# Power dissipated in capacitor

1. Oct 12, 2007

### indigojoker

a charged capacitor has capacitance C and charge Q. a resistor R is then connected what is the power dissipated right after the connection?

V=Q/C
so P=V^2/R=Q^2/(RC^2)
is this right?

what is the power dissipated in the resistor when the energy stored in the capacitor has decreased to half the initial value?

Using P=V^2/R, we need to find V when U=1/2U_o

This means, U=(1/4)Q^2/C

I'm not actually sure of how to get the voltage, to get the power dissipated. any ideas?

2. Oct 12, 2007

### Staff: Mentor

The first part looks correct, but remember that the power is dissipated in the resistor, not in the capacitor. The energy comes from the stored charge in the capacitor.

The equation that you need to add for the 2nd part is the energy stored on the capacitor, E = 1/2 CV^2. Does that help?

3. Oct 12, 2007

### indigojoker

do you mean:

(1/4)Q^2/C=1/2 CV^2

and then solve for V?

4. Oct 15, 2007

### indigojoker

i'm not sure how he stored energy in a capacitor helps on the second part.

5. Oct 15, 2007

### Staff: Mentor

When you charge up a capacitor, you are storing energy in the charge separation, and the amount of energy is related to the capacitor voltage.

6. Oct 15, 2007

### indigojoker

so the power dissipated in the resistor when the energy stored in the capacitor has decreased to half the initial value is just 1/2(1/2 CV^2)?

7. Oct 15, 2007

### Staff: Mentor

Just subtract the final energy from the initial energy, and divide that by the time it takes for the energy change to happen. Can you show us that equation?

8. Oct 18, 2007

### indigojoker

well, i am looknig for the the power dissipated in the resistor at the instant when the energy stored int he capacitor has decreased to half the initial value.

I was thinking something like this.

U_o=0.5 Q^2/C and we know that since C stays the same, then when the energy is half of original energy, we get Q-> Q/Sqrt(2) to get 0.5U_o

So if Q-> Q/Sqrt(2), then plugging into C=Q/V, we get that V=> V/Sqrt(2) in order to get the capacitance to not change.

thus, we get that P=V^2/R --> P=1/2 (V^2/R) when using the new V

ideas?