energy of capacitors?
606031
It's O.K. to stipulate there's no resistance - but not that the system can reach a static equilibrium state with no energy dissipation at all. The two assumptions are incompatible.\n\nIn the absence of resistance, the inductance, which can never be exactly zero even for a straight wire, would limit the current and also cause its direction to oscillate back and forth between the two capacitors until exactly 1/2 the original energy is lost through EM radiation. \n\nIf somehow there were no radiation, the system would keep oscillating forever with 1/2 the original energy being held in the magnetic field at the moment in each cycle when the capacitors were each exactly half charged. The rest of the time the capacitors would hold between 50% and 100% of the original energy.
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
697895
E = mc^2 = 1kg * (3*10^8)^2 = 9^10^16 Joules
365027
The capacity of a capacitor is measured in farads (F). Ususally they range from several picofarads (pF) to several microfarads (µF). For the measurement, look on the capacitor itself or if it is a disk capacitor then there is a system for figuring that out (look it up).
855082
This from http://en.wikipedia.org/wiki/Dissipation_factor\n\n"In capacitors, dissipation factor (DF) is the ratio of a capacitor's resistance (R) to its capacitive reactance (Xc). DF is usually expressed as a percentage."\n\nThe capacitor's resistance R as computed from the DF is the ESR, as it includes losses other that those due to the conductivity of the dielctric.
885785
I don't know what methods you are using in class, but here is the way I'd start:\nProblem 7.63:\nAt t<0, you have a current source feeding a resistor network.  You can calculate the initial voltage on the capacitor.\nAt t=0, the circuit is now a capacitor (with an initial voltage) discharging through 40 k-ohms in parallel with 10 k-ohms.  You can calculate the voltage after 400 micro-sec.\nAt t=400 micro-sec, the capacitor is now discharging through 40 k-ohms.  You know the voltage at the start of this stage, so you can calculate the voltage 1 ms later.\nAt t=1.400 ms, the capacitor is again discharging through a parallel resistor network.  Knowing the voltage on the capacitor at  the begining of this stage, you can calculate the voltage on the capacitor 0.2 ms later.\nKnowing the voltage on the capacitor at t = 1.600 ms, Vo can be found.  The 5 k-ohm and the 35 k-ohm resistors form a divider network.\n\nThere you have it, just look at the circuit at the various stages and calculate the voltage on the energy storage element.
329547
If there are 3 capacitors, Ca, Cb, Cc then use the following:\n\nSeries\nTotal Capacitance = 1 / (1/Ca + 1/Cb + 1/Cc)\n\nParalel\nTotal Capacitance = Ca + Cb + Cc
906536
These are a published value.  The actual voltages vary as you can see from this quote from a power plant site \n\n"POWER PLANT CAPABILITY: The two hydroelectric power generators boast a generating capacity of 92,000 kilowatts. The estimated annual energy output is 205 million kilowatt-hours."\n\nAnd none of the values quoted are a multiple of 11.  It appears that the source your using just happened to have values that were multiples of 11.\n\nA quick look on Google listed quite a number of plants that produced values that had 13800kv to 205 million kv.\n\nChance that the ones you looked at were multiples.\n\nBest of luck in your project.
1015451
Power = work done per unit of time.\n\nP=W/T
796281
That all depends on what the application is. If you put 50 volts on the 40 volt capacitor it could potentially explode. Put if you only have 40 volts on it no difference.
772410
electrical energy-------> heat + light\ndoes that help?
903593
Crystal Energy et Active H\n\nSouriez à la vie et la vie vous sourira. ^---^
349042
It depends on the circuit you ar designing and again the part of the circuit. \n\nFor example if it is a power supply smoothing capacitor it would be a large value in micro farrads. \n\nBut if you want a capacitor for removing high frequency spikes, then it would need a small capacitor in pico or nano farrads.\n\nYou have to work with the formulae for the part of the circuit you are dealing with, and there are loads of formulas.
466996
approx. 1600000000Kw hrs of electricity per hour you kept it on\n\nthat would cost a lot of money by the way and you might blow a fuse
1131325
Maybe this can help...all you have to do is a search in Yahoo or Google...\n\nhttp://www.think-energy.com/ThinkEnergy/11-14/activities/EnergyTrans.aspx\n\nTry searching for "energy" or "kinetic energy"
348003
Have a look at http://science.howstuffworks.com/power.htm
90631
energy because power wich is in mw multipiled by time wich is actualy in seconds or hour equals energy wicj is in joules