Current Electricity 5 Question 17
17. A copper wire having cross-sectional area of $0.5 mm^{2}$ and a length of $0.1 m$ is initially at $25^{\circ} C$ and is thermally insulated from the surrounding. If a current of $1.0 A$ is set up in this wire, (a) find the time in which the wire will start melting. The change of resistance with the temperature of the wire may be neglected. (b) What will this time be, if the length of the wire is doubled?
(1979)
Melting point of copper $=1075^{\circ} C$, Specific resistance of copper $=1.6 \times 10^{-8} \Omega m$, Density of copper $=9 \times 10^{3} kg / m^{3}$, Specific heat of copper $=9 \times 10^{-2} cal / kg^{\circ} C$
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Solution:
- (a) $I^{2} R t=m S \Delta \theta=m(S \Delta \theta)$
$$ I^{2} \rho _1 \cdot \frac{l}{A} \quad t=A l \rho _2(S \Delta \theta) $$
$$ \begin{aligned} & \text { Here, } \quad \rho _1=\text { specific resistance } \\ & \text { and } \quad \rho _2=\text { density } \\ & \therefore \quad t=\frac{A^{2} \rho _2(S \Delta \theta)}{I^{2} \rho _1} \end{aligned} $$
Substituting the values, we have
$$ \begin{aligned} t & =\frac{\left(0.5 \times 10^{-6}\right)^{2}\left(9 \times 10^{3}\right)\left(9 \times 10^{-2} \times 4.18 \times 1050\right)}{(1.0)^{2}\left(1.6 \times 10^{-8}\right)} \\ & =55.5 s \end{aligned} $$
(b) From Eq. (i) we can see that time is independent of length of wire.
