Total Energy of Camera Capacitor

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SUMMARY

The discussion focuses on calculating the total energy produced by a camera flash capacitor powered by a 300V battery, which generates 5000W of light for 0.005 seconds. The total energy produced is determined using the formula for power, resulting in 25 joules. Subsequently, the capacitance can be calculated using the potential energy equation U=(1/2)CV^2. If the battery voltage is reduced to 120V, the maximum power output of the flash attachment can be recalculated based on the new voltage.

PREREQUISITES
  • Understanding of electrical energy concepts, specifically potential energy (PE = qV)
  • Familiarity with capacitance calculations (C = q/V)
  • Knowledge of power and energy relationships (Power = Energy/Time)
  • Proficiency in using the energy stored in a capacitor formula (U = (1/2)CV^2)
NEXT STEPS
  • Calculate the total energy produced by a capacitor with different voltage inputs
  • Explore the relationship between capacitance and energy storage in capacitors
  • Investigate the effects of varying voltage on power output in electrical circuits
  • Learn about the design and specifications of camera flash systems
USEFUL FOR

Electrical engineering students, photographers interested in flash technology, and anyone involved in designing or optimizing capacitor-based energy storage systems.

JumpinJohny
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Homework Statement


A flash attatchment for a professional camera stores energy in a capactior. When a picture is taken, all of the charge is converted to energy, and the capacitor is fully discharged.

A.Assume the battery charging the capacitor is a 300V battery. When the light flashes, it produces 5000W of light for a time of .005s. Find the total energy produced by the flash.

B.Find the capacitance.

C.If the battery were replaced with a battery with a potential difference of 120V, how much light power could the flash attachment produce?


Homework Equations



C= q/V
PE = qV
U=(1/2)CV^2

The Attempt at a Solution


I really don't know. At first I thought I'd just solve for capacitance(which you can't anyway), until I looked at part B. I'm assuming you need to solve for the potential energy of the flash, but I'm not sure how to go about doing that.
 
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I am pretty sure in that in solving for the total energy, you just need to find the potential energy. Electric potential energy would be PE=qV. There is no charge stated in the problem, so I'm not sure where to go from there.
 
For A) Power is a measure of Energy per Time. Think about what a Watt is and how it relates to the time given, 0.005s.

For B) Once you've answered part A, it should be easy to solve for Capacitance using the equation for U.
 

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