HW: What is energy used? What is wattage?

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SUMMARY

The discussion centers on calculating the energy used by a camera flash capacitor rated at 850uF with a potential difference of 280V. The correct formula for energy stored in a capacitor is E = 1/2 * C * V^2, resulting in an energy of 33J. The power or wattage, calculated using the formula P = E/t, yields a value of 8500W when the flash duration is 3.9E-9 seconds. The initial miscalculation arose from incorrectly using charge instead of energy in the power calculation.

PREREQUISITES
  • Understanding of capacitor ratings and units (microfarads, volts)
  • Familiarity with the formula for energy in a capacitor (E = 1/2 * C * V^2)
  • Knowledge of power calculation (P = E/t)
  • Basic concepts of electric charge (Q = C * V)
NEXT STEPS
  • Study the derivation and applications of the energy formula for capacitors (E = 1/2 * C * V^2)
  • Learn about the relationship between charge, voltage, and capacitance in electrical circuits
  • Explore practical applications of capacitors in flash photography and other electronic devices
  • Investigate advanced power calculations in electrical engineering, including transient analysis
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Students in physics or electrical engineering, electronics hobbyists, and anyone interested in understanding the principles of energy storage and power calculations in capacitors.

JerseyGirl
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Cutnell Physics 7th Ed pg 600 # 39

Homework Statement



A camera flash has a capacitor.
Capacitor has a rating of: 850uF
The potential difference between the places is: 280V

a) Determine Energy used to produce flash.
b)If flash lasts 3.9E-9 S, what is the power or wattage of the flash?

Homework Equations



V=C/F
W=VA
A=C/S

The Attempt at a Solution



a) (280V)(850uF)= 0.238C

b) (0.238 C)/(3.9E-3 S)= 61.03A
W=?

The back of the book has a) =33J b) 8500W
Where did I go wrong?

Thanks in advance.
 
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Energy = 1/2*C*V^2
Power = E/t
 

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