is capacitor linear devics?
965616
yes it is a linear device.\nsimilar to it resistor & inductor r also linear devices.
475454
he discovered the capacitor and defined capacitance.\nhe gave the formula.\nCHARGE(Q)=C(capactiance)* voltage(v)\nthe unit for measuring capacitance of a capacitor is FARAD.\ncapacitor is a linear component that stores charge.\nit consist of two metal plates separated by a dielectric material eg->glass,mica,fibre..etc,\n\ncapacitance of a capacitor is 1 farad if it can store a charge of  1 coloumb when i volt of current flows through it.
1035397
Polarized capacitors are useful for single direction systems. The gate only opens one way and the lead is always singular in its transfer of power. Because of this the power is at full when it crosses the board and the Resistance is must be manipulated to make sure the power does not overload the system. As an example, most of the older laser systems worked off a polarized capacitor, when activated the capacitor discharged and all the power drained to a single pole. Thus the laser would fire in one burst until the entire capacitor was discharged. \n\nNon-polarized or dual gated systems are omni directional and therefore omni systemic. Meaning the power flow can be multi directional, however because of the multi directional flow the power is decreased more quickly and dispersed over a wider area. Thus it cannot be as useful due to resistance across the board. However non-polarized capacitors work well within the atmosphere of amplifiers, and other more commercially accepted applications. However remember that all capacitors take time to recharge, and must be fully discharged before they can be recharged. Damage could occur when attempting to ram charge any kind of capacitor.
129691
See my answer to your next question.  For phase lead, C is in series and R in shunt.
667723
Debye materials can be thought of as a simple RC circuit where the amount of polarization is re-\nlated to the voltage across the capacitor. The “source” driving the circuit is the electric field. For\na step in source, there may be a constant polarization. As the frequency goes to infinity, the polar-\nization goes to zero (leaving just the constant residual high-frequency term).
827005
It's linear alright. x = - 3/7. A little tricky because it's a line parallel to the y axis.
159857
this si not a physics question. this is a basic IC that perhaps still and certainly at one time was available at radio shack. the ic 741 behaves as ANY op amp and it is electronics 101.......happy homeworking.
282019
yes, using a static capacitor
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
93256
Try this website. Scroll down to "The Circuit" ---\n\nhttp://www.play-hookey.com/ac_theory/ac_rc_series.html
960408
http://www.eas.asu.edu/~holbert/ece201/kvl.html
792406
let the initial charge on the capasitor be q. Then for charging the equation is given by:\nQ=q(e)^(-t/T);  where e is base of natural log. 't' is the time and 'T' is time constant which is probably 1/(RC)\n\nfor discharging\nQ=q(1-(e)^(-t/T));\n\nnow the graph can be easily plotted by taking Q on 'y' axis and 'q' on x-axis.
1178298
Read this: http://en.wikipedia.org/wiki/Linear_equations
947723
Fringe capacitance is probably the capacitance due to the stray electric fields at the edges of a capacitor.  Think of the parallel plate capacitor, where the electric field lines in the center are straight.  The edges of the plates have fringing fields.  The capacitance due to the field at the edges is the fringe capacitance.  In the case of interconnect capacitance, the metal lines on the chip will cross over each other and form parallel plate capacitors, with edges.
745241
in a CE amplifier, when base voltage increase, base current increases. It causes an increase in collector current also. The collector current causes a voltage drop in the collector resistor. Because the output is situated below the collector resistance (with reference to Vcc) the output voltage will decrease as voltage drop across collector resistor increase. Thus it produces a 180 phase shift. (input positive, output negative and vice versa)\n\nYou can verify with this equation: Vcc=IcRc+Vc  (Vc=Vout)\nWhen  Ic increase, Vc must decrease because Vcc is constant.\n\nYou can analyze CB and CC circuits using the same manner by writing the voltage drop equations along a line or a loop. Then find the relation between input voltage and output voltage. You will see that there is no phase shift. (Input increase, output also increase)
245581
Energy of a capacitor = (CV^2)/2\nC = capacitance in farads\nV = voltage in volts\nenergy in joules of course
1050520
Think about the simpliest capacitor. It is clear that two metal plates that don't touch won't let a DC current pass, right? But the spacing is very thin so that the plate's charges can effect one another. When you first put current (even DC) to a discharged capacitor, it charges up and for a brief time it does conduct current. How long? Well it will charge to 2/3 of the total voltage in RC seconds (where R is the resistance of the circuit). So when AC current is "fast" enough, the cap is constantly in a charge/discharge state and current flows. The effective resistance (reactance) of the cap is 1/(2*pi*F*C). So the resistance goes down as the capacitance and the frequency rises.\n\nThe inductor, on the other hand is just a coil of wire. Again, it is easy to imagine that it will conduct DC. But when you first apply current (even DC) to a coil, the magnetic field will interact with the coil and produce an opposing voltage. This only occurs when the magnetic field is forming or collapsing (remember, magnetic fields that do not move do not produce electricity). So right when you change the current, the inductor will "resist" the change, but if the current doesn't change, it is just a wire. The reactance of a coil is 2*pi*F*L, so the resistance increases with frequency and inductance.\n\nNow, combining inductors and capacitors is another interesting subject, but you didn't ask about that....