# Homework Help: Charging a capacitor: energy dissipated

1. Aug 31, 2007

### iamquantized

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.

2. Aug 31, 2007

### Staff: Mentor

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. Aug 31, 2007

### iamquantized

No matter what your R values, the energy dissipated in R will be 1/2*CV^2 if you let t goes to infinity.

4. Sep 1, 2007

### premagg

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. Sep 1, 2007

### iamquantized

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.