Calculating Energy (capcitor study)

[Alphonse]

Study guide asked/stated: A 100 microF capacitor has 100 V across it terminals, A 100 ohm
resistor is placed across these terminals. How much energy is lost when the voltage is dropped to 50V?

I answered with w= 1/2 CV^2, 1/2*100*10E-6*100^2
= .5 J
Book says w = 25 * 10-6 * 10^4.
= .25 J

(next step is to calculate energy at 50V). Where did the 25 come from?

Hope this is the right forum (and not already asked and answered).

 

[berkeman]

Study guide asked/stated: A 100 microF capacitor has 100 V across it terminals, A 100 ohm
resistor is placed across these terminals. How much energy is lost when the voltage is dropped to 50V?

I answered with w= 1/2 CV^2, 1/2*100*10E-6*100^2
= .5 J
Book says w = 25 * 10-6 * 10^4.
= .25 J

(next step is to calculate energy at 50V). Where did the 25 come from?

Hope this is the right forum (and not already asked and answered).

The delta energy should be a subtraction... final E versus initial E. Does that help?

 

[Alphonse]

Beckeman, thanks for your reply. I agree the final answer is computed from the difference
of the two. My problem is in initial computation of energy.

Alphonse

 

[berkeman]

Study guide asked/stated: A 100 microF capacitor has 100 V across it terminals, A 100 ohm
resistor is placed across these terminals. How much energy is lost when the voltage is dropped to 50V?

I answered with w= 1/2 CV^2, 1/2*100*10E-6*100^2
= .5 J
Book says w = 25 * 10-6 * 10^4.
= .25 J

(next step is to calculate energy at 50V). Where did the 25 come from?

Hope this is the right forum (and not already asked and answered).

I agree that the book's answer appears not to be what the question is asking. Any chance you copied the question down wrong above? Your calculation looks correct to me, and the number you will get at half voltage does not seem to be related to the book's number. Maybe ask the prof?