Calculating Energy Stored and Dissipated in a Capacitor Discharge

In summary, the conversation discusses proving the equality between the initial energy stored in the capacitor and the eventual energy dissipated by the light bulb in a DC circuit. The capacitance, voltage, and resistance values are given and used to calculate the initial energy stored in the capacitor and the power dissipated by the bulb. The discharge time of the capacitor is also mentioned, with a question about the assumption of constant power during discharging. The conversation concludes with a request for help in calculating the discharge time and a reminder to only focus on the ideal case.
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NkaujHelp
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Homework Statement


Prove that the total energy initially stored in the electric field inside the capacitor is equal to the total electrical energy eventually dissipated by the bulb. A 5 V DC power source charged the capacitor. The capacitor was then connected in series with a light bulb with a resistance of 20 ohms and an ideal ammeter. When connected, the ideal ammeter read the current as 20 A. Twenty seconds later, the ammeter read the current as 0.05 A.
The capacitance was calculated to be 1.25 F.

Homework Equations


V=IR
C=Q/V
P=IdealtaV = (I^2)(R)
U=(1/2)CV^2

The Attempt at a Solution


By using U=(1/2)CV^2 and plugging in all the values given above, the total energy initially stored in the electric field inside the capacitor came to be 15.6 J. The power dissipated as heat by the bulb came out to be P=RI^2=(.25A)^2(20 ohms)=1.25 Watts. Converting watts into joules or power into energy, I took 1.25 Watts*25 seconds = 31.25J. I followed and checked the units to be correct and the units came out right. The 25 second is the number when calculated turns out to be the length of time the capacitor completely discharges and the light bulb goes out. So this is only an ideal case. I really don't need any information on non-ideal cases about this. It's just going to confuse me. But I just need help on if my values came out to be right. The energy dissipated as heat by the bulb is way more than the energy provided by the capacitor. I don't see how that is possible.
 
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  • #2
NkaujHelp said:
The power dissipated as heat by the bulb came out to be P=RI^2=(.25A)^2(20 ohms)=1.25 Watts. Converting watts into joules or power into energy, I took 1.25 Watts*25 seconds = 31.25J.
Is it correct to assume that the power of 1.25 W remains constant during the discharging of the capacitor?
The 25 second is the number when calculated turns out to be the length of time the capacitor completely discharges and the light bulb goes out.
Please show how you calculated 25 s for the discharge time. Have you studied the equations for describing the discharge of a capacitor in an RC circuit?
 
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1. What is the definition of energy stored in a capacitor?

The energy stored in a capacitor is the potential energy that is stored in the electric field between the capacitor's plates when a voltage is applied to it.

2. How is the energy stored in a capacitor calculated?

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 applied to the capacitor.

3. What factors affect the energy stored in a capacitor?

The energy stored in a capacitor is affected by the capacitance, which is determined by the distance between the plates and the dielectric material used, as well as the voltage applied to the capacitor.

4. What happens to the energy stored in a capacitor when it is connected to a circuit?

When a capacitor is connected to a circuit, the energy stored in it is released and used to power the circuit. As the capacitor discharges, the voltage across it decreases and the energy is converted into other forms, such as heat or electromagnetic radiation.

5. Can the energy stored in a capacitor be increased?

Yes, the energy stored in a capacitor can be increased by either increasing the capacitance or the voltage applied to it. However, there are limits to how much energy a capacitor can store, as it can only hold a certain amount of charge before it reaches its breakdown voltage and can no longer store any more energy.

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