A capacitor can be likened to two plates arranged face to face.

Generally, we can say that a capacitor consists of two conductors separated by an insulator (also referred to as dielectric).

This insulation may be air, glass, plastic, ...

Capacitors are used in all electronic set up.

For example, use of variable capacitors in the tuning of radio circuits.

Subjected to a voltage \(U\), a capacitor has the property of charge and conserve an electrical charge \(q\), proportional to \(U\) :

\(q=CU\)

where \(C\) is the capacitance of the capacitor.

It is an energy reservoir.

One can compare the capacitor to a tank that fills and empties, or a lung that inflates and deflates ...

This energy is returned during discharge of the capacitor (electronic flash, for example).

These phenomena of charge and discharge are not instantaneous ; these are transient phenomena.

With the notation of the following figure :

\({u_C} = \frac{q}{C}\;\;\;\;;\;\;\;\;i = \frac{{dq}}{{dt}} = \dot q\)

The power received by the capacitor (algebraically) is :

\({p_C} = {u_C}i = {u_C}.\frac{{dq}}{{dt}} = C{u_C}.\frac{{d{u_C}}}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{2}Cu_C^2} \right)\)

The energy \(E_C\) stored by the capacitor is deduced :

\({p_C} = \frac{{d{E_C}}}{{dt}}\;\;\;\;\;so\;\;\;\;\;{E_C} = \frac{1}{2}Cu_C^2 = \frac{1}{2}\frac{{{q^2}}}{C}\)

The charges carried by the capacitor plates is a continuous function of time.

Charging a capacitor by a voltage step :

We choose :

At \(t<0\) :

\(e(t)=0\)

And at \(t>0\) :

\(e(t)=E=cste\)

So, for \(t>0\) :

\({u_C} = \frac{q}{C}\;\;\;;\;\;\;i = \frac{{dq}}{{dt}}\;\;\;\;;\;\;\;\;{u_R} = Ri = RC\frac{{d{u_C}}}{{dt}}\)

And :

\({u_C} + {u_R} = E\;\;\;\;\;so\;\;\;\;\;{u_C} + RC\frac{{d{u_C}}}{{dt}} = E\)

Or :

\(\;\;\frac{{d{u_C}}}{{dt}} + \;\frac{1}{{RC}}\;{u_C} = \frac{{d{u_C}}}{{dt}} + \;\frac{1}{\tau }\;{u_C} = \;\frac{1}{\tau }E\;\;\;\;(with\;\tau = RC)\)

If the capacitor is discharged at \(t<0\) :

\({u_C} = E(1 - {e^{ - t/\tau }})\;\;\;\;;\;\;\;\;{u_R} = E - {u_C} = E{e^{ - t/\tau }}\)

\(\tau\) is the time constant of the circuit (RC) : it gives the magnitude of the charging time of the capacitor.

The charge is continuous at \(t=0\) ; the intensity is discontinuous (to pass from \(0\) to \(E/R\) from \(t=0^-\) to \(t=0^+\)).

Energy Aspect :

What happens to the energy supplied by the generator ?

According to the conservation of energy, we will show that :

\({E_G}={E_R}+{E_C}\)

With :

• \(E_G\) : energy supplied by the generator

• \(E_R\) : energy dissipated by Joule effect in \(R\)

• \(E_C\) : energy stored by \(C\)

We calculate :

• \({E_G} = \int_0^\infty {Ei(t)dt} = \frac{{{E^2}}}{R}\int_0^\infty {\;{e^{ - t/\tau }}dt} = \frac{{{E^2}}}{R}\tau = C{E^2}\)

• \({E_R} = \int_0^\infty {R{i^2}(t)dt} = \frac{{{E^2}}}{R}\int_0^\infty {\;{e^{ - 2t/\tau }}dt} = \frac{{{E^2}}}{{2R}}\tau = \frac{1}{2}C{E^2}\)

• \({E_C} = \frac{1}{2}C{E^2}\)

Whence :

\({E_G} = {E_R} + {E_C}\)

Discharge of the capacitor :

The following figure shows the results.

All winding carrying a current creates a magnétic field proportional to intensity \(i\).

When the intensity \(i\) depends on the time, it appears across the coil a self-induction emf (phenomenon of induction).

