Calculating Energy (capcitor study)

In summary, the conversation was about calculating the energy lost when the voltage across a 100 microF capacitor is dropped from 100V to 50V with a 100 ohm resistor in place. One person used the formula w= 1/2 CV^2 to get a result of .5 J, while the book's answer was given as w = 25 * 10-6 * 10^4, equaling .25 J. The discrepancy between the two answers was discussed and it was suggested to ask the professor for clarification.
  • #1
Alphonse
2
0
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).
 
Physics news on Phys.org
  • #2
Alphonse said:
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?
 
  • #3
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
 
  • #4
Alphonse said:
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?
 

FAQ: Calculating Energy (capcitor study)

What is energy?

Energy is the ability to do work. It is a physical quantity that is measured in joules (J).

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It is made up of two conductive plates separated by an insulating material, known as a dielectric.

How do you calculate the energy stored in a capacitor?

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage across the capacitor.

What factors affect the energy stored in a capacitor?

The energy stored in a capacitor is affected by its capacitance and the voltage applied to it. The higher the capacitance and voltage, the more energy the capacitor can store. The type of dielectric material used in the capacitor can also affect its energy storage capacity.

How is the energy stored in a capacitor used in practical applications?

The energy stored in a capacitor can be used to power electronic devices, such as cameras, flashes, and electronic flashlights. It can also be used in energy storage systems, such as in electric cars and renewable energy sources like solar panels. In addition, capacitors are used in many electronic circuits for filtering and smoothing out voltage fluctuations.

Similar threads

Replies
2
Views
2K
Replies
5
Views
4K
Replies
2
Views
8K
Replies
7
Views
2K
Replies
10
Views
8K
Back
Top