Problem 3.2 Source:
Problem 25.23 - page 134 - HALLIDAY, RESNICK,
WALKER, Jearl - Book: Fundamentos de Física - Vol 3 - Ed. LTC - 8ª edição - 2009.
In the circuit shown in Figure 03-02.1, V = 9.0 volts, C2 = 3 µF,
C4 = 4 µF
and all capacitors are initially discharged. When the S key is closed
a total charge of 12 µC passes through the point a and a total charge of
8 µC passes through the b point.
a) what is the value of C3 ?
b) what is the value of C1 ?
Figure 03-02.1
Solution of the Problem 3.2
Item a
Let's start analyzing the circuit by the capacitor C4 since we know
its value of 4 µF and also its charge, q4 = 8 µC. Let's remember
the equation that relates charge, capacitance and voltage, or:
q4 = C4 V4
Doing the calculations with the data that we have and based
in the above equation, we can calculate the
potential difference between the terminals of the C4:
V4 = q4 / C4 =
8 x 10-6 / 4 x 10-6 = 2 volts
Note that the voltage on C3 is the same as on C4.
As by the point a pass 12 µC and by the point b pass 8 µC, then
it is obvious that the charge of C3 is the difference of these two charges,
q3 = 4 µC. Therefore, using the previous equation we can calculate the value of
C3, or:
C3 = q3 / V4 = 4 x 10-6 / 2 = 2 x 10-6 F = 2 µF
Item b
As we know that by point a pass a charge of 12 µC
then C1 and C2 also passes that same charge.
In this way, q2 = 12 µC.
Since the value of C2 = 3 µF is given and we have the charge, then
it is easy to calculate the voltage across the terminals of this capacitor, or:
VC2 = q2 / C2 = 4 volts
We must remember that in a series of capacitors, the charge on
each capacitor in the circuit is the same. Therefore C1 has the same charge of the
C2, or q1 = 12 µC. We lack to know what the tension of
C1. But if we add the voltages of VC4 and
VC2 we will have a total of 6 volts. As the
source is 9 volts, so by making the difference of the two, we have VC1 = 3 volts.
Therefore, knowing the load and the voltage, we can calculate the value of C1,
or:
