Unformatted text preview: hen two capacitors are connected in parallel, the equivalent capacitance CP is given by
CP = C1 + C2 (Equation 20.18), where C1 and C2 are the individual capacitances. Therefore,
CP is greater than either C1 or C2. Thus, when the capacitors are connected in parallel, the
greater capacitance leads to a smaller reactance (C is in the denominator in Equation 23.2),
which in turn leads to a greater current. As a result, the current delivered by the generator
increases when the second capacitor is connected in parallel with the first capacitor.
The capacitance of a parallel plate capacitor is given by C = κε0A/d (Equation 19.10),
where κ is the dielectric constant of the material between the plates, ε0 is the permittivity of
free space, A is the area of each plate, and d is the separation between the plates. When the
capacitor is empty, κ = 1, so that C = κ Cempty. Thus, the capacitance increases when the
dielectric material is inserted.
SOLUTION Using Equation 23.1 to express the current as Irms = Vrms/XC and Equation 23.2
to express the capacitive reactance as XC = 1/(2π f C), we have for the current that
I rms = V rms
XC = V rms b 1 / 2π f C =
gV rms 2π f C Applying this result to the case where the empty capacitor C1 is connected alone to the
generator and to the case where the “full” capacitor C2 (which contains the dielectric
material) is connected in parallel with C1, we obtain I1, rms = V 2π f C1
1444 rms
2444
3
C1 alone and I P, rms = V 2π f CP
1444rms
2444
3
C1 and C2 in parallel Dividing the two expressions gives I P, rms
I 1, rms = V rms 2 π f C P
V rms 2 π f C1 = CP
C1 According to Equation 20.18, the equivalent capacitance of the two capacitors in parallel is
CP = C1 + C2, so that the result for the current ratio becomes 1246 ALTERNATING CURRENT CIRCUITS I P, rms
I 1, rms = C1 + C 2
C1 = 1+ C2
C1 Since the capacitance of a filled capacitor is given by Equation 19.10 as C = κε0A/d, we find
that
I P, rms
κε A / d
= 1+ 0
= 1+κ
I 1, rms
ε0A/d
Solving for IP, rms gives b gb b
gg I P, rms = I 1, rms 1 + κ = 0.22 A 1 + 4 .2 = 1.1 A 7. REASONING The capacitance C is related to the capacitive reactance XC and the
frequency f via Equation 23.2 as C = 1/(2π fXC). The capacitive reactance, in turn is related
to the rmsvoltage Vrms and the rmscurrent Irms by XC = Vrms/Irms (see Equation 23.1). Thus,
the capacitance can be written as C = Irms/(2π fVrms). The magnitude of the maximum charge
q on one plate of the capacitor is, from Equation 19.8, the product of the capacitance C and
the peak voltage V.
SOLUTION a. Recall that the rmsvoltage Vrms is related to the peak voltage V by Vrms = V
2 . The capacitance is, then,
C= I rms
2 π f V rms = 3.0 A b 2 π 750 Hz = FJ
g1402V I
GK
H 6.4 × 10 −6 F b. The maximum charge on one plate of the capacitor is c b
hg q = CV = 6.4 × 10 −6 F 140 V = 8. 9 .0 × 10 −4 C REASONING AND SOLUTION Equations 23.1 and 23.2 indicate that the rms current in a
capacitor is I = V / X C , where V is the rms voltage and X C = 1 / 2 π f C . Therefore, the
current is I = V 2 π f C . For a single capacitor C = C1 , and we have b I = V 2 π f C1 g Chapter 23 Problems 1247 For two capacitors in series, Equation 20.19 indicates that the equivalent capacitance can be
obtained from C –1 = C1–1 + C 2–1 , which can be solved to show that C = C1 C2 / C1 + C2 .
The total series current is, then,
C1 C 2
I series = V 2 π f C = V 2 π f
C1 + C 2 c F
G
H h I
J
K The series current is onethird of the current I. It follows, therefore, that I series
I V 2π f
= FC C I
G +C J
C
H K=
1 2 1 C2 2 V 2 π f C1 C1 + C 2 = 1
3 or C1
C2 =2 For two capacitors in parallel, Equation 20.18 indicates that the equivalent capacitance is
C = C1 + C 2 . The total current in this case is c I parallel = V 2 π f C = V 2 π f C1 + C2 h The ratio of Iparallel to the current I in the single capacitor is
I parallel
I 9. c V 2 π f C1 + C 2 = V 2 π f C1 SSM REASONING
respectively, h= C1 + C2
C1 = 1+ C2
C1 = 1+ 1
=
2 3
2 The individual reactances are given by Equations 23.2 and 23.4,
1
2π f C Capacitive reactance XC = Inductive reactance X L = 2π f L When the reactances are equal, we have X C = X L , from which we find
1
= 2π f L
2π f C or 4 π 2 f 2 LC = 1 The last expression may be solved for the frequency f. 1248 ALTERNATING CURRENT CIRCUITS Solving for f with L = 52 × 10−3 H and C = 76 × 10−6 F, we obtain SOLUTION
f= 1
2π LC = 1
2 π (52 × 10 –3 H) (76 × 10 –6 = 8.0 × 10 1 Hz
F) 10. REASONING The rms voltage Vrms across the inductor is given by Vrms = I rms X L
(Equation 23.3), where Irms is the rms current in the circuit, and XL is the inductive
reactance. The inductor is the only circuit element connected to the generator, so the rms
voltage across the inductor is equal to the rms generator voltage: Vrms = 15.0 V.
SOLUTION Solving Equation 23.3 for XL, we obtain
XL = Vrms
I rms = 15.0 V
= 24.6 Ω
0.610 A 11. REASONING AND SOLUTION We know that
V = IXL = I(2π fL) = (0.20 A)(2π)(750 Hz)(0.080 H) = 75 V 12. REASONING The curre...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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