This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ________________________________
93. REASONING A capacitor with a capacitance C stores a charge q when connected between
the terminals of a battery of voltage V, according to q = CV (Equation 19.8). This equation
can be used to calculate the voltage if q and C are known. In this problem we know that the
two capacitors together store a total charge of 5.4 × 10−5 C. This is also the charge stored by
the equivalent capacitor, which has the equivalent capacitance CP of the parallel
combination, or CP = C1 + C2 (Equation 20.18). Thus, we can determine the battery voltage
by using Equation 19.8 with the given total charge and the equivalent capacitance obtained
from Equation 20.18.
SOLUTION With q = qTotal and C = CP = C1 + C2 Equation 19.8 becomes qTotal = CPV = ( C1 + C2 ) V or V= qTotal C1 + C2 = 5.4 ×10−5 C
= 9.0 V
2.0 ×10−6 F + 4.0 ×10−6 F 1116 ELECTRIC CIRCUITS 94. REASONING The equivalent capacitance CS of a set of three capacitors connected in
1
1
1
1
series is given by
(Equation 20.19). In this case, we know that the
=
+
+
CS C1 C2 C3 equivalent capacitance is CS = 3.00 µF, and the capacitances of two of the individual capacitors in this series combination are C1 = 6.00 µF and C2 = 9.00 µF. We will use
Equation 20.19 to determine the remaining capacitance C3. SOLUTION Solving Equation 20.19 for C3, we obtain 1
1
1
1
=
−−
C3 CS C1 C2 C3 = or 1
1
1
1
−−
CS C1 C2 Therefore, the third capacitance is
1
= 18 µ F
1
1
1
−
−
3.00 µ F 6.00 µ F 9.00 µ F
______________________________________________________________________________
C3 = 95. SSM REASONING AND SOLUTION Let C0 be the capacitance of an empty capacitor.
Then the capacitances are as follows, according to the discussion following Equation 19.10:
C1 = 3.00C0 and C2 = 5.00C0 The series capacitance of the two is 1
1
1
=
+
Cs 2.50C0 4.00C0 or Cs = 1.54C0 Now κC0 = 1.54C0 or
κ = 1.54
______________________________________________________________________________
96. REASONING When capacitors are connected in parallel, each receives the entire voltage
V of the battery. Thus, the total energy stored in the two capacitors is 1 C1V 2 + 1 C2V 2 (see
2
2
Equation 19.11b). When the capacitors are connected in series, the sum of the voltages
across each capacitor equals the battery voltage: V1 + V2 = V . Thus, the voltage across each Chapter 20 Problems capacitor is series is less than the battery voltage, so the total energy,
less than when the capacitors are wired in parallel. 1
2 1117 C1V1 + 1 C2V2 , is
2
2 2 SOLUTION
a. The voltage across each capacitor is the battery voltage, or 60.0 V. The energy stored in
both capacitors is Total energy = 1 C1V 2 + 1 C2V 2 =
2
2
= 1
2 ( 2.00 × 10 −6 1
2 ( C1 + C2 )V 2 ) F + 4.00 × 10−6 F ( 60.0 V ) = 1.08 × 10−2 J
2 b. According to the discussion in Section 20.12, the total energy stored by capacitors in
series is Total energy = 1 CSV 2 , where CS is the equivalent capacitance of the series
2
combination: 1
1
1
1
1
=
+
=
+
−6
CS C1 C2 2.00 × 10 F 4.00 × 10−6 F
Solving this equation yields CS = 1.33 × 10
Total energy = 1
2 −6 (20.19) F. The total energy is (1.33 × 10−6 F) ( 60.0 V )2 = 2.39 × 10−3 J ______________________________________________________________________________
97. 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 8.0µF capacitors are in series. Using Equation 20.19, 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 or Cs = 4.0 µ F This 4.0µF capacitance is in parallel with the 4.0µF capacitance already shown in the text
diagram. Using Equation 20.18, 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 8.0µF capacitance is between the 5.0 and the 6.0µF capacitances and in series with
them. Equation 20.19 can be used, then, to determine the equivalent capacitance between A
and B in the text diagram: 1118 ELECTRIC CIRCUITS 1
1
1
1
=
+
+
or
Cs = 2.0 µ F
Cs 5.0 µ F 8.0 µ F 6.0 µ F
______________________________________________________________________________
98. REASONING To find the equivalent capacitance of the three capacitors, we must first,
following CP = C1 + C2 + C3 + L (Equation 20.18), add the capacitances of the two parallel
capacitors together. We must then combine the result of Equation 20.18 with the remaining
1
1
1
1
capacitance in accordance with
=+
+
+ L (Equation 20.19). As Equation
CS C1 C2 C3
20.19 shows, combining capacitors in series decreases the overall capacitance, and the
resulting equivalent capacitance CS is smaller than any of the capacitances being added.
Therefore, the way to maximize the overall equivalent capacitance is to choose the largest
capacitance (C1) to be connected in series with the parallel combination C23 of the smaller
capacitances.
SOLUTION When C2 and C3 are connected in parallel, their equivalent capacitance C23 is,
from CP = C1 + C2 + C3 + L (Equation 2...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details