Deriving the damped circuit equation and energy dissipated through a resistor.

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SUMMARY

The discussion focuses on deriving the damped circuit equation and calculating the energy dissipated through a resistor in an RC circuit. The energy stored in a capacitor is expressed as 0.5CE², where E is the voltage, and the instantaneous power dissipated in a resistor R during capacitor discharge is analyzed. The equation E = R dq/dt + q/C is solved to show that the charge q(t) can be expressed as q(t) = EC(1 - e^(-t/RC)). Both problems require calculus for their solutions.

PREREQUISITES
  • Understanding of RC circuit theory
  • Familiarity with calculus, specifically integration and differentiation
  • Knowledge of energy storage in capacitors
  • Ability to manipulate exponential functions
NEXT STEPS
  • Study the derivation of the damped circuit equation in detail
  • Learn about energy dissipation in resistive circuits
  • Explore the application of calculus in electrical engineering problems
  • Review the behavior of exponential decay in RC circuits
USEFUL FOR

Students preparing for exams in electrical engineering, particularly those focusing on circuit analysis and energy dissipation in capacitors and resistors.

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Homework Statement



I figure I will just combine these two questions into one topic.

1) The energy stored in a capacitor is .5CE^2, where E is the voltage. Wat is the instantaneous power dissipated in a resistor R through which this capacitor discharges? Show that the total energy dissipated through the resistor is exactly .5CE^2

2) E=R dq/dt + q/C Show that the solution to this equation is q(t)=EC (1-e^(-t/RC))

Homework Equations



I=dq/dt=-Q/RC e^(-t/RC)

Voltage across a resistor in a series RC circuit: V=Ee^(-t/RC)

Both problems require calculus.

The Attempt at a Solution



For number 1... P=i^2 R ? I'm not sure what to do with this.

The second question I am just totally clueless about. I've never seen a derivative taken where you end up with e when you didn't start with e. I think this may be another example of "hey, we are going to give you this question to struggle with even though you have never learned the math. Good luck on the exam."

This is my last ditch attempt before my exam tomorrow.
 
Last edited:
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I'll write the solution for 2) as you say it's out of your reach.

[tex]E=R \ dq/dt + q/C[/tex]

[tex]E - {q \over C} = R \ {dq \over dt}[/tex]

[tex]dt = R \ {dq \over (E - {q \over C})}[/tex]

[tex]\int_0^{\infty} {dt} = \int_0^{\infty} R\ { dq \over {(E - {q \over C}) } }[/tex]

[tex]t = -{RC} \ ln {(E - {q \over C})} +K[/tex]

Then with some passages you arrive to the final solution.
 

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