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Unformatted text preview: 0.18),
C23 = C2 + C3 (1) When the equivalent capacitance C23 is connected in series with C1, the resulting equivalent
capacitance CS is, according to Equation 20.19 and Equation (1),
−1 −1 −1 −1
1
1
1 1 1
1
1
CS = = +
= +
=
+ = 36 µ F
C C C C C +C 67 µ F 45 µ F + 33 µ F 23 2
3 S
1
1
______________________________________________________________________________ 99. REASONING
a. When capacitors are wired in parallel, the total charge q supplied to them is the sum of
the charges supplied to the individual capacitors, or q = q1 + q2. The individual charges can
be obtained from q1 = C1V and q2 = C2V, since the capacitances, C1 and C2, and the voltage
V are known.
b. When capacitors are wired in series, the voltage V across them is equal to the sum of the
voltages across the individual capacitors, or V = V1 + V2. However, the charge q on each
capacitor is the same. The individual voltages can be obtained from V1 = q/C1 and
V2 = q/C2. Chapter 20 Problems 1119 SOLUTION
a. Substituting q1 = C1V and q2 = C2V (Equation 19.8) into q = q1 + q2, we have
q = q1 + q2 = C1V + C2V = ( C1 + C2 ) V
= ( 2.00 ×10−6 F + 4.00 ×10−6 F ) ( 60.0 V ) = 3.60 ×10−4 C
b. Substituting V1 = q/C1 and V2 = q/C2 (Equation 19.8) into V = V1 + V2 gives V = V1 + V2 = q
q
+
C1 C2 Solving this relation for q, we have V 60.0 V = 8.00 × 10−5 C
1
1
1
1
+
+
C 1 C2 2.00 ×10−6 F 4.00 ×10−6 F
______________________________________________________________________________
q= = 100. REASONING AND SOLUTION The 7.00 and 3.00µF capacitors are in parallel.
According to Equation 20.18, the equivalent capacitance of the two is
7.00 µF + 3.00 µF = 10.0 µF. This 10.0µF capacitance is in series with the 5.00µF
capacitance. According to Equation 20.19, the equivalent capacitance of the complete
arrangement can be obtained as follows: 1
1
1
–1
=
+
= 0.300 ( µ F )
C 10.0 µ F 5.00 µ F C= or 1
0.300 ( µ F ) –1 = 3.33 µ F The battery separates an amount of charge
Q = CV = (3.33 × 10–6 F)(30.0 V) = 99.9 × 10–6 C
This amount of charge resides on the 5.00 µ F capacitor, so its voltage is
V5 = (99.9 × 10 –6 C)/(5.00 × 10 –6 F) = 20.0 V The loop rule gives the voltage across the 3.00 µ F capacitor to be
V3 = 30.0 V – 20.0 V = 10.0 V
This is also the voltage across the 7.00 µ F capacitor, since it is in parallel, so V7 = 10.0 V .
______________________________________________________________________________ 1120 ELECTRIC CIRCUITS 101. SSM REASONING The foil effectively converts the capacitor into two capacitors in
series. Equation 19.10 gives the expression for the capacitance of a capacitor of plate area
A and plate separation d (no dielectric): C0 = ε 0 A / d . We can use this expression to
determine the capacitance of the individual capacitors created by the presence of the foil.
Then using the fact that the "two capacitors" are in series, we can use Equation 20.19 to
find the equivalent capacitance of the system.
SOLUTION Since the foil is placed onethird of the way from one plate of the original
capacitor to the other, we have d1 = (2 / 3)d , and d 2 = (1/ 3)d . Then
C1 = ε0 A
(2 / 3)d and C2 = ε0 A
(1/ 3)d = = 3ε 0 A
2d 3ε 0 A
d Since these two capacitors are effectively in series, it follows that 1
1
1
1
1
3d
d
=
+
=
+
=
=
Cs C1 C2 3ε 0 A / ( 2d ) 3ε 0 A / d 3ε 0 A ε 0 A
But C0 = ε 0 A / d , so that d / ( ε 0 A ) = 1/ C0 , and we have 1
1
=
or
Cs = C0
Cs C0
______________________________________________________________________________
102. REASONING AND SOLUTION Charge is conserved during the reequilibrium.
Therefore, using q0 and qf to denote the initial and final charges, respectively, we have
q10 + q20 = 18.0 µ C = q1f + q2f (1) After equilibrium has been established the capacitors will have equal voltages across them,
since they are connected in parallel. Thus, Vf = q1f/C1 = q2f/C2, which leads to
q1f = q2f(C1/C2) = q2f (2.00 µ F)/(8.00 µ F) = 0.250 q2f
Substituting this result into Equation (1) gives
18.0 µ C = 0.250 q2f + q2f or q2f = 14.4 µ C Chapter 20 Problems 1121 Hence,
–6 –6 Vf = q2f/C2 = (14.4 × 10 C)/(8.00 × 10 F) = 1.80 V
______________________________________________________________________________
103. SSM REASONING The charge q on a discharging capacitor in a RC circuit is given by
Equation 20.22: q = q0 e – t / RC , where q0 is the original charge at time t = 0 s. Once t (time
for one pulse) and the ratio q / q0 are known, this expression can be solved for C.
SOLUTION Since the pacemaker delivers 81 pulses per minute, the time for one pulse is
1 min
60.0 s
×
= 0.74 s/pulse
81 pulses 1.00 min) Since one pulse is delivered every time the fullycharged capacitor loses 63.2% of its
original charge, the charge remaining is 36.8% of the original charge. Thus, we have
q = (0.368)q0 , or q / q0 = 0.368 .
From Equation 20.22, we have q
= e – t / RC
q0
Taking the natural logarithm of both sides, we have,
q
t
ln = −
q RC 0 Solving for C, we find −t
−(0.74 s)
=
= 4.1 × 10−7...
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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

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