# Capacitor - capacitance

The capacitance (or ideal capacitor) of videomodels represents in 3D the behavior of the linear circuit theory capacitance.

The circuit theory allows to model each real component by a set of basic ideal components (or elements ) interconnected.
A two terminals component is called dipole . It imposes a relationship linking the voltage between its terminals and the current flowing therethrough.
In the important case of linear circuits, the dipole elements are only five:
element : at all times, the element imposes only :
resistance proportionality between voltage U and current I
capacitance proportionality between voltage U variation and current I
inductance proportionality between voltage U and current I variation
voltage source the voltage U (for any current)
current source the current I (for any voltage)

The capacitance is characterized at all times by law :

$$I = C \cdot \frac{dU}{dt}$$

This element is involved in the modeling of many real components, especially capacitor which is the real component whose behavior is supposed to match the best than the capacitance.

Very often C coefficient, also called capacitance is simply considered as a constant independent of U and I . We say then that the capacitance is linear.

The capacitance symbol (like capacitor) consists of two parallel segments (see figure below).

The capacitance is here represented in 3D by cyan parallel bars or plates, joined by dotted lines symbolizing the existence of an electric field between the plates. Note that the electric field is expressed in volts per meter and its representation is only "relative" on videomodels because there are no indications of distances on a circuit diagram.

If we look for the capacitor on the videomodel below, we can see the permanent link between the current and the voltage variation.

• Select the image at time 16ms by clicking if necessary to the left or to the right of curves
• One see on the left of curves that at this time the voltage V4 is 1.79V and its derivative is -232V/s.
• At the same time, the current IR3 is -51.1mA.
• It is possible to use the calculator to compute the capacitance :
• To open calculator ( Cal= )
• Copy the following lines to the left column of the calculator
dV4_dt = -232;V/s
IR3 = -51.1m;A
C = IR3 / dV4_dt;F
• Click the right column for the result C=220µF.

What is :

• At what time do we observe the most negative current IR3 ?
On the videomodel, we measure 14ms
• Why ?
The switch has just been opened
• How to predict this current from components values ?
IR3 = V4 / (5Ω + 30Ω)
• Idem for the most positive current.
Rép: ...
• What is the voltage across the charged capacitor ?
2.32V
• Deduce this voltage from component values.
2.4V * 30Ω / (30Ω + 1Ω)
• When the capacitor is discharging, how long do we measure 37% of the starting voltage? (we have 1/exp(1) = 37%)