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Charge and Discharge of a Capacitor :


Let us examine the circuit of Figure 3 in which the capacitor is represented by its graphic symbol.


As soon as the capacitor (C) is connected to the battery, there is the phenomenon already analyzed previously, namely that a number of electrical charges pass from one armature to another.

charging current of the capacitor.

The charging current persists until the amount of electricity reached on the capacitor plates generates, between them, a potential difference equal to the voltage of the battery. The capacitor is then said charged.

Once the capacitor is charged, no current flows in the circuit, since the voltage created across (C) is equal to but opposite to the voltage of the battery.

The discharge of the capacitor can easily be observed. Simply remove the capacitor and connect it, for example, to the terminals of a resistor, as shown in Figure 4


The voltage present across the capacitor circulates a current in the resistor R which, in the conventional direction, is directed from the positive armature to the negative armature.

This current due to the electrical charges accumulated on the capacitor plates only lasts a short time. It stops when the charges present in excess on a frame have joined the frame on which they are lacking. This operation carried out, the capacitor is said to be discharged and the current created by this discharge is called the discharge current of the capacitor.

If the capacitor once removed from its charging circuit (Figure 3) is not connected to a resistor, it keeps on its frames accumulated charges.

The capacitor would remain charged indefinitely and the dielectric between its frames was a perfect insulator. In practice, this never happens and the dielectric gradually passes the electrical charges from one armature to another, which slowly discharges the capacitor.

The most important thing to remember from what we have just seen is :

that a capacitor, after having charged, prevents any subsequent circulation of the current supplied by a battery.


We have just seen how we can charge a capacitor by means of a battery and then discharge it into a resistor. It is easy to understand that during these two operations, electrical energy is involved.

For this, it suffices to observe the thermal effect generated in the resistor by the discharge current of the capacitor. This thermal effect is necessarily at the cost of an electrical energy consumption.

This energy consumed was evidently supplied by the capacitor who himself had received it from the battery.

If the capacitor is able to yield this energy to the resistance, it is because it has not dissipated it before but stored.

The capacitor has the property of storing electrical energy, so it is a conservative element of energy unlike the resistor which is a dissipative element.

Let us now see how it is possible to quantify the electrical energy stored by a capacitor and how it has been stored.


To charge a capacitor, the battery must move a quantity of electricity (Q) from one armature to another of this component. This quantity of electricity is determined by the product of the voltage (V) of the battery by the capacitance (C) of the capacitor. To produce this phenomenon, a certain energy (W) is provided by the battery and this energy is equal to the product of the quantity of electricity (Q) by the voltage (V).

However, this energy (W) is not totally stored by the capacitor, in fact, the capacitor stores only half of the energy (W) provided by the battery. The second half is dissipated as heat in the internal resistance of the cell and possibly in other resistors of the circuit.

To be convinced of this, consider Figure 5 where the resistor (R) represents the total resistance of the circuit, that is to say the internal resistance of the battery plus that of the electrical connections and armatures of the capacitor.


At the moment when the capacitor is connected to the circuit, the voltage Vc at its terminals is zero, so the entire voltage V of the battery is found at the terminals of the resistor R (Vr = V). Then as the load of the capacitor, the voltage Vc increases, while Vr decreases. When the charge is complete, all the voltage of the battery is found at the terminals of C, while Vr is zero.

The voltages Vc and Vr thus have a similar but opposite appearance since one is increasing and the other decreasing. Since the electric current (I) passes through both R and C, the quantity of electricity (Q) supplied by the cell to the circuit divides well into two equal parts between R and C. The energy Wc stored by the capacitor is therefore equal to the energy Wr dissipated in the resistance.

The total energy W supplied by the stack is equal to Wr + Wc but like Wr = Wc, we also have Wr = Wc / 2.

The energy provided by a battery for charging a capacitor is given by the formula W = Q x V ; this allows us to quantify the energy actually stored by the capacitor :

Wc = Q x V / 2

Or knowing that the quantity of electricity accumulated by a capacitor is given by the product Q = C x V, we can replace Q in the preceding formula by its value and we obtain :

Wc = C x V² / 2

All the energy Wc stored by the capacitor is then completely restored by it during its discharge.

