# Circuit formula problem

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.   Related Advanced Physics Homework Help News on Phys.org
mjsd
Homework Helper
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 $$P=I^2 R$$

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
Homework Helper
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....
$$\displaymath{U = \frac{1}{2}CV^2}$$