Resistance , current and temperature

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SUMMARY

The discussion centers on the relationship between current flow, resistance, and temperature change in resistors. It establishes that power consumption is calculated using the formula \(P = I^2 \times R\), leading to energy production in watt-hours. The temperature increase of the resistor after one hour depends on the specific heat of the material and the heat transfer mechanisms involved, including conduction, convection, and radiation. A rough estimate can be made by converting watts to BTUs, but this approach is complicated by the resistor's material properties and heat dissipation characteristics.

PREREQUISITES
  • Understanding of electrical power calculations (P = I^2 * R)
  • Knowledge of specific heat capacities of materials
  • Familiarity with heat transfer mechanisms (conduction, convection, radiation)
  • Basic principles of thermodynamics
NEXT STEPS
  • Research specific heat capacities of common resistor materials, such as tungsten
  • Learn about heat transfer principles in electrical components
  • Explore methods for calculating temperature changes in resistive materials
  • Investigate the impact of convection on thermal management in electronic devices
USEFUL FOR

Electrical engineers, physics students, and anyone involved in thermal management of electronic components will benefit from this discussion.

adool_617
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i have a question and i will be glad if anyone could help

you know if a current flow in resistance , a power is consumed =(I^2)*R

and if this current flows for an hour , this will produce energy and it's quantity is watt*hour

my question is , if the temperature of the resistor was 20 degrees at the start ( time=zero)

after 1 hour what will be the temperature
 
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Hi adool_617, welcome to PF. The temperature depends on a couple important factors:

- The specific heat of the resistor material, which connects dissipated energy to temperature change.
- The heat transfer details of the system (conduction, convection, and radiation), which govern the rate that energy is removed from the resistor.

So more information is needed to answer the question.
 
I agree with the answer provided...that question is asked here regulalry,,,it's NOT simple...

for a really rough cut approximate answer you could convert watts to BTU's and if you know the specific heat of the resistor do a crude calculation...a kwh is about 3413 BTU's...
pick some specific heat fropm a physics book or http://en.wikipedia.org/wiki/Specific_heat#Table_of_specific_heat_capacities

Maybe use tungeston as a surrogate...

But the problem with this approach is that likely a resistor is an internal resistance material analogous to tungestion, coated with ceramic...and that will change the heat disipation characteristics, perhaps dramatically.

And once the resistor begins to heat, convection begins which slows subsequent temperature rise...just as if a fan were applied...
 

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