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Circuit formula problem

  • Thread starter pari786
  • Start date
29
0
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.:confused: :confused: :confused:
 

Answers and Replies

mjsd
Homework Helper
725
3
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 [tex]P=I^2 R[/tex]
 
24
0
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 [tex]P=I^2 R[/tex]
Hi, Do you calculate the energy by this formula;

Q=CV
Q= (0.25 * 10^-6) ( 50)
Q = 0.0000125 C
 
mjsd
Homework Helper
725
3
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....
[tex]\displaymath{U = \frac{1}{2}CV^2}[/tex]
 
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