Finding kilowatt hours from parallel capacitors

AI Thread Summary
To calculate the cost of charging a bank of 2900 parallel-connected capacitors (5.00 µF each) to 45,000 V, one must first determine the energy stored using the formula E = 1/2 CV², which results in energy measured in joules. The conversion from joules to kilowatt hours is done using the factor 1 Joule = 2.77E-7 kWh. The total charge for one capacitor is calculated as Q = CV, resulting in 0.225 C for a single capacitor. After finding the total energy for all capacitors, the cost can be calculated by multiplying the total kWh by the unit cost of $0.03. Understanding the relationship between voltage, capacitance, and energy is crucial for solving this problem.
CIERAcyanide
Messages
9
Reaction score
0
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!
 
Physics news on Phys.org


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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top