How much energy is dissipated by the 25 ohm resistor?

[pari786]

Question:

A 0.25 x 10^-6 F capacitor is charged to 50V. It is then connected in series with a 25 ohm resistor and a 100 ohm resistor and alowed to discharge completely. How much energy is dissipated by the 25 ohm resistor ?


Attempt:

I found a formula in the textbook related to this question:

The resistors dissipate energy at the rate:

PR = I^2R = (Delta V)^2 / R


I don't know what to do next ... and if I'm using the right formula. Please somebody help.

 

[mjsd]

ok... conservation of energy
ask yourself, how much energy was stored in capacitor initially, then how do u think the energy are dissipated between the 25 and 100 ohm resistor? remember everything is in series... and you already have $$P=I^2 R$$

 

[ada15]

ok... conservation of energy
ask yourself, how much energy was stored in capacitor initially, then how do u think the energy are dissipated between the 25 and 100 ohm resistor? remember everything is in series... and you already have $$P=I^2 R$$

Hi, Do you calculate the energy by this formula;

Q=CV
Q= (0.25 * 10^-6) ( 50)
Q = 0.0000125 C

 

[mjsd]

Hi, Do you calculate the energy by this formula;

Q=CV
Q= (0.25 * 10^-6) ( 50)
Q = 0.0000125 C

Your Q means "charge" not energy

formula for energy stored in a capacitor is given in most books or can easily googled...
$$\displaymath{U = \frac{1}{2}CV^2}$$