PARENTHESIS : It is interesting to see which laws vary the voltage across the capacitor and the current flowing in the circuit during the capacitor charge. These patterns are reported respectively Figures 6-b and 6-c ; while Figure 6-a, gives the electric circuit taken as an example :


At the initial moment t0, where the electrical connection is established, a current I = V / R equal to that which would circulate continuously, if we had not a capacitor but a simple wire having no current, flows from the cell to the capacitor resistance (Figure 6-c). Just after t0, the capacitor starts charging and the voltage Vc at its terminals increases (Figure 6-b). As a result, the current (I) starts to decrease until it vanishes when (C) is charged : the voltage across (C) is maximum.

The variations of the current and the voltage have a so-called exponential pace, the equations of such curves are as follows :

I = V / R . (e- t / RC)

Vc = V (1 - e - t / RC)

equations in which (e = 2.72 approximate, represents the base of natural or Neperian logarithms and the value given by the product RC (Resistance in ohm and capacity in farad) constitutes the time constant of the circuit measured in seconds. It follows that in any time interval equal to RC (from 0 to RC, from RC to 2 RC, etc.), the value of the current (charge or discharge) always decreases in the same ratio of 2 72.

Example : Let's take the interval from 0 to RC, so (t = RC)

        I = V / R . e- t / RC

or     t = RC              I = V / R . e- RC / RC = V / R . e- 1

         e-1 = 1 / e                   I = V / R . 1 / e = V / R / e = V / R / 2,72

The current (I) decreased by 2.72 times since at t = 0 its value was V / R and at time t = RC, its value is V / R / 2.72.

It follows that theoretically the current never vanishes and that the charging or discharging time of the capacitor is infinitely large. However, in practice, we find that after a time equal to 5 times the constant RC, the current is 0.7% of its initial value and we can consider that the charge (or discharge) of the capacitor is complete.


We will now analyze how the capacitor stores energy.

Suppose we load an air capacitor and imagine that one of the positive charges present in excess on the positive armature is detached from it and is in the dielectric (Figure 7-a).


This load is pushed back by the positive armature whereas on the contrary, it is attracted by the negative armature. On this load therefore acts a force having the direction given by the arrow in Figure 7-a. This same force would act on any other positive charge detaching from the positive armature.

If we trace the paths followed by a certain number of loads, we obtain the different trajectories represented by arrow lines (figure 7-b).

These lines are called lines of force because the force that determines the displacement of the positive charges acts along it. The set of lines of force delimits the area of the space in which an electric charge is subjected to a force. The area thus determined represents an electric force field or simply an electric field.

Any positive charge in the field is subject to force and tends to move. This force accomplishes a work which is given by the product of the intensity of the force by the length of the displacement of the load. All work is obtained at the cost of energy consumption. In the case of the capacitor, the energy consumed to produce the work is the electrical energy stored by the capacitor.

We therefore understand that the energy stored by the capacitor decreases each time a charge is detached from the positive armature and passes on the negative one. At each transfer, a small amount of energy is transformed into work done by the force that causes the movement of the load. If all the charges are detached from their armature, all the energy stored by the capacitor is transformed into work and the latter is completely discharged.

In fact, no load can be detached from the positive armature since the dielectric is an almost perfect insulator. On the other hand, these charges can move at the same time as the armature.

Let's analyze the consequences of a combination of the two frames, consequences that we know but to which no precise answer has been given.

We assume that the capacitor of Figure 8-a has a capacity of 3 µF and is charged by a 4-volt battery. The quantity of electricity (Q) present on its plates is given by the formula Q = C x V, that is to say 3 µF x 4 V = 12 µC.


If now the capacitor is disconnected from the battery, it obviously keeps the amount of electricity of 12 µC and a voltage between its armatures of 4 V.

Imagine that the positive armature approaches the negative armature and that the distance between them is halved (Figure 8-b).

