ssrs barcodelib From the impedance diagram, Fig. 9-16, 5 jXL Z in Software

Maker Quick Response Code in Software From the impedance diagram, Fig. 9-16, 5 jXL Z

From the impedance diagram, Fig. 9-16, 5 jXL Z
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XL 5 tan 808 28:4 
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Then 28:4 ! 30 10 , whence ! 945:2 rad/s and f 150:4 Hz. Z 5 j28:4 
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Fig. 9-16
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Fig. 9-17
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SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
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[CHAP. 9
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At what frequency will the current lead the voltage by 308 in a series circuit with R 8  and C 30 mF
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From the impedance diagram, Fig. 9-17, 8 jXC Z Then 308 XC 8 tan 308 4:62  or f 1149 Hz
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1 4:62 2f 30 10 6
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A series RC circuit, with R 10 , has an impedance with an angle of 458 at f1 500 Hz. Find the frequency for which the magnitude of the impedance is (a) twice that at f1 , (b) one-half that at f1 .
From 10 jXC Z1 458, XC 10  and Z1 14:14 . q 28:28 2 10 2 26:45  (a) For twice the magnitude, 10 jXC 28:28 8 or XC
Then, since XC is inversely proportional to f , 10 f 2 26:45 500 or f2 189 Hz
(b) A magnitude Z3 7:07  is impossible; the smallest magnitude possible is Z R 10 .
A two-element series circuit has voltage V 240 08 V and current I 50 608 A. Determine the current which results when the resistance is reduced to (a) 30 percent, (b) 60 percent, of its former value.
Z V 240 08 4:8 I 50 608 608 2:40 j4:16  80:28 
30% 2:40 0:72 I1
Z1 0:72 j4:16 4:22 80:28 A
240 08 56:8 4:22 80:28
60% 2:40 1:44 I2
Z2 1:44 j4:16 4:40 70:98 A
70:98
240 08 54:5 4:40 70:98
For the circuit shown in Fig. 9-18, obtain Zeq and compute I.
For series impedances, Zeq 10 Then 08 4:47 63:48 12:0 j4:0 12:65 18:438 
V 100 08 7:91 Zeq 12:65 18:438
18:438 A
Evaluate the impedance Z1 in the circuit of Fig. 9-19.
Z V 20 I 608 10:0 j17:3 
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
Fig. 9-18
Fig. 9-19
Then, since impedances in series add, 5:0 j8:0 Z1 10:0 j17:3 or Z1 5:0 j9:3 
Compute the equivalent impedance Zeq and admittance Yeq for the four-branch circuit shown in Fig. 9-20.
Using admittances, 1 j0:20 S j5 1 Y2 0:05 j0:087 S 5 j8:66 Y1 Then and Y3 1 0:067 S 15 1 Y4 j0:10 S j10 58:08 S
Yeq Y1 Y2 Y3 Y4 0:117 j0:187 0:221 Zeq 1 4:53 Yeq 58:08 
Fig. 9-20
The total current I entering the circuit shown in Fig. 9-20 is 33:0 current I3 and the voltage V.
V IZeq 33:0
13:08 A. Obtain the branch
13:08 4:53 58:08 149:5 45:08 V   1 I3 VY3 149:5 45:08 08 9:97 45:08 A 15
Find Z1 in the three-branch network of Fig. 9-21, if I 31:5 V 50:0 60:08 V.
24:08 A for an applied voltage
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
[CHAP. 9
Y Then whence Y1 0:354
I 0:630 V
36:08 0:510 j0:370 S 1 1 10 4:0 j3:0
0:510 j0:370 Y1 458 S and Z1 2:0 j2:0 .
Fig. 9-21
The constants R and L of a coil can be obtained by connecting the coil in series with a known resistance and measuring the coil voltage Vx , the resistor voltage V1 , and the total voltage VT (Fig. 9-22). The frequency must also be known, but the phase angles of the voltages are not
Fig. 9-22
Fig. 9-23
known.
Given that f 60 Hz, V1 20 V, Vx 22:4 V, and VT 36:0 V, nd R and L.
The measured voltages are e ective values; but, as far as impedance calculations are concerned, it makes no di erence whether e ective or peak values are used. The (e ective) current is I V1 =10 2:0 A. Then Zx 22:4 11:2  2:0 Zeq 36:0 18:0  2:0
From the impedance diagram, Fig. 9-23, 18:0 2 10 R 2 !L 2 11:2 2 R2 !L 2 where ! 260 377 rad/s. Solving simultaneously, R 4:92  L 26:7 mH
In the parallel circuit shown in Fig. 9-24, the e ective values of the currents are: Ix 18:0 A, I1 15:0 A, IT 30:0 A. Determine R and XL .
CHAP. 9]
SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS
The problem can be solved in a manner similar to that used in Problem 9.14 but with the admittance diagram. The (e ective) voltage is V I1 4:0 60:0 V. Then Yx Ix 0:300 S V Yeq IT 0:500 S V Y1 1 0:250 S 4:0
Fig. 9-24 From the admittance diagram, Fig. 9-25, 0:500 2 0:250 G 2 B2 L 0:300 2 G2 B2 L which yield G 0:195 S, BL 0:228 S. Then R i.e., XL 4:39 . 1 5:13  G and jXL
Fig. 9-25
1 j4:39  jBL
Obtain the phasor voltage VAB in the two-branch parallel circuit of Fig. 9-26.
By current-division methods, I1 4:64 120:18 A and I2 17:4 AYB may be considered. Choosing the former, VAB VAX VXB I1 20 I2 j6 92:8 120:18 104:4 30:18 A. Either path AXB or path 59:98 11:6 59:98 V
Fig. 9-26
Fig. 9-27
In the parallel circuit shown in Fig. 9-27, VAB 48:3
By voltage division in the two branches: VAX and so or VAB VAX
308 V.
Find the applied voltage V.
j4 1 j8:66 V V VBX V 4 j4 1 j 5 j8:66   1 j8:66 1 V V VBX 1 j 5 j8:66 0:268 j1 1058 48:3 308 50:0 1358 V
V 0:268 j1 VAB 1:035
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