Application de la charge d’un condensateur
In this exercise, we study an application of the charging of a capacitor in the domain
of meteorology. To measure the amount of humidity relative to air (noted by h %), we
can use a detector where the principle is equivalent to a capacitor of capacitance C (in
pF = 10-12 F) varying with the relative humidity h: C = 0.4 h + 105. For that, the
student sets up the circuit of figure 1.
A. theoretical study of an RC series circuit.
1. Using the law of addition voltage, establish the differential equation governing the
variation of the voltage uC = uBM across the extremities of the capacitor.

t

(K)
A
i
E
q
C
Fig. 1
R
B
M
2. The solution of this differential equation has the form: uC = A (1 - e ). Determine A and .
3.  is called time constant. Explain why  is a time magnitude.
4. A good estimation of the time constant of an RC series circuit is obtained, by plotting the
tangent to the curve at the origin, as indicated in the figure 2. Justify this method and
determine the equation of this tangent.
5. What is the minimum time, in terms of , needed so that the charge of the capacitor reaches
Fig. 2
the steady state?
B. Determination of the relative humidity of air.
To measure the capacitance C of the detector, we connect it in series to a (LFG)
generator delivering a rectangular signal (0; +E), of variable frequency and a resistor
(R) of resistance R = 10 M. We use an oscilloscope to visualize the voltage uC
across the capacitor. We obtain the adjacent waveform.
1. Determine, from figure 3:
a) the characteristics (voltage E and period T) of the signal delivered by the
generator. Is this period sufficient to measure exactly ? Why?
SV = 2 V/div
Fig. 3
b) The time constant . Deduce the value of the capacitance C of the detector.
Sh = 1 ms/div
2. What is the value of air humidity h?
A-1- uAM = uAB + uBM = Ri + uC = E
Expected answers
du
dq
¼ ; i = + dt  C dtC
¼
 E = RC
du C
+ uC
dt
½
st
Differential equation of 1 order
t
t

du C
A 
 e  ; by substitution in the
2- uC = A (1- e ) is a solution of the differential equation;
dt

differential equation  A = E and  = RC. ½
t
3- the power of the exponential must not has unit   ant t have same unit  second ¼

t
t

du C
E
E 

e )t = 0 =
4- uC = E (1 – e ) 
.
) t 0  slope of the tangent =
RC
RC ½
dt
E
This tangent cut u = E at t =   u =
t is the equation of the tangent. At O.
RC
At t =   U = E. ½
5- The capacitor becomes pratically chareged for t  5, ¼


t
since for t = 5  uC = E(1 – e  ) = 0.99 E ≈ E.
¼
B- 1-a- E = 6  2 = 12 V ½ ; T/2 = 8  1 = 8 ms then T = 16 ms; ½ Yes, since 5 < T/2.
b-  = 1.2  1 = 1.2 ms.
½ ;  = RC  C = 1.2  10-10F = 120 pF. ½
2- 120 = 0.4 h + 105  h = 37.5% ½
¼