Homework Hint
Assignment (1-81)
Read (20.12-20.13), Do PROBLEMS (87, 88, 97) Ch. 20
87.
REASONING
Our approach to this problem is to deal with the arrangement in parts. We will combine separately those
parts that involve a series connection and those that involve a parallel connection.
SOLUTION
The 24, 12, and 8.0-µF capacitors are in series. Using the reciprocal equation for capacitors, we can
find the equivalent capacitance for the three capacitors:
1
1
1
1
=
+
+
Cs 24 µ F 12 µ F 8.0 µ F
Cs = 4.0 µ F
or
This 4.0-µF capacitance is in parallel with the 4.0-µF capacitance already shown in the text diagram.
Taking into consideration the fact that capacitors in parallel add, we find that the equivalent capacitance
for the parallel group is
Cp = 4.0 µ F + 4.0 µ F = 8.0 µ F
This 8.0-µF capacitance is between the 5.0 and the 6.0-µF capacitances and in series with them.
Finally, the reciprocal equation for capacitors in series can be used, then, to determine the equivalent
capacitance between A and B in the:
1
1
1
1
=
+
+
Cs 5.0 µ F 8.0 µ F 6.0 µ F
or
Cs = 2.0 µ F
__________________________________________________________________________________________
88.
REASONING AND SOLUTION
The equivalent capacitance of the circuit is
1/C = 1/(4.0 µF) + 1/(6.0 µF) + 1/(12.0 µF)
or
C = 2.0 µF
The total charge provided by the battery is, then,
Q = CV = (2.0 × 10–6 F)(50.0 V) = 1.0 × 10–4 C
This charge appears on each capacitor in a series circuit, so the voltage across the 4.0-µF capacitor is
V1 = Q/C1 = (1.0 × 10–4 C)/(4.0 × 10–6 F) = 25 V
____________________________________________________________________________________________________________
97.
REASONING
The time constant of a simple RC circuit is given by the equation
resistance and C is the capacitance in the circuit.
•
τ = RC,
where R is the
The two resistors are wired in parallel, so we can obtain the equivalent resistance by using the
reciprocal equation:
[RT]-1 = [R1 ]-1 + [R2]-1 + … + [Rn ]-1
•
The two capacitors are also wired in parallel, and their equivalent capacitance is given by the
equation:
CT = C1 + C2 + … + Cn
Therefore, the time constant (τ ) is the product of the equivalent resistance and
equivalent capacitance.
SOLUTION
The equivalent resistance of the two resistors in parallel is:
1
1
1
3
=
+
=
RP 2.0 kΩ 4.0 kΩ 4.0 kΩ
or
R P = 1.3 kΩ
The equivalent capacitance is:
CP = 3.0 µ F + 6.0 µ F = 9.0 µ F
Thus, the time constant for the charge to “build up” is:
(
)(
)
τ = RP CP = 1.3 × 103 Ω 9.0 × 10−6 F = 1.2 × 10−2 s