The capacitance of a capacitor is given by the ratio of it’s charge to the P.D. across it’s plate. i.e.
capacitance formula =
Explanation When Capacitor In Parallel
Let three capacitors C1 , C2 & C3 are connected in parallel respectively by connected across P.D. now, charge on the capacitor C1
Q1 = C1V
If, C is the equivalent capacitance, of the circuit, the total charge
Q = C.V = Q1 + Q2 + Q3
= C1V + C2V + C3V
Q = V (C1 +C2 +C3)
C = C1 + C2 + C3
OR, Q = VC
C = Q/V
Units or capacitance
It is necessary to be able to define the “size” of a capacitor. The capacitance of a capacitor is a measure of its ability to store charge, and the basic unit of capacitance is the Farad, named after Michael Faraday.
A capacitor with a capacitance of one Farad is too large for most electronics applications, and components with much smaller values of capacitance are normally used. Three prefixes (multipliers) are used, µ (micro), n (nano), and p (pico):
CAPACITANCE UNITS PREFIXES AND MULTIPLIERS
|µ||10-6 (millionth)||1000000µF = 1F|
|n||10-9 (thousand-millionth)||1000nF = 1µF|
|p||10-12 (million-millionth)||1000pF = 1nF|
Capacitor charging and discharging
It is also possible to look at the voltage across the capacitor as well as looking at the charge. After all, it is easier to measure the voltage on it using a simple meter. When the capacitor is discharged there is no voltage across it. Similarly, once it is fully charged no current is flowing from the voltage source and therefore it has the same voltage across it as the source.
In reality, there will always be some resistance in the circuit, and therefore the capacitor will be connected to the voltage source through a resistor. This means that it will take a finite time for the capacitor to charge up, and the rise in voltage does not take place instantly. It is found that the rate at which the voltage rises is much faster at first than after it has been charging for some while. Eventually, it reaches a point when it is virtually fully charged and almost no current flows. In theory, the capacitor never becomes fully charged as the curve is asymptotic. However in reality it reaches a point where it can be considered to be fully charged or discharged and no current flows.
Similarly, the capacitor will always discharge through a resistance. As the charge on the capacitor falls, so the voltage across the plates is reduced. This means that the current will be reduced and in turn the rate at which the charge is reduced falls. This means that the voltage across the capacitor falls in an exponential fashion, gradually approaching zero.
The rate at which the voltage rises or decays is dependent upon the resistance in the circuit. The greater the resistance the smaller the amount of charge which is transferred and the longer it takes for the capacitor to charge or discharge.
So far the case when a battery has been connected to charge the capacitor and disconnected and a resistor applied to charge it up to have been considered. If an alternating waveform, which by its nature is continually changing is applied to the capacitor, then it will be in a continual state of charging and discharging. For this to happen a current must be flowing in the circuit. In this way, a capacitor will allow an alternating current to flow, but it will block a direct current. As such capacitors are used for coupling an AC signal between two circuits that are at different steady-state potentials.
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