Calculating Energy Stored and Dissipated in a Capacitor Discharge

  • Thread starter Thread starter NkaujHelp
  • Start date Start date
  • Tags Tags
    Capacitor Energy
Click For Summary
SUMMARY

The discussion focuses on calculating the energy stored in a capacitor and the energy dissipated by a light bulb during discharge. A capacitor with a capacitance of 1.25 F was charged with a 5 V DC power source and connected to a 20-ohm light bulb. The initial energy stored in the capacitor was calculated to be 15.6 J, while the energy dissipated by the bulb over 25 seconds was calculated as 31.25 J. The discrepancy in energy values raised questions about the assumption of constant power during discharge.

PREREQUISITES
  • Understanding of capacitor discharge in RC circuits
  • Familiarity with Ohm's Law (V=IR)
  • Knowledge of energy equations for capacitors (U=(1/2)CV^2)
  • Ability to calculate power dissipation (P=I^2R)
NEXT STEPS
  • Study the equations governing capacitor discharge in RC circuits
  • Learn about energy conservation in electrical systems
  • Explore the effects of non-ideal components in capacitor circuits
  • Investigate the relationship between current decay and time in capacitor discharges
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone interested in understanding capacitor behavior in circuits.

NkaujHelp
Messages
16
Reaction score
0

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.
 
Physics news on Phys.org
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?
 
  • Like
Likes   Reactions: CWatters

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
14K
Replies
6
Views
1K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K