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1
1
1
=
+
L
L1 L2 L= or L1 L2
L1 + L2 = 0
b.030 H g0.060 H g=
b
0.030 H + 0.060 H 0 . 020 H 17. REASONING AND SOLUTION We begin by calculating the impedance of the circuit
using Z = c R2 + X L – XC XC = h. We have
2 1
1
=
= 28.8 Ω
2π f C
2 π 1350 Hz 4 .10 × 10 −6 F b g
c h 1252 ALTERNATING CURRENT CIRCUITS XL = 2π fL = 2π (1350 Hz)(5.30 × 10 H) = 45.0 Ω
–3 16
45
b .0 Ω g+ b .0 Ω − 28.8 Ω g = 22 .8 Ω
2 Z= 2 The current is therefore,
I = V/Z = (15.0 V)/(22.8 Ω) = 0.658 A
Since the circuit elements are in series, the current through each element is the same. The
voltage across each element is
VR = IR = (0.658A)(16.0 Ω) = 10.5 V
VC = IXC = (0.658 A)(28.8 Ω) = 19.0 V
VL = IXL = (0.658 A)(45.0 Ω) = 29.6 V 18. REASONING For a series RCL circuit the total impedance Z and the phase angle φ are
given by
Z = R2 + ( X L − X C ) 2 (23.7) tan φ = XL − XC
R (23.8) where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.
In the present case, there is no capacitance, so that X C = 0 Ω . Therefore, these equations
simplify to the following:
2
Z = R2 + X L (1) tan φ = XL
R (2) We are given neither R nor XL. However, we do know the current and voltage when only
V
the resistor is connected and can determine R from these values using R = rms
I rms
(Equation 20.14). In addition, we know the current and voltage when only the inductor is
V
connected and can determine XL from these values using X L = rms (Equation 23.3). Since
I rms
the generator frequency is fixed, this value for XL also applies for the series combination of
the resistor and the inductor. Chapter 23 Problems 1253 SOLUTION With only the resistor connected, Equation 20.14 indicates that the resistance
is
V
112 V
= 224 Ω
R = rms =
I rms 0.500 A With only the inductor connected, Equation 23.3 indicates that the inductive reactance is
XL = Vrms
I rms = 112 V
= 2.80 × 102 Ω
0.400 A a. Using these values for R and XL in Equations (1) and (2), we find that the impedance is
2
Z = R2 + X L = ( 224 Ω )2 + ( 2.80 × 102 Ω ) 2 = 359 Ω b. The phase angle φ between the current and the voltage of the generator is tan φ = XL
R or 2.80 × 102 Ω XL = tan −1 = 51.3° R 224 Ω φ = tan −1 19. SSM REASONING The voltage supplied by the generator can be found from Equation
23.6, V rms = I rms Z . The value of I rms is given in the problem statement, so we must obtain
the impedance of the circuit. SOLUTION The impedance of the circuit is, according to Equation 23.7, Z= R 2 + ( X L – X C ) 2 = ( 275 Ω ) 2 + (648 Ω – 415 Ω ) 2 = 3.60 × 10 2 Ω The rms voltage of the generator is
V rms = I rms Z = ( 0.233 A )( 3.60 × 10 2 Ω ) = 83.9 V
20. REASONING The phase angle is given by tan φ = (XL – XC)/R (Equation 23.8). When a
series circuit contains only a resistor and a capacitor, the inductive reactance XL is zero, and
the phase angle is negative, signifying that the current leads the voltage of the generator.
2
The impedance is given by Equation 23.7 with XL = 0 Ω, or Z = R 2 + X C . Since the 1254 ALTERNATING CURRENT CIRCUITS phase angle φ and the impedance Z are given, we can use these relations to find the
resistance R and XC.
SOLUTION Since the phase angle is negative, we can conclude that only a resistor and a
capacitor are present. Using Equations 23.8, then, we have
tan φ = − XC or R b g X C = − R tan −75. 0 ° = 3. 73 R According to Equation 23.7, the impedance is
Z = 192 Ω = 2
R2 + XC Substituting XC = 3.73R into this expression for Z gives
192 Ω = b g = 14.9 R
192
b Ωg =
R= R 2 + 3.73 R 2 2 or 192
b Ω g= 14.9 R
2 2 2 14 .9 49 .7 Ω Using the fact that XC = 3.73 R, we obtain
XC = 3.73 (49.7 Ω) = 185 Ω 21. SSM REASONING We can use the equations for a series RCL circuit to solve this problem provided that we set X C = 0 Ω since there is no capacitor in the circuit. The
current in the circuit can be found from Equation 23.6, Vrms = I rms Z , once the impedance of
the circuit has been obtained. Equation 23.8, tan φ = ( X L − X C ) / R , can then be used (with
X C = 0 Ω) to find the phase angle between the current and the voltage. SOLUTION The inductive reactance is (Equation 23.4)
X L = 2π f L = 2π (106 Hz ) ( 0.200 H ) = 133 Ω
The impedance of the circuit is
2
Z = R 2 + ( X L – X C ) 2 = R 2 + X L = (215 Ω) 2 + (133 Ω) 2 = 253 Ω a. The current through each circuit element is, using Equation 23.6, Chapter 23 Problems I rms = Vrms
Z = 1255 234 V
= 0.925 A
253 Ω b. The phase angle between the current and the voltage is, according to Equation 23.8 (with
X C = 0 Ω),
tan φ = XL − XC
R = XL
R = 133 Ω
= 0.619
215 Ω φ = tan –1 (0.619) = 31.8° or 22. REASONING The average power dissipated is that dissipated in the resistor and is
2
P = I rms R , according to Equation 20.15b. We are given the current Irms but need to find the
resistance R. Since the inductive reactance XL is known, we can find the resistance from the
impedance, which is Z = 2
R 2 + X L , according to Eq...
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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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