Charging a capacitor: energy dissipated

In summary: Thus, in order to minimise the energy lost, one can charge the capacitor for few time constants only.
  • #1
iamquantized
28
0
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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  • #2
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...
 
  • #3
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...
 
  • #4
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.
 
  • #5
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.
 

Related to Charging a capacitor: energy dissipated

1. How does a capacitor store energy?

A capacitor stores energy by storing electric charge on its plates. When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, and the energy is stored in this electric field.

2. Why does a capacitor need to be charged?

A capacitor needs to be charged in order to store energy. When a capacitor is charged, it can then be used as a power source to release this stored energy when needed.

3. How is a capacitor charged?

A capacitor can be charged by connecting it to a power source, such as a battery or power supply, and allowing current to flow into the capacitor. The amount of charge that can be stored on the capacitor depends on its capacitance, which is determined by its size and the materials used.

4. How is energy dissipated when a capacitor is discharged?

When a capacitor is discharged, the stored energy is released in the form of an electric current. This current flows from one plate to the other, and the energy is dissipated as heat in the circuit. The amount of energy dissipated is proportional to the capacitance and the voltage of the capacitor.

5. What factors affect the energy dissipated when charging a capacitor?

The energy dissipated when charging a capacitor is affected by its capacitance, voltage, and the resistance in the circuit. A higher capacitance or voltage will result in more energy being stored, while a higher resistance will result in less energy being dissipated. The time it takes to charge the capacitor also affects the amount of energy dissipated, with a longer charging time resulting in less energy being dissipated.

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