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Explain that magniture of current which can melt a fuse wire is independent of its length. |
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Answer» Solution :Assume that the fuse wire is a cylindrical object of radius r and length l. LET resistivity OFITS material be `rho`. RESISTANCE of the fuse wire can be written as `R=rho l/(pir^2)` Hence rate of heat generation `H=I^2R rArr H=I^2(rhol/(pir^2))` As the temperature of fuse wire increases its rate of radiation increases. Rate of radiation for temperature T can be written as `(DeltaU)/(Deltat)=e sigma(2pirl)T^4` In equation (ii) e is emissivity of the material `sigma` is the stefan-Boltzmann constant `2pirl` is the surface of fuse wire and T is the absolute temperature . Usually the size of fuse wireis small and wo can easily neglect the rate of absorption of heat . To attain a steady STATE temperature , the rate of electrical heating BECOMES equal tothe rate of radiation. `H=(DeltaU)/(Deltat)` `I^2rho l/(pir^2)=e sigma(2pirl)T^4` `rArr I^2rho=2 pi^2e sigmar^3T^4` |
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