(Reflected wave) adding voltage / subtracting current

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SUMMARY

The discussion centers on the principles of reflected voltage and current in electrical circuits, specifically addressing why reflected voltage is added to input voltage while reflected current is subtracted from input current. It emphasizes that the behavior of reflected waves is dictated by boundary conditions at the reflecting surface, particularly in cases involving perfect conductors. The voltage drop across a perfect conductor is zero, necessitating adjustments in current to maintain equilibrium, which influences the resulting magnetic field in the reflected wave.

PREREQUISITES
  • Understanding of electrical circuit theory
  • Familiarity with wave propagation in transmission lines
  • Knowledge of boundary conditions in electromagnetic theory
  • Concept of reflected waves in electrical engineering
NEXT STEPS
  • Study the principles of transmission line theory
  • Learn about boundary conditions in electromagnetic fields
  • Research the behavior of voltage and current reflections in open and short circuits
  • Examine the impact of perfect conductors on wave propagation
USEFUL FOR

Electrical engineers, students of circuit theory, and professionals involved in signal integrity and transmission line design will benefit from this discussion.

nomisme
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why we add reflected voltage to the input voltage but subtract current from the input current?

Also why the reflected current is shown on the other side of the load, traveling back to the input source but not on the same path where it travels from?

Does voltage wave reflect the same way as current does?
 
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nomisme said:
why we add reflected voltage to the input voltage but subtract current from the input current?

Also why the reflected current is shown on the other side of the load, traveling back to the input source but not on the same path where it travels from?

Does voltage wave reflect the same way as current does?

It's all to do with the Boundary Conditions at the reflecting surface. If the surface is a perfect conductor, the voltage drop across the surface must be Zero and the currents will 'adjust' to ensure that. This will account for the resulting H field in the reflected wave, I think.
 

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