Finding kilowatt hours from parallel capacitors

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SUMMARY

The discussion focuses on calculating the energy stored in a bank of 2900 parallel-connected capacitors, each with a capacitance of 5.00 µF, charged to 45,000 V. The relevant formulas include E = 1/2 CV² and E = Q²/2C, where E represents energy in joules. The conversion from joules to kilowatt hours is established using the factor 1 Joule = 2.77E-7 kWh. The final cost of charging the capacitors is determined by multiplying the total energy in kWh by the unit cost of $0.03 per kW·h.

PREREQUISITES
  • Understanding of capacitance and the formula E = 1/2 CV²
  • Knowledge of energy conversion from joules to kilowatt hours
  • Familiarity with the concept of charge (Coulombs) in capacitors
  • Basic principles of parallel circuits in electrical engineering
NEXT STEPS
  • Research the detailed application of the formula E = 1/2 CV² for energy calculations in capacitors
  • Learn about the conversion process from joules to kilowatt hours in electrical energy calculations
  • Explore the implications of connecting capacitors in parallel versus series
  • Investigate the cost analysis of charging capacitors based on energy consumption
USEFUL FOR

Electrical engineers, students studying circuit theory, and anyone involved in energy storage and cost analysis of capacitors.

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Please Help! Finding kilowatt hours from parallel capacitors

Homework Statement



A parallel-connected bank of 5.00 µF capacitors is used to store electric energy. What does it cost to charge the 2900 capacitors of the bank to 45,000 V, assuming a unit cost of $0.03 per kW·h?


Homework Equations



CV = Q
1 Joule = 2.77E-7 kWh
1 Volt = 1 Joule per Coulomb

The Attempt at a Solution


I'm not sure how to convert from Volts to Joules and then to kilowatt hours. I tried CV=Q to find the Coulombs, from where I wanted to find the number of Joules. If my thinking is correct, the 45,000 Volts means there are 45,000 Joules per Coulomb. When i multiplied 5E-6 times 45,000 to find Q my result was that the charge was only .225 C for one capacitor, which seems low. I'm not sure where to go from here, and I'm very unsure of how to convert from Volts to Joules to Kilowatt hours. I know at the end I need to remember to multiply my final result by 2900 for the number of capacitors. I also don't know how the fact they're connected by parallel plays into this.

Please help!
 
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Are you familiar with the formula E=1/2CV^2 or E=Q^2/2C? Those are the formulas for the energy stored in a capacitor.
 


Okay, i'll try that and see if i can work out the right answer
 
Last edited:


The units for E is in joules correct?
 


That depends on what you use for C, V, and Q. If you use standard SI units, E will come out in joules.
 

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