Thevenin Equiv with just Capacitors

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SUMMARY

The discussion centers on calculating the Thevenin equivalent voltage (Vth) in a circuit composed solely of capacitors. The participant successfully determined the equivalent capacitance to be 10F but expressed confusion regarding the Vth, initially estimating it at 22.5V based on the voltage division principle. However, the provided answer key states Vth as 15V, leading to speculation about a potential error in the problem statement regarding the voltage source. The correct approach to find Vth involves analyzing the circuit configuration and applying the principles of capacitor behavior in steady-state conditions.

PREREQUISITES
  • Understanding of Thevenin's theorem
  • Knowledge of capacitor circuits and equivalent capacitance
  • Familiarity with voltage division in electrical circuits
  • Basic circuit analysis techniques
NEXT STEPS
  • Study Thevenin's theorem applications in capacitor circuits
  • Learn about voltage division specifically for capacitors
  • Explore steady-state analysis of RC circuits
  • Review common mistakes in circuit analysis involving capacitors
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Electrical engineering students, circuit designers, and anyone seeking to deepen their understanding of Thevenin equivalents in capacitor networks.

kostoglotov
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Homework Statement



TWB5rRu.png


imgur link: http://i.imgur.com/TWB5rRu.png

Homework Equations

The Attempt at a Solution



I can find the equivalent capacitance, 10F, but I'm not really sure how to find the equiv. Thev. voltage. I figured that with the voltage source present, half the voltage will be lost across that first capacitor (since the next two parallel caps have an equivalence of 5F as well), so therefore at the very least in the Thev equiv circuit we should be seeing 22.5 V across the terminals. But the answer in the back gives Vth as 15V, less than 22.5.

How does one find the Vth?
 
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Hmm. I wonder if at some point the problem was changed slightly without the answer key being updated. If the voltage source was 30 V rather than 45 V, then the given answer would make sense. I can't see the open circuit voltage being anything other than 22.5 V.
 
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