A wire of resistance 10 Ω is drawn out so that its length is thrice its original length. Calculate its new resistance (resistivity and density of the wire remain unchanged).

Given,
Resistance of the wire, R = 10 Ω

Length of the new wire is thrice of it's initial length.

The volume of wire remains same in both cases.
So, Volume = Area of cross-section x Length

V = A' L' = AL

and

\frac{A'}{A} = \frac{L}{L'} = \frac{L}{3 L} = \frac{1}{3}

[

∵

L' = 3L]

Ratio of the resistances are,

∴

\frac{R'}{R} = \frac{ρ \frac{L'}{A'}}{ρ \frac{L}{A}} = \frac{L'}{L} \times \frac{A}{A'} = \frac{3}{1} \times \frac{3}{1} = 9

Therefore,

R' = 9 R = 9 \times 10 = 90 Ω .

196 Views

Electricity

Hope you found this question and answer to be good. Find many more questions on Electricity with answers for your assignments and practice.

Science

Browse through more topics from Science for questions and snapshot.