more,

Reistance

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