Understanding Capacitor Discharge: Uncovering the Direction of Current Flow

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SUMMARY

The discussion centers on the behavior of current flow in capacitor circuits, specifically when analyzing the discharge of capacitors C1 and C2, where C2 is charged to 100V and C1 to 0V. Participants confirm that the direction of current flow can be arbitrary, as long as the final voltage calculations, such as V2(t), yield correct results. A negative value for V2(t) indicates that the assumed direction of current flow was incorrect, necessitating a reversal of the current arrow in the diagram. This highlights the importance of understanding the conventions in circuit analysis rather than strictly adhering to initial assumptions.

PREREQUISITES
  • Basic understanding of capacitor behavior and charging/discharging principles.
  • Familiarity with voltage and current direction conventions in circuit diagrams.
  • Knowledge of transient analysis in electrical circuits.
  • Experience with interpreting and solving differential equations related to circuit analysis.
NEXT STEPS
  • Study the principles of capacitor discharge and time constants in RC circuits.
  • Learn about Kirchhoff's voltage law and its application in circuit analysis.
  • Explore the use of simulation tools like LTspice for visualizing current and voltage in circuits.
  • Investigate the mathematical techniques for solving differential equations in electrical engineering contexts.
USEFUL FOR

Electrical engineers, students studying circuit theory, and anyone involved in analyzing capacitor circuits and current flow dynamics.

tamtam402
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In this scenario, C2 is charged to 100V and C1 to 0V. Wouldn't the current go in the other direction, unless I misunderstood something? The only reason I can see why the current is displayed in this direction is because we're asked to find V2(t), and doing it this way you'd find a negative V2 and it would "help" us see that we can have the rest of the equation equal to V2(t).
 

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yes, you are correct. It really doesn't matter which way you draw voltage and current, as long as you get the right answer (negative if your arrow points the "wrong" way)
 
agreed. You will get -v2(t) result. Typically diagrams are drawn with arbitrary direction of current. And once you find the value to be negative it simply implies the original direction of arrow is wrong and simply needs to be reversed.
 

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