Energy Stored in Capacitor at t=870μs: Calculations & Answers

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Discussion Overview

The discussion revolves around calculating the energy stored in a capacitor at a specific time (t = 870 microseconds) given an initial voltage and a current waveform. Participants explore the relationship between voltage, current, and energy in capacitors, addressing potential errors in calculations and interpretations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equation for voltage as v(t) = 50*e^-2000t + 30 and attempts to calculate energy using w(t) = 1/2cv^2.
  • Another participant points out a potential sign error in the integral used to calculate the change in voltage and questions its relation to energy change.
  • Clarification is sought regarding the interpretation of the definite integral limits in the context of voltage and energy calculations.
  • Participants discuss the initial voltage of the capacitor, noting a discrepancy between the calculated voltage at t=0 (80 volts) and the stated initial voltage (-20 V).
  • There is a challenge regarding whether the energy should be calculated as w(870 microseconds) - w(0 seconds) or simply w(870 microseconds).

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating energy and the implications of the initial voltage. The discussion remains unresolved with multiple competing interpretations of the calculations and their relationships.

Contextual Notes

There are ambiguities in the definitions of the integral limits and the relationship between voltage change and energy change that remain unaddressed. The initial conditions and the implications of the current waveform on the calculations are also not fully clarified.

carpaltunnel
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Homework Statement



The initial voltage on the 0.5 microF capacitor shown in the figure is -20 V. the capacitor current has the waveform shown in the figure. https://i.imgur.com/P8jwI7s.jpg?1

How much energy is stored in the capacitor at t = 870 micro seconds.

Homework Equations



v(t) = 1/c * integral from t to t0 i dtao + v(t0) EDIT: "integral from t to t0" is a definite integral with t on the top and t0 on the bottom
w(t) = 1/2cv^2

The Attempt at a Solution



got v(t) = 50*e^-2000t + 30

after plugging into w(t) tested the following answers:

https://imgur.com/3P8ipf4
 
Last edited:
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Your integral calculated the change in voltage (except, you have a sign error) How does that relate to the change in energy?
 
"from t to t0" is admittedly ambiguous but i mean that t is on the top of the definite integral and t0 is on the bottom. voltage relates to energy by the equation w(t) = 1/2cv^2. i plugged in my equation for v(t) and the time as 870 microseconds. please let me know what else is unclear/wrong. thank you for your reply.
 
carpaltunnel said:
got v(t) = 50*e^-2000t + 30
What does that give for t=0?
carpaltunnel said:
voltage relates to energy by the equation w(t) = 1/2cv^2.
But that's different from change in voltage and change in energy.
 
"what does that give for t=0?"

80 volts. what are you implying?

"but that's different from change in voltage and change in energy"

what are you implying? the expressions for v(t) and w(t) i have provided are the capacitor voltage and energy stored in a capacitor, respectively, at a particular instance for a time. i don't think the answer is w(870 microseconds) - w(0 seconds). please let me know if i am wrong. thank you for your reply.
 
carpaltunnel said:
80 volts.
But:
carpaltunnel said:
The initial voltage on the 0.5 microF capacitor shown in the figure is -20 V.

carpaltunnel said:
i don't think the answer is w(870 microseconds) - w(0 seconds)
Yes, you're right, I misread the question.
 

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