ssrs barcodelib resistor, (b) a load Z 1 j, and (c) a load Z 1 j. Find P in each of the three cases. in Software

Make Quick Response Code in Software resistor, (b) a load Z 1 j, and (c) a load Z 1 j. Find P in each of the three cases.

resistor, (b) a load Z 1 j, and (c) a load Z 1 j. Find P in each of the three cases.
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(a) P V 2 =R 1=1 1 W p (b) and (c) jZj j1 jj 2. p I V=jZj 1= 2. P RI 2 0:5 W
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Obtain the complete power information for a passive circuit with an applied voltage v 150 cos !t 108 V and a resulting current i 5:0 cos !t 508 A.
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Using the complex power
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AC POWER
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[CHAP. 10
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Fig. 10-13    150 5:0 p 108 p 508 375 608 187:5 j342:8 2 2
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S Veff I eff
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Thus, P 187:5 W, Q 324:8 var (inductive), S 375 VA, and pf cos 608 0:50 lagging.
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A two-element series circuit has average power 940 W and power factor 0.707 leading. mine the circuit elements if the applied voltage is v 99:0 cos 6000t 308 V.
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p The e ective applied voltage is 99:0= 2 70:0 V. 940 70:0 Ieff 0:707 Then, 19:0 2 R 940, from which R 2:60
. Z R jXC Finally, from 2:60 1=!C, C 64:1 mF. Substituting in P Veff Ieff cos , or Ieff 19:0 A
Deter-
For a leading pf,  cos 1 0:707 458, and so XC R tan 458 2:60
where
Find the two elements of a series circuit having current i 4:24 cos 5000t 458 A, power 180 W, and power factor 0.80 lagging.
p The e ective value of the current is Ieff 4:24= 2 3:0 A: 180 3:0 2 R tor. The impedance angle is  cos From the power triangle,
Then,
R 20:0
0:80 36:878, wherefore the second element must be an induc-
2 Q Ieff XL tan 36:878 180 P
XL 15:0
CHAP. 10]
AC POWER
Finally, from 15.0=5000L, L 3:0 mH.
Obtain the power information for each element in Fig. 10-14 and construct the power triangle.
Fig. 10-14 p The e ective current is 14.14/ 2 10 A. P 10 2 3 300 W Qj6
10 2 6 600 var inductive q S 300 2 600 200 2 500 VA Q j2
10 2 2 200 var capacitive pf P=S 0:6 lagging
The power triangle is shown in Fig. 10-15.
Fig. 10-15
A series circuit of R 10
and XC 5
has an e ective applied voltage of 120 V. the complete power information.
Z Then:
2 P Ieff R 1152 W 2 Q Ieff XC 576 var (capacitive)
Determine
p 102 52 11:18
Ieff
120 10:73 A 11:18 q 1152 2 576 2 1288 VA
and pf 1152=1288 0:894 leading
Impedances Zi 5:83 59:08
and Z2 8:94 63:438
are in series and carry an e ective current of 5.0 A. Determine the complete power information.
ZT Z1 Z2 7:0 j3:0
Hence, PT 5:0 2 7:0 175 W ST QT 5:0 2 3:0 75 var (inductive) pf 175 0:919 lagging 190:4
q 175 2 75 2 190:4 VA
AC POWER
[CHAP. 10
Obtain the total power information for the parallel circuit shown in Fig. 10-16.
Fig. 10-16 By current division, I5 17:88 18:438 A I4 26:05 12:538 A  2   17:88 26:05 2 PT p 5 p 4 2156 W 2 2   17:88 2 QT p 3 480 var (capacitive) 2 q ST 2156 2 480 2 2209 VA pf Alternate Method 4 5 j3 2:40 j0:53
9 j3 p p Then, P 42:4= 2 2 2:40 2157 W and Q 42:4= 2 2 0:53 476 var (capacitive). Zeq 2156 0:976 leading 2209
Then,
10.10 Find the power factor for the circuit shown in Fig. 10-17.
With no voltage or current speci ed, P, Q, and S cannot be calculated. the cosine of the angle on the equivalent impedance. Zeq 3 j4 10 3:68 36:038
13 j4 pf cos 36:038 0:809 lagging However, the power factor is
Fig. 10-17
10.11 If the total power in the circuit Fig. 10-17 is 1100 W, what are the powers in the two resistors
By current division, I1;eff Z2 10 p 2 2 42 I2;eff Z1 3
CHAP. 10]
AC POWER
and so
I 2 3 P3
6 21;eff P10
I2;eff 10 5
Solving simultaneously with P3
P10
1100 W gives P3
600 W, P10
500 W.
10.12 Obtain the power factor of a two-branch parallel circuit where the rst branch has Z1 2 j4
, and the second Z2 6 j0
. To what value must the 6-
resistor be changed to result in the overall power factor 0.90 lagging
Since the angle on the equivalent admittance is the negative of the angle on the equivalent impedance, its cosine also gives the power factor. Yeq 1 1 0:334 36:848 S 2 j4 6 pf cos 36:848 0:80 lagging
The pf is lagging because the impedance angle is positive. Now, for a change in power factor to 0.90, the admittance angle must become cos 1 0:90 25:848. Then,   1 1 1 1 1 0 j Yeq 2 j4 R 10 R 5 requires 1=5 tan 25:848 1 1 10 R or R 3:20
10.13 A voltage, 28.28 608 V, is applied to a two-branch parallel circuit in which Z1 4 308 and Z1 5 608
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