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Unformatted text preview: 42. REASONING AND SOLUTION The current in an RCLcircuit is
I= V c R2 + X L − XC h
2 Rearranging terms
(XL − XC)2 = (V/I)2 − R2
Using Equations 23.2 and 23.4 for XC and XL, respectively, we obtain 1
2π f L −
=
2π f C 2 V 2 −R =
I 2 26.0 V 2 − (108 Ω ) = 149 Ω
0.141 A Multiplying by f leads to
2πf L − (149 Ω)f − 1/(2πC) = 0
2 ALTERNATING CURRENT CIRCUITS 1270 or
2πf (5.42 × 10 H) − (149 Ω)f − 1/[(2π (0.200 × 10 F)] = 0
2 –3 –6 We can solve this quadratic equation for the frequencies. We obtain
f1 = 3.11 × 10 3 Hz f2 = and 7 .50 × 10 3 Hz 43. SSM REASONING AND SOLUTION The rms voltage can be calculated using V = IXC,
where the capacitive reactance XC can be found using XC = 1
1
=
= 54 Ω
2π f C 2π (3.4 × 103 Hz)(0.86 ×10−6 F) The voltage is, therefore, V = IX C = ( 35 × 10 −3 A )( 54 Ω ) = 1.9 V 44. REASONING Since, on the average, only the resistor consumes power, the average power
2
dissipated in the circuit is P = I rms R (Equation 20.15b), where Irms is the rms current and R
is the resistance. The current is related to the voltage V of the generator and the impedance Z
of the circuit by I rms = Vrms / Z (Equation 23.6). Thus, the average power can be written as
P = V 2 R / Z 2 . The impedance of the circuit is (see Equation 23.7) Z = R 2 + ( X L − X C ) 2 2
= R 2 + X C , since there is no inductor in the circuit. Therefore, the expression for the average power becomes
P= V 2R
V 2R
=2
2
Z2
R + XC SOLUTION The capacitive reactance is XC = 1
1
=
= 2400 Ω
2π f C 2π ( 60.0 Hz ) 1.1 × 10−6 F ( ) The average power dissipated is (120 V )2 ( 2700 Ω )
V 2R
P= 2
=
= 3.0 W
2
R + X C ( 2700 Ω )2 + ( 2400 Ω )2 (23.2) Chapter 23 Problems c 45. REASONING From Figure 23.11 we see that V02 = V L − VC h+ V
2 2
R 1271 . Since VL = 0 V (L = 0 H), and we know V0 and VR, we can use this equation to find VC.
SOLUTION Solving the equation above for VC gives
VC = 45
24
b V g− b V g =
2 2
V 02 − V R = 2 38 V 46. REASONING AND SOLUTION At very high frequencies the capacitors behave as if they
were replaced with wires that have zero resistance, while the inductors behave as if they
were cut out of the circuit. The drawings below show the circuits under this condition.
Circuit I behaves as if the two resistors are in parallel, and the equivalent resistance can be
–1
obtained from R P = R –1 + R –1 as R P = R / 2 . Circuit II behaves as if the two resistors are
in series, and the equivalent resistance is RS = R + R = 2 R . In either case, the current is the
voltage divided by the resistance. Therefore, the ratio of the currents in the two circuits is
I circuit I
I circuit II
R = b g=
V / bR g
2 V / R/2 4 R R Circuit I
High frequency R Circuit II
High frequency 47. SSM WWW REASONING The rms current can be calculated from Equation 23.3,
I rms = V rms / X L , provided that the inductive reactance is obtained first. Then the peak value
of the current I 0 supplied by the generator can be calculated from the rms current I rms by
using Equation 20.12, I 0 = 2 I rms . 1272 ALTERNATING CURRENT CIRCUITS SOLUTION At the frequency of f = 620 Hz , we find, using Equations 23.4 and 23.3, that
X L = 2 π f L = 2 π ( 620 Hz)(8.2 × 10 –3 H) = 32 Ω
I rms = V rms
XL = 10.0 V
= 0.31 A
32 Ω Therefore, from Equation 20.12, we find that the peak value I0 of the current supplied by the
generator must be b g I 0 = 2 I rms = 2 0.31 A = 0.44 A 48. REASONING The rms current in an inductor is Irms = Vrms/XL, according to Equation 23.3.
The inductive reactance is XL = 2π f L, according to Equation 23.4. Applying these
expressions to both generators will allow us to obtain the desired current. SOLUTION Using Equations 23.3 and 23.4, we find that the current in an inductor is
I rms = V rms
XL = V rms
2π f L Applying this result to the two generators gives V rms
I1 =
2π f L
14 2413
4
4 and V rms
I2 =
2π f 2 L
14 244
4
3 Generator 1 Generator 2 Dividing the equation for generator 2 by the equation for generator 1, we obtain
V rms I2
I1 = 2π f 2 L
f
=1
V rms
f2
2π f 1 L or I 2 = I1 f1
f2 b = 0.30 A 1.5
g kHz =
6.0 kHz 0.075 A Chapter 23 Problems 1273 49. REASONING AND SOLUTION
a. Equation 20.14 gives 120 V
V
=
=
R
240 Ω I= 0 .50 A b. Since there is no inductor, Equations 23.6 and 23.7 apply with X L = 0 Ω. Therefore, the
current is I = V / Z = V /
V I= F1 I
+G
2
Hπ f C J
K
2 R2 2
R 2 + X C . Using Equation 23.2 for X C , we find = 120 V L
O
1
P
240
b Ω g+ M b .0 Hz g10.0 × 10 F h
2
M 60 c
P
Nπ
Q
2 = 0 . 34 A 2 –6 c. When an inductor is present, Equations 23.6 and 23.7 give the current as
I =V / Z =V / c R2 + X L – XC h.
2 This expression reduces to I = V/R when X L = X C . Using Equations 23.2 and 23.4 for the reactances, we find that
2π f L = 1
2π f C or L= 1
1
=
=
2
2
2
4π f C
4 π 60.0 Hz 10.0 × 10 –6 F b 2 gc h 0 . 704 H Yes, it is possible to return the current to the value calculated in part (a). 50. REASONING For a series RCL circuit the total impedance Z and the phase an...
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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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