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So, whenever you excite a discharged capacitor with a signal that "starts up", i.e. is zero before a given instant of time, the capacitor is subjected to a transient response that dies off after some time … EDIT BTW, to answer explicitly the title question, the basic relationship describing a capacitor is this: $$ q(t) = C \cdot v(t) $$ where \$q\$ is the charge "stored" in the capacitor, \$C\$ is the capacitance …

If you have just one big capacitor, once it fails, you are left with a nonworking system. Moreover, a bigger cap can do more damage to nearby components if it fails spectacularly. … A single, big capacitor can be stressed mechanically more heavily when subjected to vibrations. …

The plates of the capacitor are metallic, hence most of the electrons of the outer shells of the atoms are in conduction band, not in the valence band. … Actually after 3-5 time constants you may consider the capacitor charged. …

Note the symbols on that schematic: both Cin1 and Cin2 are electrolytic caps. So, even for high voltage ones, you can get them at reasonable price (10uF-100uF, 400V electrolytic caps are fairly common …

The "trick" is to remember that if you define reactances as: \[ X_L = \omega L \qquad X_C = \frac{1}{\omega C} \] then the impedance for an inductor and a capacitor are: \[ Z_L = j X_L = j \omega L …

If by "trickle charge" you mean "leaving the cell connected to a charge that in a form of another provides the cell with some power (e.g. a small constant current)" the answer is: ABSOLUTELY NOT! You …

During high current periods the capacitor will act as the primary power source, while during low current periods the battery will be the primary power source and recharge the capacitor. … When dimensioning the capacitor it is important to know the battery’s internal resistance and the load profile. With this information it’s quite simple to dimension a suitable capacitor. …

Just the outline of the solution: By KCL the currents of the two current sources add up at the upper node and go into the capacitor. … This will give you \$v_c(6\mu s)\$, assuming the capacitor is discharged at \$t=0\$ (i.e., assuming \$v_c(0)=0\$), otherwise the problem is not well specified. …

You can double-check the design of that power supply using the information in the Texas Instruments LM317 datasheet. At page 10 you find this: which is followed by a detailed design procedure. Th …

That's not a protection device, if you intend "protection" as something that prevents the chip to die, but it serves as a so-called bypass capacitor. … The bypass capacitor reduces this risk in a way similar a filter capacitor after a bridge rectifier reduces the ripple of a rectified voltage. …

It is a matter of history, sadly. Capacitors where built well before the International System of measurement (SI) was established and some prefixes weren't used so much (e.g. nano). For example, usua …

To give further insight: think of a single capacitor as a tank circuit, where the capacitor may resonate with its parasitic inductance. … Assuming a simple model where a capacitor has a single resonant frequency, paralleling different caps, with different resonant frequencies, allows us to "combine" the different frequencies responses so …

That situation assumes an ideal capacitor and an ideal inductor, which are only mathematical abstractions. Any real component suffers from some parasitic effect that will alter its ideal behavior. … As a first level approximation consider this: a real capacitor has some leakage current, which can be modeled by a resistor in parallel to the ideal capacitor, whereas an inductor has some wire resistance …

You don't tell us explicitly, but is seems that is some kind of homework or some textbook explanation. I guess that \$I\$ and \$I_L\$ are not the total currents in those branches, but only the transie …

If we model the setup with a capacitor in parallel with the load, and let \$i\$ be the current into the load and \$v\$ the voltage across the two, the circuit equations are: $$ \left\{ \begin{aligned} …