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Unformatted text preview: voltage by π /2 radians, or 90°. Thus, the current in the circuit obeys the relation
I(t) = I0 sin (2π ft − π / 2), where I0 is the peak current. Chapter 23 Problems 1259 SOLUTION
–4
a. The instantaneous value of the voltage at a time of 1.20 × 10 s is ( )( ) V (t ) = V0 sin 2π f t = ( 32.0 V ) sin 2π 1.50 × 103 Hz 1.20 × 10−4 s = 29.0 V Note: When evaluating the sine function in the expression above, be sure to set your
calculator to the radian mode.
b. The inductive and capacitive reactances are ( )( ) X L = 2π f L = 2π 1.50 × 103 Hz 7.20 × 10−3 H = 67.9 Ω (23.4) 1
1
=
= 16.1 Ω
3
2π f C
2π 1.50 × 10 Hz 6.60 × 10−6 F (23.2) XC = ( )( ) Since XL is greater than XC, the current lags the voltage by π /2 radians. Thus, the
instantaneous current in the circuit is I(t) = I0 sin (2π ft − π /2), where I0 = V0/Z. The
impedance Z of the circuit is Z= R2 + ( X L − X C ) 2 = ( 0 Ω )2 + ( 67.9 Ω − 16.1 Ω )2 = 51.8 Ω (23.7) The instantaneous current is I= V0
Z ( sin 2π f t − 1 π
2 ) ( )( ) 32.0 V 3
−4
1 = sin 2π 1.50 × 10 Hz 1.20 × 10 s − 2 π = 51.8 Ω −0.263 A 28. REASONING The rms voltages across the inductor L and the capacitor C are given,
respectively, by VL, rms = I rms X L (Equation 23.3) and VC, rms = I rms X C (Equation 23.1),
where Irms is the rms current in the circuit, X L = 2π f L (Equation 23.4) is the inductive
1
(Equation 23.2) is the capacitive reactance of
2π f C
the capacitor. We know the frequency f of the generator, but we are not given the rms
current in the circuit. We will make use of Equations 23.3 and 23.1 to eliminate the
unknown current Irms, and then solve for the rms voltage across the inductor. reactance of the inductor, and X C = 1260 ALTERNATING CURRENT CIRCUITS SOLUTION Solving Equation 23.1 for Irms gives I rms = VC, rms
XC . Substituting this relation into Equation 23.3, we obtain
VL, rms = I rms X L = VC, rms X L (1) XC Substituting Equations 23.2 and 23.4 for the reactances into Equation (1) yields
VL, rms = VC, rms X L
XC = VC, rms 2π f L
1 2π f C = VC, rms ( 2π f ) LC
2 Therefore, when the rms voltage across the capacitor is VC, rms = 2.20 V, the rms voltage
across the inductor is
VL, rms = VC, rms ( 2π f )2 LC VL, rms = ( 2.20 V ) 2π ( 375 Hz ) 2 (84.0 × 10−3 H )(5.80 × 10−6 F ) = 5.95 V 29. REASONING AND SOLUTION With only the resistor in the circuit, the power dissipated
2
2
is P1 = V0 /R = 1.000 W. Therefore, V0 = (1.000 W) R. When the capacitor is added in
series with the resistor, the power dissipated is given by P2 = I2V0 cos φ = 0.500 W, where
cos φ is the power factor, with cos φ = R/Z2, and I2 = V0/Z2. The impedance Z2 is
2
Z 2 = R 2 + X C . Substituting yields,
P2 = V02R/Z22 = (1.000 W) R2/(R2 + XC2) = 0.500 W
Solving for XC gives, XC = R. When the inductor is added in series with the resistor, we
have P3 = V0I3 cos φ = 0.250 W, where I3 = V0/Z3 and cos φ = R/Z3. The impedance Z3 is
2
Z 3 = R 2 + X L . Thus,
2 2 2 P3 = (1.000 W) R /(R + XL ) = 0.250 W
Solving for XL, we find that XL = R 3 . Finally, when both the inductor and capacitor are
added in series with the resistor we have Chapter 23 Problems P4 = V02 R R2 + ( X L − X C ) 2 = (1.000 W ) R 2 ( R2 + R 3 − R ) = 2 1261 0.651 W 30. REASONING The inductance L and the capacitance C of a series RCL circuit determine
the resonant frequency f0 according to
f0 = 1 (23.10) 2π LC As we see from Equation 23.10, the smaller the inductance L, the larger the resonant
frequency, and the larger the inductance, the smaller the resonant frequency. Therefore, in
part (a) we will use the largest frequency to determine the minimum inductance, and in part
(b) we will use the smallest frequency to find the maximum inductance. We note that
1 MHz = 1×106 Hz.
SOLUTION
a. Squaring both sides of Equation 23.10 and solving for L, we obtain
f 02 = 1
4π LC or 2 L= 1
4π f 02C
2 (1) The minimum inductance is obtained when the resonant frequency is greatest, so Equation
(1) gives
L= 1 ( 4π 2 9.0 × 106 Hz ) (1.8 ×10−11 F )
2 = 1.7 × 10 −5 H b. Using Equation (1) once more, this time with the smallest resonant frequency, yields the
maximum inductance:
L= 1
4π 2 ( 4.0 ×10 6 Hz ) (1.8 ×10
2 −11 F ) = 8.8 × 10 −5 H 31. SSM REASONING The resonant frequency f0 of a series RCL circuit depends on the
1
inductance L and capacitance C through the relation f 0 =
(Equation 23.10). Since
2π LC
all the variables are known except L, we can use this relation to find the inductance. 1262 ALTERNATING CURRENT CIRCUITS SOLUTION Solving Equation 23.10 for the inductance gives L= 1
1
=
2
4π f0 C 4π 2 690 ×103 Hz ( 2 ) ( 2.0 ×10−9 F)
2 = 2.7 × 10−5 H 32. REASONING Only the resistor, on average, consumes power. Therefore, the average
power delivered to the circuit is equal to the average power delivered to the resistor. The
2
average power is given by P = I rms R (Equation 20.15b), where Irms is the rms current in the
circuit and R is the resistance. According to Equation 23.6, the current is given by
Irms = Vrms...
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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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