Investigating Fuses: Explaining Why They Melt

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SUMMARY

Fuses melt due to the heat generated by electrical current exceeding the melting point of the fuse wire. The heat generated can be calculated using the formula dh=i²Rdt, where R is determined by the resistivity of the material, expressed as R=ρ(l/A). This relationship highlights the importance of electrical resistivity, thermal conductivity, and the physical properties of the fuse wire, including its thermal expansion, in understanding fuse operation.

PREREQUISITES
  • Electrical resistivity and its impact on resistance
  • Basic principles of thermal conductivity
  • Understanding of thermal expansion in materials
  • Knowledge of circuit components and their functions
NEXT STEPS
  • Research the relationship between electrical resistivity and temperature changes in materials
  • Explore the principles of thermal conductivity in electrical components
  • Study the effects of thermal expansion on electrical circuits
  • Learn about different types of fuses and their specific applications
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Students in physics, electrical engineers, and anyone interested in the principles of electrical safety and circuit design.

O_oSam
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I have just started a Physics project which I have already missed the deadline for >_<

I have a very minimal set of results, I basically took a power supply>smoothing unit and put them in a circuit with a component holder and two resistors (in parrallel so I didnt blow the ammeter) and took readings.

Was wondering if I could get some helpful information etc. on fuses. Can anyone give me a detailed explanation of why fuses melt, I was hoping to tie in Electrical resistivity [I'm aware it changes as fuse heats] and possible thermal conductivity?

Does the thermal expansion of a wire have any relevance?


Thanks

Sam
 
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Im not sure if I should tell you this, because its obviously something you should do yourself.

Fuses melt because the heat generated in them increases the temperature of the fuse wire beyond the melting point. The heat generated is equal to dh=i^2Rdt and R is related to resistivity by the expression R=\rho \frac{l}{A}, where A is the cross sectional area, \rho is the resistivity and l is the length of the wire under consideration.
 

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