If the positive armature comes into contact with the negative armature, ie if it moves a distance d, all the energy stored by the capacitor is transformed into work. We therefore understand that if the positive armature moves by d / 2, the energy stored is reduced by half. However, as the two frames are isolated, the amount of electricity can not decrease; as a result, the voltage between frames is reduced by half and takes the value of 2 volts (Figure 8-b). The value of the capacitor increases and becomes :

C = Q / V = 12 µC / 2 = 6 µF

We have thus just given a concrete explanation to the fact that the capacitance of a capacitor increases when the distance between its armatures decreases and that, in particular, it doubles when the distance is reduced by half.

At the beginning of this lesson, we brought the reinforcement closer while leaving the capacitor connected to the battery (figure 1) and we saw that the battery was providing an additional charge current. We can now say that this current serves to keep the voltage across the capacitor equal to the voltage supplied by the battery.

Now introduce between the armatures of the charged capacitor of Figure 8-a, but not connected to the battery, a solid dielectric (Figure 8-c). If this dielectric has a relative dielectric constant of 2, capacitance of the capacitor is doubled. The introduction of this dielectric places the capacitor under the same conditions as in Figure 8-b, after bringing its reinforcements closer together.

In this case too, half of the stored energy is transformed into work, but since there is no displacement of reinforcement, this work is necessarily produced differently. To explain this, we must remember the polarization principle of the dielectric shown in Figure 1-a and 1-b (polarization of the dielectric). The eccentricity of the electron orbits is determined by the intensity of the field which tends to attract the electrons towards the positive armature of the capacitor. The work done by this force on each electron is very small, however the number of electrons of the dielectric being considerable, the sum of the different forces results in the consumption of half of the energy stored by the capacitor.


We now know that the electric field is characterized by lines of force of which we already know the direction and the direction of their action (figure 7-b), but to be complete on the electric field, it is also necessary to know its intensity.

The intensity of the electric field acting in the dielectric of a capacitor is obtained by dividing the tension existing between its armatures by the distance which separates them.


Note : Do not confuse this symbol with that of the electromotive force of a battery (f.e.m.) hence the presence of the arrow on the symbol indicating that it is a vector.

The unit of the electric field strength is the volt per meter (symbol V / m).

It is easy to understand that the intensity of the electric field increases when the tension between the armatures increases or when the distance between them decreases. This fact has a consequence of primary practical importance. However, the intensity of the electric field can not increase indefinitely and, arriving at a certain threshold, the value of the intensity becomes such that the electric charges can cross the dielectric from one reinforcement to the other.

This passage of current is manifested in the form of a violent electric discharge, a kind of flash which perforates the dielectric and establishes an irreversible contact between the armatures of the capacitor. The capacitor is then short-circuited and becomes unusable. We can consider this discharge of the capacitor or breakdown as an instantaneous transformation into heat of all the energy stored by it.

The value of the field strength for which there is breakdown is the dielectric strength of the material constituting the insulator. This value is different for each type of material.

Each dielectric material is therefore characterized not only by its relative dielectric constant but also by its dielectric strength.

The table in Figure 9 gives the dielectric strength of the materials already listed in Figure 1.1 for their relative dielectric constant.


Dielectric strength kV / cm

Dry air


Capacitor special paper (KRAFT)

200 à 400


600 à 1800

Titanate of Magnesia

50 à 100

Rutile, Rutile Zirconia, Calcium Titanate

40 à 80

Titanates and Barium Zirconates

40 à 60

Polystyrene (Styroflex)


Polytetrafluoroethylene (PTFE, Teflon)

400 à 800

Polymonochlorotrifluoretylene (PCFTE)

1000 à 2000

Polyethylene terephthalate (Polyester, Mylar)

1000 à 2000

Aluminum electrolytic

environ 10 000

Tantalum electrolyte

environ 10 000

  Fig. 9. - Dielectric strength of different materials

For the dielectric materials commonly used, the value of the dielectric strength is extremely high and for this we use as unit not the V / cm but the kV / cm as in figure 9.

For example, for a capacitor whose two armatures are 1 cm apart and having polystyrene as dielectric, the breakdown occurs for a voltage of the order of 400 kV.

