the formula for energy stored in a capacitor is (1/2)*cv^2
given v=20v,
c =10*10^-6 f
by using this information you can easily calculate the energy
A capacitor that is suddenly connected to a battery will charge to the battery voltage. The time to do this is dependent on the current capacity of the battery and wiring, and the capacitance of the capacitor. This represents an instantaneous short circuit, which lasts for a (usually) very short time - but damage could be done if there was no resistance. A charged capacitor that is suddenly disconnected from a battery will hold that voltage. The length of time it will hold is dependent on how much leakage current there is.
The units of capacitance are called farads. A one farad capacitor is a capacitor with 1 volt potential difference with 1 coulomb of charge on the capacitor, C = Q/V or Q=CV So the charge held on your capacitor is Q = CV = 9Volts * 0.40*10-6Farads=3.6*10-6 Coulombs
You COULD use a huge capacitor, like a bank of 1-farad car stereo caps, to run a motor, because a cap is basically a battery that doesn't hold very much electricity. And if you did it, you'd control the voltage to the motor the same way you would with a battery--resistors, potentiometers, voltage regulators, whatever you like.I wouldn't do it, though, because a capacitor doesn't hold enough electricity to let the motor run for very long.Quick note: single phase AC motors all have "run capacitors" and "start capacitors" but those are something else entirely.CommentI don't think that you will ever find a 1-F (one farad) capacitor! A farad is a HUGE amount of capacitance! Practical capacitors are expressed in picofarads (10-12 farad) or microfarads (10-6 farad), never in farads! Bear in mind that capacitance is directly proportional to the area of its plates -so imagine how huge a 1-F capacitor would have to be!Running a motor from a capacitor is completely impractical. A capacitor will completely discharge in 5 CR, where Cis the capacitance of the capacitor and R is the resistance connected across its plates.So, let's pretend from one moment that you can find a 1-F capacitor and let's say the motor has a armature resistance of, say, 50 ohms (quite high!), then the capacitor would completely discharge in 5 x 1 x 50 = 250 s, or a little over 4 minutes!!!! However, if we used a capacitor with a realistic value of capacitance (in microfarads), then the discharge time would be measured in milliseconds! So, not much use for driving a motor!!!
The charge in a capacitor is between the plates. The dielectric is only an insulator that allows the plates to be very close without touching and discharging the charge. There is no battery in a capacitor.
In a battery or capacitor.
A capacitor can't deliver electrical energy 'constantly'. Much like a battery, electrical energy can be stored in it, and then delivered later. The capacitor only stores some definite amount of energy ... equal to 1/2 of its capacitance multiplied by the square of the voltage to which it's charged ... and later, after it has delivered that amount of energy, it delivers no more without being charged again.
A capacitor that is suddenly connected to a battery will charge to the battery voltage. The time to do this is dependent on the current capacity of the battery and wiring, and the capacitance of the capacitor. This represents an instantaneous short circuit, which lasts for a (usually) very short time - but damage could be done if there was no resistance. A charged capacitor that is suddenly disconnected from a battery will hold that voltage. The length of time it will hold is dependent on how much leakage current there is.
A capacitor could be two parallel plates close together but unconnected, and then the plates are connected to either side of a battery. Current flowing causes charge to build up on the plates, positive on one plate and negative on the other, until the voltage across the capacitor is equal to the battery voltage. The amount of charge that has flowed in, divided by the voltage, is called the capacitance, measured in Farads.If the battery was one volt, and the charge was one coulomb (i.e. one amp for one second), then the capacitor has a capacitance of one Farad.Usually capacitors are measured in microfarads or picofarads.
The simplest capacitor is just two parallel metal plates, not touching. When a battery is connected across the plates, the plates become charged, with electric charges sitting facing each other, positive ones on the positive plate and negative on the negative. When the battery is removed, the charges stay where they are so the capacitor is a way to store electric charge and energy, a bit like a rechargeable battery. Supposing the battery was 1 v and the charge is +1 coulomb on one plate and -1 coulomb on the other. That means the capacitor has a capacitance of 1 Farad. The amount of charge a capacitor can store is given by the formula Q = CxV in other words the charge is the capacitance times the voltage. So a large capacitance can store more charge for the same voltage. With the 2-plate capacitor the capacitance increases if the plates are bigger and also if they are closer together. Larger capacitance can be produced by using two sets of interleaved plates. Each set has all its plates connected together, and there is dielectric insulation between all the plates.
The capacitor will hold the charge, until it leaks off due to resistances in the dielectric or external.
The units of capacitance are called farads. A one farad capacitor is a capacitor with 1 volt potential difference with 1 coulomb of charge on the capacitor, C = Q/V or Q=CV So the charge held on your capacitor is Q = CV = 9Volts * 0.40*10-6Farads=3.6*10-6 Coulombs
The time constant is equivalent to 1/(R*C); since C (the capacitance of the capacitor) is not changing, yes, the charging and discharging times will be the same, provided the Thevenin resistance is the same as well - if you charge a capacitor using a AA battery, then remove the battery, and discharge through a resistor, you have changed the Thevenin resistance, thus the discharge time will NOT be equal.
Usually a tiny fraction of a second. Actually it will depend on the characteristics of the the capacitor, and of the remaining circuit (mainly, any resistor in series). The "time constant" of a capacitor with a resistor in series to charge from 0 to a fraction of (1 - 1/e), about 68%, of its final value. This time is the product of the resistance and the capacitance. After about 5 time constant, you can consider the capacitor completely loaded for all practical purposes - i.e., it will be at the same voltage as the battery.
If the capacitor is charged then the battery will explode.
LC means a circuit contain capacitor(C) and inductor(L). They are connected in series. L is inductance, units is Henry(H). C is capacitance, units is Faraday(F) don't have battery.
A car battery does not need a capacitor as it's regulating capacity (when in good condition) is well above any capacitor you can fit into your car.
when a capacitor reaches it, it acts as a battery