Charging a capacitor: energy dissipated

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In a circuit with a resistor R and capacitor C connected to a voltage source V, the energy stored in the capacitor is Es=1/2*CV^2, while the resistor dissipates an equal amount of energy Ed=1/2*CV^2 when fully charged. To reduce energy dissipation by the resistor, one can charge the capacitor for a shorter duration, ideally only for a few time constants, which minimizes heat losses. However, this approach results in the energy stored in the capacitor being less than Es, as the energy dissipated by the resistor equals the energy stored. Ultimately, the energy losses in the resistor are unavoidable if the charging time approaches infinity.
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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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