how do you know how many volts and amps a capacitor discharges?

Question

the point of this question is that i don't know how to figure out the discharge of the capacitor by knowing the farads on it the coulombs and the rated voltage please help

Answer 1

the amount of voltage initially on a capacitor will be whatever voltage you charge it to. It will never be any higher unless there's other circuitry involved. The current (Amps) will flow at a rate decided by the resistance it is emptied into. To use the common water pipe analogy, the voltage is the amount of water pressure, and the amperage is the size of the pipe.

A capacitor is like a water tower. There's lots of voltage, because the water's trying to get to the ground. If you put a small hole in the water tower, you have a high resistance path to ground. Your voltage will slowly fall as more of the water drains out. the current will start fairly slow, and gradually fall along with the voltage(there's less pressure pushing it).
If you were to blast a big hole in it with a stick of dynamite, you would have a huge amount of flow(current/Amps), but the voltage would fall quickly, and the current will follow.

it all relates to ohms law. Voltage = Current * Resistance

or you could re-arrange it to calculate your current, based on the voltage you've charged the capacitor to, and the resistance you're draining it through. Current = Voltage/Resistance

I knew a physics teacher who would demonstrate this to his classes by shorting a very large capacitor, which was charged to 12V with a screwdriver. It would make a huge shower of sparks, and often partially weld the screwdriver to the top because of the huge currents. He'd then charge it up again and would ask if anyone wants to put their finger on top of it. Nobody would of course, but he would then do it himself and get no shock at all. The resistance of skin is high, so with only 12V there's almost no current flow, even though there's a huge current capability if you give it a low resistance.

If you want the capacitors you're playing with to last a long time, though, you're best charging and discharging through a resistor. Very quick charges and discharges create hot spots inside the capacitor, and will eventually damage them.

Answer 2

Let's call the sum of the internal resistance of the cap and the external resistor it's discharging through R, and the capacitance C.
When R is connected to a charged cap, the voltage on C decreases at a rate proportional to the current, that is
dV/dt = I/C
Since I in turn depends on V (I=V/R) this means that the voltage follows a logarithmic decay curve, whose equation is
V = Vi * (e^(-t/(R*C)))
where Vi is the voltage the cap is initially charged to.
(Note e is the base of natural logarithms = 2.71828)
R*C is called the "time constant" of the discharge and is in seconds when R is in ohms and C is in Farads. So if you discharge a 1 uF cap through 1 Kohm, the time constant is 0.001 s, and every 0.001 s the voltage decreases by a factor of 1/e, thus in steps of 0.001 s from 0, the voltage is Vi, Vi/e, Vi/e^2, etc.
The charge in the cap in Coulombs is Vi*C, and Vi shouldn't exceed the rated voltage or the cap may fail. Charge is equal to current * time; thus the time integral of the discharge current = Vi*C. Similarly the time integral of the discharge voltage is Vi*C/R. When you short a charged cap, the instantaneous current can be very high, especially since most caps have a very low internal resistance. So be careful doing this; you might actually blow out an internal circuit path in the cap. Incidentally, this rapid discharge capability means a high instantaneous power, which is put to use in such consumer items as camera flashes and strobes, and capacitor discharge (CD) ignition systems.

Answer 3

A capacitor will have the rated charge on it only when charged to the rated volatge. It can be charged to any voltage less than this with correspondingly fewer electrons.

The discharge amperage will be a function of the external load and the internal resistance of the capacitior. Since the voltage falls all during the discharge, things get fairly complicated for knowing the second-to-second value of the amperage on discharge.

Answer 4

Capacitors stick to this easy formula: q=CV. What this means is the fee (in coulombs) on a capacitor equals the capacitance elevated via way of the voltage for the time of it. The voltage a cap can manage is chic on it is actual construction. on account that a cap is unquestionably 2 steel plates separated via way of an insulator, the max voltage is a function of the hollow between the plates. The dialectric withstanding voltage of the dialectric is severe additionally. the present means is set via the "effective series resistance" of a capacitor. No caps are desirable, the all exhibit some series resistance. The decrease it extremely is the less complicated. This resistance dissipates potential consistent with the situation-unfastened equation P = I^2* R the place i^2 denotes raised to the potential of two. As you will see, potential dissipation is straight away proportional to the resistance.

Answer 5

It depends on the resistor through which the capacitor discharges. Here is a nice website that allows you to enter in various parameters and see how it affects the discharge curve.