IIT JEE , AIEEE Forum - Physics , Electricity

kasirajan.1990

A copper wire having a cross sectional area of 0.5 mm²and a length of 0.1 m is intially at 25 C and is thermally insulated from its surroundings .If a current of 10 A
is set up in the wire........
(i) find the time in which the wire will start melting .The change in resistance with
the temperature of the wire may be neglected .
(ii) what will this time be if length of the wire is doubled ?
(for Cu , density = 9 x 10³ kg/m³
specific heat = 378 J/kg C
melting point = 1075 C
electrical resistivity = 1.6 x 10^-8 ohm - metre )

Ank999

use the formula
R = R0 (1+adt)

iitkgp_bipin

Here we have to neglect the variation of resistance with the temperature as stated in problem.

Initially it is at 25 ⁰C and at melting point its temperature becomes 1075 ⁰C.
Change in temperature,

T = 1050 ⁰C

Electrical heat developed due to the passage of current raises the temperature causing the the thermal heat to develop.

Resistivity,

= 1.6 x 10^-8
Length, l = 0.1 m
Area, A = 0.1x10^-4 m²
Resistance, R =

l/A

Electrical heat developed = i²Rt (where t is the time elapsed)

Density, d = 9 x 10³ kg/m³
Volume = Al
Mass, m = dAl

Thermal heat = mC(

T)

Equating these two forms of heat :

mC(

T) = i²Rt

(dAl)C(

T) = i²(

l/A)t

t = dA²C(

T) / i²

Putting these values :

t = 5581406.25 sec = 1550.39 hours

Since its expression is independent of length it would remain same even if its length is doubled.

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