If the distance between the armatures is only 1 mm, the same breakdown occurs at 40 kV. There are capacitors where the thickness of the dielectric is only a few thousandths of a millimeter, so we understand that their breakdown occurs even for low voltages, voltages that we encounter in electrical or electronic circuits where capacitors are used. For this reason, each capacitor carries a voltage indication called the operating voltage : a value that must not be exceeded, otherwise the component may be damaged by a breakdown.

Remember that a capacitor is characterized not only by its capacitance but also by its operating voltage.

Even air can lose its dielectric properties after a breakdown, so you note that the air has a dielectric strength that is 21 kV / cm for dry air.

The lightning that we observe during storms is the manifestation of the breakdown of the air. Indeed, electric charges accumulate in clouds that behave like the frames of a capacitor.

It thus establishes an electric field between two clouds which are at different potentials or between a cloud and the earth. When the intensity of the electric field exceeds the rigidity of the air, which moreover strongly decreases when the air is humid, there is an electric discharge between the clouds or between the earth and the cloud.

The second case is very dangerous and to prevent the electric shock from causing damage to the homes endangering the lives of its occupants, the buildings are protected by a lightning rod.

Since the lightning rod is the highest point in the building, it is exposed to electric shock. The lightning rod, connected to the earth, transmits the electric shock. To facilitate contact with the earth, note the presence of an iron plate buried in the ground (required).


For the time being, we have considered circuits with only one capacitor, but these components, such as resistors, can form different groups.


Figure 10 shows a parallel array of two capacitors called for circumstance C1 and C2.


C1 and C2 each have an armature connected to the positive pole of the cell and the other armature connected to the negative pole of the same battery, so that at the terminals of each capacitor, there is the same voltage.

This last characteristic is common to any parallel grouping as it has already been said during the analysis of the groups of resistances.

Since between the frames of C1 and C2 we apply the same voltage, each capacitor is charged with a quantity of electricity all the greater as its capacity is high.

Let's see how we can determine the equivalent capacity (Ceq) presented by such a circuit, and this knowing the value of C1 and C2.

For this purpose, let us imagine bringing the two capacitors together until their armature connected to the same pole is in contact, as shown in Figure 11-a. This is possible insofar as the armatures are connected to the same electrical potential.


The two capacitors thus combined constitute a single capacitor called Ceq in Figure 11-b. This capacitor has the same dielectric, the same distance between frames as C1 and C2. The only difference lies in the increase of the surface of the frames.

Given the formula giving the capacity of a capacitor :

C = e0 x er x (S / d)

We know that if the surface S increases, capacitance S of the capacitor also increases and this in the same proportions.

In the case of Figure 11-a, the surface S of Ceq is equal to the sum of the surfaces of C1 and C2. We therefore deduce that the capacitance of the equivalent capacitor Ceq of Figure 11-b is equal to the sum of the capacitances of C1 and C2.

The capacity equivalent to two or more capacitors connected in parallel is equal to the sum of the capacitances of each capacitor :

Ceq = C1 + C2 + C3 + ....


Consider the capacitors C1 and of Figure 12. To facilitate our explanations, the frames of these capacitors are called A, B, C and D.


When charging C1 and C2, the armature A is positively charged and the armature D negatively.

The armatures B and C not connected to the battery together with the conductor that connects them, a simple metal body. This body is charged by induction with the signs shown in figure 12. On the armature B appears a quantity of equal electricity but of sign opposite to that present on A, while on C appears an amount of equal electricity but of opposite sign to that on D.

If we call (+ Q) the quantity of electricity present on A, we have (- Q) on B, and if we call (- Q) the quantity of electricity present on D, we have + Q on C.

Capacitors C1 and C2 thus store the same amount of electricity Q.

As in any serial assembly, the voltage V supplied by the battery is divided into two voltages V1 and V2 respectively across capacitors C1 and C2. In Figure 13 is reported the same circuit with the different voltages present.