In receptor convention, this emf \(e\) is written (assuming ideal coil, that is to say without resistance) :

\(e = -L\frac{{di}}{{dt}}\)

The role of a self-induction coil is to oppose any change in the current in a circuit (Lenz's law).

In particular, the intensity of the current in a coil is necessarily continuous.

Voltage, power and energy :

With the notation of the following figure :

\({u_L} = L\frac{{di}}{{dt}} + ri\)

The power received by the coil (algebraically) is (ideal coil here, either \(r = 0\)) :

\({p_L} = {u_L}i = L\frac{{di}}{{dt}}.i = \frac{d}{{dt}}\left( {\frac{1}{2}L{i^2}} \right)\)

The energy \(E_L\) stored by the coil is deduced :

\({p_L} = \frac{{d{E_L}}}{{dt}}\;\;\;\;\;so\;\;\;\;\;{E_L} = \frac{1}{2}L{i^2}\)

The intensity of the current flowing through a coil is a continuous function of time.

Response to a step voltage (ideal coil) :

We choose :

At \(t<0\) :

\(e(t)=0\)

And at \(t>0\) :

\(e(t)=E=cste\).

Then :

\({u_R} = Ri\;\;\;;\;\;\;{u_L} = L\frac{{di}}{{dt}} = \frac{L}{R}\frac{{d{u_R}}}{{dt}}\)

And :

\({u_L} + {u_R} = E\;\;\;\;\;so\;\;\;\;\;\frac{L}{R}\frac{{d{u_R}}}{{dt}} + {u_R} = E\)

Or :

\(\;\;\frac{{d{u_R}}}{{dt}} + \;\frac{R}{L}\;{u_R} = \frac{{d{u_R}}}{{dt}} + \;\frac{1}{\tau }\;{u_R} = E\;\;\;\;(with\;\tau = \frac{L}{R})\)

If the current is zero at \(t<0\) :

\({u_R} = E(1 - {e^{ - t/\tau }})\;\;\;\;;\;\;\;\;{u_L} = E - {u_R} = E{e^{ - t/\tau }}\)

\(\tau\) is the time constant of the circuit (RL) : it gives the order of magnitude of the established duration of permanent regime (\(i=E/R\)).

The coil then behaves as a single wire.

The current (and voltage \(u_R\)) is continuous at \(t=0\) ; the tension \(u_L\) is discontinuous (pass from \(0\) to \(E\) from \(t=0^-\) to \(t=0^+\)).

Energy aspect :

We will show that :

\(E_G=E_R+E_L\)

With :

• \(E_G\) : energy supplied by the generator

• \(E_R\) : energy dissipated by Joule effect in \(R\)

• \(E_L\) : energy stored in the coil

Indeed :

\(E = {u_L} + {u_R} = Ri + L\frac{{di}}{{dt}}\;\;\;\;so\;\;\;\;Ei = R{i^2} + Li\frac{{di}}{{dt}} = R{i^2} + \frac{d}{{dt}}\left( {\frac{1}{2}L{i^2}} \right)\)

Hence the result :

\(\int_0^\infty {Ei} dt\; = \int_0^\infty {R{i^2}} dt + \frac{1}{2}LI_{perm}^2\;\;\;\;\;\;\;\;\;(with\;{I_{perm}} = \frac{E}{R})\)

Differential equation circuit (RLC) :

The Kirchhoff's Voltage Law (KVL) gives :

\({u_L} + {u_R} + {u_C} = L\frac{{di}}{{dt}} + Ri + \frac{1}{C}q = e(t)\)

Yet :

\(i = \dot q\;\;\;;\;\;\;{u_C} = \frac{q}{C}\;\;\;so\;\;\;i = C\frac{{d{u_C}}}{{dt}}\;\)

So :

\(LC\frac{{{d^2}{u_C}}}{{d{t^2}}} + RC\frac{{d{u_C}}}{{dt}} + {u_C} = e(t)\)

Is :

\(\frac{{{d^2}{u_C}}}{{d{t^2}}} + \frac{R}{L}\frac{{d{u_C}}}{{dt}} + \frac{1}{{LC}}{u_C} = \frac{1}{{LC}}e(t)\)

Or :

\(\frac{{{d^2}{u_C}}}{{d{t^2}}} + 2\sigma {\omega _0}\frac{{d{u_C}}}{{dt}} + \omega _0^2{u_C} = \omega _0^2e(t)\;\;\;\;\;\;with\;\;\;\;\omega _0^2 = \frac{1}{{LC}}\;\;and\;\;2\sigma {\omega _0} = \frac{R}{L}\)

The different regimes :

Solutions are sought from the homogeneous equation of the form \(e^{rt}\), with \(r\) is a complex number.

