How Does Increasing Core Permeability Affect an RL Circuit's Performance?

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SUMMARY

Increasing the core permeability in an RL circuit affects its performance by maintaining the same steady-state current and voltage while increasing the magnetic energy due to a higher B-field. When substituting a core with a relative permeability of μr = 100 with one of μr = 1000, the force required to pull the core out increases by a factor of ten, resulting in lost energy that must be compensated by the power supply. The calculations of this energy growth depend on the specific dimensions of the magnetic circuit.

PREREQUISITES
  • Understanding of RL circuit fundamentals
  • Knowledge of magnetic permeability and its effects
  • Familiarity with magnetic fields (B-field and H-field)
  • Basic principles of energy conservation in electrical circuits
NEXT STEPS
  • Study the effects of core materials on inductance in RL circuits
  • Learn about magnetic energy calculations in electrical circuits
  • Explore the relationship between magnetic fields and circuit performance
  • Investigate the impact of different dielectric materials on capacitor performance
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Electrical engineers, circuit designers, and students studying electromagnetism and circuit theory will benefit from this discussion.

cnh1995
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In an RC dc circuit at steady state, if the permitivitty of the capacitor is increased by inserting a different dielectric, additional charging current will flow and the capacitor will accept more charge (due to increase in capacitance). What will happen if the inductance in an RL circuit is increased similarly (at the steady state),i.e. by increasing the permeability of the core? Here, the final value of the current is same for both the cores.
 
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cnh1995 said:
What will happen if the inductance in an RL circuit is increased similarly (at the steady state),i.e. by increasing the permeability of the core?
Say you have some solenoid ( cylindric core ) with some relative permeability, μr = 100. You pull it out from within the coil, and are using some force to do so ( F100 ).
Now you substitute it with another core ( μr = 1000 ), and putting in the core, the new core will be attracted by a force, F1000.
F1000 ≈ 10 * F100 , so all in all the result is lost energy as seen from the circuit, and that lost energy must be compensated somehow by the powersupply ( yielding more amps or volts integrated over time of replacement ).

From another point of view, you can say that switching the cores has caused:
- the steady state current is the same.
- the voltage across the RL circuit is the same.
- the H-field ( proportional to current ) is the same.
- the B-field has increased ( B = μ0 * μr * H )

As the magnetic energy tends to be proportional to B*H, the powersupply must also deliver this growth in magnetic energy, but this a hard calculation because the steady state shape of the magnetic fields will be altered by substitution of the core. The calculations of growth in magnetic energy depends on the exact dimensions of the magnetic circuit.
 

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