Calculating expended energy in a battery

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SUMMARY

The discussion focuses on calculating the energy expended by a 42 V battery to charge two capacitors, with capacitances of 3.62 µF and 6.11 µF, connected in parallel. The total capacitance is determined by summing the individual capacitances, resulting in a total of 9.73 µF. The energy stored in the capacitors can be calculated using the formula U = 0.5 * C * V², where C is the total capacitance and V is the voltage. The correct application of this formula will yield the energy required from the battery.

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  • Understanding of capacitor theory and parallel circuits
  • Familiarity with the formula for energy stored in a capacitor (U = 0.5 * C * V²)
  • Basic knowledge of electrical units (voltage, capacitance, energy)
  • Ability to perform calculations involving scientific notation
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Homework Statement


How much energy must a 42 V battery expend to fully charge a 3.62x10-6 F and a 6.11x10-6 F capacitor when they are placed in parallel?


Homework Equations



capacitor in parallel therefore total capacitance is the sum...

The Attempt at a Solution


I thought that you could use U=Vq but apparently you can't because I was getting the wrong answers.. this is the same question that I tried to do before but with different values. HELP :)
 
Physics news on Phys.org
When in parallel, what is the total capacitance?

How do you find the energy stored in a capacitor?
 
Thank you so much!
 

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