How much energy is dissipated by the 25 ohm resistor?

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SUMMARY

The energy dissipated by a 25 ohm resistor in a circuit with a 0.25 µF capacitor charged to 50 V is calculated using the formula for energy stored in a capacitor, which is (1/2)CV². Initially, the energy stored is (1/2)(0.25 x 10^-6 F)(50 V)², resulting in 0.000625 Joules. Upon complete discharge, all this energy is dissipated through the resistors in the circuit, confirming that the 25 ohm resistor dissipates the same amount of energy as the capacitor initially stored.

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  • Understanding of capacitor charging and discharging principles
  • Familiarity with Ohm's Law and power equations (P=IV, P=I²R, P=V²/R)
  • Basic knowledge of series circuits and energy conservation
  • Ability to perform calculations involving capacitance and voltage
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  • Study the derivation and application of the energy stored in a capacitor formula: E = (1/2)CV²
  • Learn about series and parallel resistor combinations and their impact on total resistance
  • Explore the concept of power dissipation in resistors and its implications in circuit design
  • Investigate practical applications of capacitors in electronic circuits
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iceman887
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Homework Statement



A 0.25 UF capacitor is charged to 50 V. It is then connected in series with a 25 ohm resistor and a 100 ohm resistor and allowed to discharge completely.

Homework Equations


P=I V=I^2R=(V^2)/R


The Attempt at a Solution


i'm not sure how to approach this problem. it hink there is formula i need that missing for calculating energy
 
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I can't remember exactly how it works, but if you go to allaboutcircuits.com you will find every bit of info for this type of problem. We actually used that site as the textbook when I took Electronics and Circuit Theory courses
 
The energy stored in a capacitor is (1/2)CV^2. Initially, V=50 Volts. After the capacitor is completely discharged, the potential across it is clearly zero. so how much energy must have been dissipated?
 

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