Charging a capacitor: energy dissipated

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Discussion Overview

The discussion revolves around the energy dissipated in a resistor when charging a capacitor in a circuit with a voltage source. Participants explore how to design circuits and voltage sources to minimize energy loss in the resistor while charging the capacitor to a specific energy level.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant states that the energy dissipated in the resistor will always equal 1/2*CV^2 as time approaches infinity.
  • Another participant suggests that to reduce energy loss, the capacitor should be charged for a shorter duration, specifically for a few time constants, to minimize heat losses.
  • It is noted that while energy loss in the resistor can be less than 1/2*CV^2, the energy stored in the capacitor will also be less than this amount, indicating a balance between stored and dissipated energy.
  • Participants are encouraged to calculate the total energy dissipated by the resistor for various resistor values to better understand the implications of their designs.

Areas of Agreement / Disagreement

Participants generally agree that energy loss in the resistor is unavoidable when allowing time to approach infinity. However, there is no consensus on the best methods to design circuits or voltage sources to achieve lower energy dissipation, as different strategies are proposed.

Contextual Notes

Participants mention the need for specific calculations involving different resistor values and the implications of these calculations on energy dissipation, but do not resolve the mathematical details or assumptions involved.

iamquantized
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Consider a simple circuit where resistance R is connected in series with a capacitor C and a voltage source V. The switch is connected at t=0. After t>>RC, the capacitor is fully charged with energy Es=1/2*CV^2 and R dissipates energy of Ed=1/2*CV^2.

Question.
1) How can one design a circuit to reduce energy dissipated by R?
2) How can one design a source V(t) such that it can charge the capacitor to Es with Ed<Es.
 
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Thread moved to Homework Help.

Welcome to the PF, iamquantized. It's a great place. Please keep in mind that homework and coursework questions need to be posted here in the Homework Help forums, and not in the general technical forums.

We also require that you show us some of your own work, before we can offer much in the way of tutorial help. For this problem, you should calculate the total energy dissipated by the resistor, for several different resistor values. Just pick some simple numbers like a 1V power supply and a 1uF capacitor, and do the integration for a 1 Ohm resistor, and a 10 Ohm resistor, and a 100 Ohm resistor... you can use time=infinity for the upper bound on the integral.

Please show us what you get for each value of R, and tell us what you think it means...
 
No matter what your R values, the energy dissipated in R will be 1/2*CV^2 if you let t goes to infinity.





berkeman said:
Thread moved to Homework Help.

Welcome to the PF, iamquantized. It's a great place. Please keep in mind that homework and coursework questions need to be posted here in the Homework Help forums, and not in the general technical forums.

We also require that you show us some of your own work, before we can offer much in the way of tutorial help. For this problem, you should calculate the total energy dissipated by the resistor, for several different resistor values. Just pick some simple numbers like a 1V power supply and a 1uF capacitor, and do the integration for a 1 Ohm resistor, and a 10 Ohm resistor, and a 100 Ohm resistor... you can use time=infinity for the upper bound on the integral.

Please show us what you get for each value of R, and tell us what you think it means...
 
if u let t tending to infinity the losses of 0.5cv2 are unavoidable.
But if u want to minimise the losses then you can just charge the capacitor for few time constants only.So in less time the heat losses will be very low.This is applicable for both of your questions.
 
premagg said:
if u let t tending to infinity the losses of 0.5cv2 are unavoidable.
But if u want to minimise the losses then you can just charge the capacitor for few time constants only.So in less time the heat losses will be very low.This is applicable for both of your questions.

Although the energy loss in resistor is less than 0.5CV^2, but the energy stored in capacitor will also be less than 0.5CV^2. In fact, energy stored is still = energy dissipated by resistor.
 

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