This leads to the characteristic polynomial :

\({r^2} + 2\sigma {\omega _0}r + \omega _0^2 = 0\)

Whose discriminant is :

\(\Delta = 4\omega _0^2({\sigma ^2} - 1)\)

So :

\(\begin{array}{l}\Delta < 0\;\;\;so\;\;\;\sigma < 1\;:\;\;pseudo - periodic\;regime\; (underdamped)\;({r_1},\;{r_2}\; \in \;C) \\\Delta > 0\;\;\;so\;\;\;\sigma > 1\;:\;overdamped\;regime\;({r_1},\;{r_2}\; \in \;R) \\\Delta = 0\;\;\;so\;\;\;\sigma = 1\;:\;critically\;damped\;regime\;(r_1=r_2 = - {\omega _0}) \\\end{array}\)

It is then necessary to add a special solution, which depends on the form of \(e(t)\).

Response of the circuit (RLC) to a step voltage :

For \(t>0\) :

\(\frac{{{d^2}{u_C}}}{{d{t^2}}} + 2\sigma {\omega _0}\frac{{d{u_C}}}{{dt}} + \omega _0^2{u_C} = \omega _0^2E\)

The solution of the previous differential equation is then :

\(\begin{array}{l}\sigma < 1\;\;\;:\;\;\;{u_C} = {e^{ - \sigma {\omega _0}t}}\left[ {A\cos ({\omega _0}\sqrt {1 - {\sigma ^2}} \;t) + B\sin ({\omega _0}\sqrt {1 - {\sigma ^2}} \;t)} \right] + E \\\sigma > 1\;\;\;:\;\;\;{u_C} = {e^{ - \sigma {\omega _0}t}}\left[ {A{e^{({\omega _0}\sqrt {{\sigma ^2} - 1} )\;t}} + B{e^{({\omega _0}\sqrt {{\sigma ^2} - 1} )\;t}}} \right] + E \\\sigma = 1\;\;\;:\;\;\;{u_C} = {e^{ - {\omega _0}t}}\left[ {A + Bt} \right] + E \\\end{array}\)

After \(t \approx \frac{5}{{\sigma {\omega _0}}}\), the steady state is reached (\(u_C\approx E\)).

Energy Aspect :

The final state is :

\(i=0\) and \(u_C=E\) (the capacitor is charged).

Show that the final energy balance is written :

\({E_G} = {E_R} + {E_C}\)

With :

• \(E_G\) : energy supplied by the generator

• \(E_R\) : energy dissipated by Joule effect in \(R\)

• \(E_C\) : energy stored by \(C\)

Indeed :

\(E = Ri + \frac{1}{C}q + L\frac{{di}}{{dt}}\;\;\;;\;\;\;Ei = R{i^2} + \frac{d}{{dt}}\left( {\frac{1}{2}C{u_C}^2} \right) + \frac{d}{{dt}}\left( {\frac{1}{2}L{i^2}} \right)\)

And :

\(\int_0^\infty {Eidt} = \int_0^\infty {R{i^2}dt} + \frac{1}{2}C{E^2} + 0\;\;\;\;\;so\;\;\;\;\;{E_G} = {E_R} + {E_C}\)

One can calculate the energy supplied by the generator :

\(\int_0^\infty {Eidt} = \int_0^\infty {E.\dot qdt} = \int_0^\infty {CE\left( {\frac{{d{u_C}}}{{dt}}} \right)dt} = CE\left[ {{u_C}} \right]_0^\infty = C{E^2}\)

We deduce that the energy dissipated by Joule effect in the resistance \(R\) is :

\({E_R} = {E_G} - {E_C} = C{E^2} - \frac{1}{2}C{E^2} = \frac{1}{2}C{E^2}\)

Therefore :

\({E_R} = {E_C} = \frac{1}{2}C{E^2} = \frac{1}{2}{E_G}\)

Half of the energy supplied by the generator is dissipated by Joule effect.

We find the same result as in the case of the (RC) circuit : this is not surprising since the inductor has no influence on the initial and final states of the circuit.

Response to a square voltage :