The voltage V1 across C1 is equal to :

V1 = Q / C1

The voltage V2 across C2 is equal to :

V2 = Q / C2

As it is said previously V = V1 + V2, therefore :


We can replace capacitors C1 and C2 in Figure 13 by an equivalent capacitor that we call a Ceq (Figure 14).


(Ceq) equivalent to C1 and C2, stores the same quantity of electricity Q as C1 and C2.

We can write the relationship (2) :

V = Q x 1 / Ceq --------------) (2)

Indeed : Ceq = Q / V              V = Q / Ceq = Q x 1 / Ceq

The relations (1) and (2) are equal since they both give the value of the voltage V.

(1) = (2) ---------) Q x (1 / C1 + 1 / C2) = Q x 1 / Ceq

By simplifying (1) and (2) by Q, we get the value of Ceq :


We thus determined the value of Ceq as a function of C1 and C2. Extended to the general case of several capacitors in series, this formula becomes :


Thus, to calculate the equivalent capacitance (Ceq) of two or more capacitors in series, the following three operations are to be performed :

When only two capacitors are in series, we adopt the following formula which derives from the general formula :

Ceq = (C1 x C2) / (C1 + C2)

By a practical example encrypted, implement what we have just seen.

Or to calculate the capacity equivalent to the circuit represented figure 15.


To calculate Ceq, perform the three required operations :

1 / C1 = 1 / 5 = 0,2

1 / C2 = 1 / 10 = 0,1

1 / C3 = 1 / 2 = 0,5

1 / C1 + 1 / C2 + 1 / C3 = 0,2 + 0,1 + 0,5  =  0,8  = -------> 1 / Ceq

Ceq = 1 / 0,8 = 1,25 nF

Capacitors C1, C2 and C3 in series are therefore equivalent to a single capacitance of 1.25 nF.

To make a comparison between the two types of associations, it should be noted that in the case of capacitors in parallel, the value of the equivalent capacitor is always greater than the value of each capacitor while in the case of a series association, the value of the equivalent capacitor is in all cases less than the value of each capacitor and even better, it is smaller than the smallest capacitance.

The formulas presented are also used for the calculation of more complex circuits born from the combination of the two types of associations.

Let's see, for example, how to calculate the total capacity of the circuit shown in Figure 16.


In this circuit, we see that C1, C2 and C3 constitute a parallel group connected in series with C4.

Let us first calculate the capacitor equivalent to C1, C2 and C3 in parallel, a capacitor we call C123.

C123 = C1 + C2 + C3 = 1 µF + 5 µF + 2 µF = 8 µF

Replace in Figure 16, C1, C2 and C3 by their equivalent capacitor C123, we obtain Figure 17-a).


With Figure 17-a, we are in the presence of two capacitors (C123 and C4) connected in series. Calculation of the equivalent capacitor (Ceq) to this assembly gives the capacitor equivalent to the circuit of Figure 16 :

Ceq = C123 x C4 / C123 + C4 = 8 x 12 / 8 + 12 = 96 / 20 = 4,8 µF

In the presence of complex circuits such as that of Figure 16, it is always necessary to simplify the circuit gradually to obtain more than one capacitor whose capacity represents the overall capacity of the starting circuit.

Thus ends the analysis of the groups of capacitors.

As we have said, it is important to know that a capacitor once charged prevents any flow of current supplied by a battery.

In the next lessons, we will see that there are other generators providing currents different from that provided by a stack. With respect to these currents, the capacitors react differently. This property is used when we want to separate, in the same circuit, two different types of currents.

This property will be analyzed in detail in the next lessons.

We end this lesson with a summary table of electrical quantities for the capacitor and their unit and formula if necessary (Figure 18).

Electric quantities

Unit of measure






Absolute dielectric constant


Farad by meter

F / m






C = er x e0 x (S / d)

Quantity of electricity stored




Q = C x V

Stored energy




W = C x V² / 2

Fig. 18. - Electrical quantities relating to the capacitor.

In the next lesson, we will examine the third fundamental component of electronic circuits : inductance, as well as all the phenomena that it generates when it is introduced into a circuit, for example electromagnetism.

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