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Unformatted text preview: is known, we may use Equation 23.10,
1
f0 =
to find the inductance L, provided the capacitance C can be determined. The
2 π LC
capacitance can be found by using the definitions of capacitive and inductive reactances.
SOLUTION
a. Solving Equation 23.10 for the inductance, we have L= 1
4 π f 02 C
2 (1) where f0 is the resonant frequency. From Equations 23. 2 and 23.4, the capacitive and
inductive reactances are
1
XC =
and
X L = 2π f L
2π f C 1266 ALTERNATING CURRENT CIRCUITS where f is any frequency. Solving the first of these equations for f, substituting the result
L
into the second equation, and solving for C yields C =
. Substituting this result into
XL XC
Equation (1) above and solving for L gives L= 1
2π f 0 XL XC = 1
2 π 1500 Hz b 30
b
g b .0 Ω g5.0 Ω g= 1.3 × 10 −3 H b. The capacitance is
C= 38. REASONING L
=
XL XC 1.3 × 10 −3 H
30.0 Ω 5.0 Ω g g=
b b 8.7 × 10 −6 F We will find the desired percentage from the ratio LB / LA . The beat ( frequency that is heard is f0B − f 0A , and the resonant frequencies are f0B = 1/ 2π LBC ( ) ) and f0A = 1/ 2π LAC , according to Equation 23.10. By expressing the beat frequency in
terms of these expressions, we will be able to obtain LB / LA .
SOLUTION Using Equation 23.10 to express each resonant frequency, we find that the
beat frequency is f0B − f0A = ( 1
1
−
2π LBC 2π LAC ) Factoring out the term 1/ 2π LAC gives f0B − f0A 1
2π LBC
1
1
=
−1 =
1
2π LA C
2π LA C
2π LAC LA
−1
LB LA / LB − 1 is a positive quantity, we can solve for Noting that LB is less than LA so that
LA / LB : f0B − f0A
LA
=
+1
LB 1/ 2π L C
A ( ( ) ) Remembering that f0A = 1/ 2π LA C , we see that this result becomes Chapter 23 Problems f −f
LA
= 0B 0A + 1
LB
f0A or LA f0B − f0A
=
+ 1 LB f0A 1267 2 Taking the reciprocal of this expression reveals that
LB
1
1
=
=
= 0.977
2
2
LA f − f 7.3 kHz + 1 0B 0A + 1 630 kHz f 0A Thus, the percentage reduction in LB is (1.000 − 0.977 ) × 100% = 2.3% . 39. SSM WWW REASONING Since we know the values of the resonant frequency of the
circuit, the capacitance, and the generator voltage, we can find the value of the inductance
from Equation 23.10, the expression for the resonant frequency. The resistance can be found
from energy considerations at resonance; the power factor is given by cos φ , where the phase
angle φ is given by Equation 23.8, tan φ = ( X L – X C ) / R .
SOLUTION
a. Solving Equation 23.10 for the inductance L, we find that L= 1
1
=
=
2
2
3
4π f 0 C
4 π (1.30 × 10 Hz) 2 ( 5.10 × 10 –6 F)
2 2.94 × 10 –3 H b. At resonance, f = f 0 , and the current is a maximum. This occurs when X L = X C , so
2
that Z = R . Thus, the average power P provided by the generator is P = V rms / R , and
solving for R we find
V2
(11.0 V) 2
R = rms =
= 4.84 Ω
P
25.0 W c. When the generator frequency is 2.31 kHz, the individual reactances are
XC = 1
1
=
= 13.5 Ω
3
2 π f C 2 π (2.31 × 10 Hz)( 5.10 × 10 –6 F) X L = 2 π f L = 2 π (2.31 × 10 3 Hz)( 2 .94 × 10 –3 H) = 42 .7 Ω 1268 ALTERNATING CURRENT CIRCUITS The phase angle φ is, from Equation 23.8, φ = tan –1 X
F
G
H L – XC
R I = tan F Ω – 13.5 Ω I = 80.6°
42.7
J G 4.84 Ω J
K
KH
–1 The power factor is then given by cos φ = cos 80.6 ° = 0.163 40. REASONING
a. When a capacitor stores charge, it also stores electrical energy. The energy stored by the ( ) capacitor can be expressed as Energy = 1 q 2 / C , according to Equation 19.11c.
2
b. There is no resistance in the circuit, so no energy is lost as it shuttles back and forth
between the capacitor and the inductor. The energy removed from the capacitor when it is
completely discharged is 1
2 ( q2 / C ) . This energy is gained by the inductor. The energy
1
2 stored by an inductor is given by Energy = LI 2 (Equation 22.10), where L is the inductance and I is the current. The maximum energy stored by the inductor is
where Imax is the maximum current in the inductor. 1
2 2
LI max , SOLUTION
a. The electrical energy stored in the fully charged capacitor is ( ) 2 −6 q 2 1 2.90 ×10 C Energy = 1 = 2 = 1.17 × 10−6 J
2 C 3.60 ×10−6 F (19.11c) b. Since the energy stored by the capacitor is equal to the maximum energy stored by the
inductor, we can write 2
1q
2 C 2
1 = 2 LI max
1 24
43 1 24
4 3 Energy stored
Energy stored
by capacitor by inductor The maximum current in the inductor is or I max = q
LC Chapter 23 Problems q
=
LC I max = 2.90 × 10−6 C ( 75.0 ×10 −3 )( H 3.60 × 10 −6 F ) 1269 = 5.58 × 10−3 A 41. REASONING AND SOLUTION At resonance, the power dissipated by the circuit is
P = IV. At a nonresonant frequency, the power dissipated is
P ′ = I ′V cos φ = 1
2 P= 1
2 IV so that
I ′ cos φ = 1
2 I I
R
=
2I ′
Z′ cos φ = and We also know that
I= V
R I′ = and V
Z′ so that I
Z′
=
I′
R Combining these results yields FI F I
GJ G J
HK H K 1I
1 Z′
=
2 I′
2R
1 24
43 Z′ = R 2 so that R
Z Finally, the power factor, cos φ, is cos φ = R
R
=
=
Z′
R2 0.707...
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 Spring '13
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