A fuse made of lead wire has an area of cross-section 0.2 mm². On short circuiting, the current in the fuse wire reaches 30 amp. How long after the short circuiting will the fuse begin to melt?
Specific heat capacity of lead = 134.4 Jkg^–1 k^–1Melting point of lead = 327 °C
Density of lead = 11340 kg/m³Resistivity of lead = 22 x 10^–8 ohm-m
Initial temperature of the wire = 20°C
Neglect heat loss.

Solution

Area of cross-section of the wire, A= 0.2 mm²

If L be the length of the wire, its resistance

R = \frac{ρL}{A} = \frac{(22 \times 10^{- 8}) L}{(0.2 \times 10^{- 6}) m^{2}}

...(i)

Heat produced in the wire in one second, H=I²R = (30)² R J.

Heat required to raise the temperature of the wire to 327°C is given by,

Q = ms ∆ T

...(ii)

= (L A d) (134.4) (307) J

Time required to melt the wire

∆ T = \frac{Q}{I^{2} R} = \frac{L Ad \times 134.4 \times 307}{I^{2} \times ρ L} \times A

[from (i) and (ii)]

= \frac{A^{2} d}{I^{2} ρ} \times 134.4 \times 307 = \frac{{(0.2 \times 10^{- 6})}^{2}}{900} \times \frac{11340}{22 \times 10^{- 8}} \times 134.4 \times 307 = 0.0945 s e c

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