the output terminsls, calculate the values of IL, ILs, and IL. 1á. The current I = 15/-30° when E = 120/10° in Fig. 15.8c. What will be the %3! current I in Fig. 15-8b if E = 110/40°? 122 25 whon F-110/0° For what yoltage

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answer number 12. write your solution pls. 

418
ELECTRICAL CIRCUITS-ALTERNATING CURRENT
three impedances: E; = 120/30°, E: = 120/-30°, 21 = 2 ohms (resistance)
Z; = 2/90° ohms (inductive reactance), Z = 10/-53.1°. CRlculate the v
of the load current It and the voltage across the load, using the superposition
theorem.
5. Referring to Fig. 15-2a, interchange the sources of emf E, and E, making
E, = 120/-30° and E, 120/30°, and calculate the values of IL, EL, and P,
8. The following particulars are given for the T-network of Fig. 15.4: E = 120/02
Z. = 40 ohms (resistance), Z, = (15.6 + j19.2) ohms (resistance-inductive resc-
tance), Z. = -j30 ohms (capscitive reactance), ZL = -j40 ohms (capacitive
resctance). Using Thévenin's theorem calculate the current through and the
voltage across the load impedance ZL.
7. Using Thévenin's theorem calculate the load current and voltage, given the follow-
ing dets: E = 120/0°, Z. = Z, = j9, Z. = 12, Zz = (3.68 - j8.76). (Refer to
Fig. 15:4.)
8. Using the dats of Prob. 7, calculate the short-circuit current I, as illustrated
in Fig. 15-5, for use in Eq. (145) (page 402); after evaluating this check the load
current as determined by Thevenin's theorem in Prob. 7.
9. The following information is given. in connection with Fig. 15-4: E = 120,0°,
Z. = 12/-90°, Z, = 12/90°, Z. = 12, ZL = 8.5/45°. Calculate the load er-
rent Is and the load voltage EL, using Thévenin's theorem.
20. Solve Prob. 9 using Norton's theorem.
11. Referring to Prob. 9, assume the same values for E, Z,, Z, and Z. If three losd
impedances Zı = S.5/45°, ZL: = 8.5/-45°, and Zz: = 8.5/0° are connected to
the output terminsls, calculate the values of I Lı, ILa, and IL:.
A8. The current I = 15/-30° when E = 120/10° in Fig. 15.8a. Whst will be the
current I in Fig. 15-85 if E = 110/40°?
18. In Fig. 15.8a the current I = 4.33 - j2.5 when E 110/0°. For what voltage
in Fig. 15.8b will the current I be 6/30°?
A. When switch S: is open and switch S, is closed in Fig. 15-9, the currents are
I = (8 + j5) and I: (4 -j3). Calculate the current I when switch S: is
open and (a) switch S: is closed to the right, and (b) switch S; is closed to the left.
16. A 48-volt battery has an internal resistance of 0.22 ohm and-is connected to a
variable-resistance load through a line resistance of 1.28 ohms. For what value
of load resistance will the load power be a maximum, snd what will be the loid
current, load power, and power loss under this condition?
16. A 228-volt constant-potential generator delivers load to a variable impedance
whose values ofR and X are readily adjustable. Assuming a "looking-bsek"
impedance of (6 + j4.5) ohms calculate, for maximum load power, (a) the losd
impedance, (b) the current, (c) the load power and power factor.
17. If the load in Prob. 16 is a variable resistor calculate, for maximum load power,
the load resistance and power.
18. If the load in Prob. 16 contains a constant inductive reactance of 3.5 ohms and a
variable resistance, calculate, for maximum load power, (a) the load resistance
and impedance, (b) the load current and power.
19. If the load in Prob. 16 contains a constant resistance of 4.8 ohms and a variavie
inductive reactance, calculate, for maximum load power, (e) the load reaetance
and impedance, (b) the lond current and power.
20. Referring to Fig. 15-13a, transfora the upper delta into an equivalent star
determine the total current I as in Example 13.
Transcribed Image Text:418 ELECTRICAL CIRCUITS-ALTERNATING CURRENT three impedances: E; = 120/30°, E: = 120/-30°, 21 = 2 ohms (resistance) Z; = 2/90° ohms (inductive reactance), Z = 10/-53.1°. CRlculate the v of the load current It and the voltage across the load, using the superposition theorem. 5. Referring to Fig. 15-2a, interchange the sources of emf E, and E, making E, = 120/-30° and E, 120/30°, and calculate the values of IL, EL, and P, 8. The following particulars are given for the T-network of Fig. 15.4: E = 120/02 Z. = 40 ohms (resistance), Z, = (15.6 + j19.2) ohms (resistance-inductive resc- tance), Z. = -j30 ohms (capscitive reactance), ZL = -j40 ohms (capacitive resctance). Using Thévenin's theorem calculate the current through and the voltage across the load impedance ZL. 7. Using Thévenin's theorem calculate the load current and voltage, given the follow- ing dets: E = 120/0°, Z. = Z, = j9, Z. = 12, Zz = (3.68 - j8.76). (Refer to Fig. 15:4.) 8. Using the dats of Prob. 7, calculate the short-circuit current I, as illustrated in Fig. 15-5, for use in Eq. (145) (page 402); after evaluating this check the load current as determined by Thevenin's theorem in Prob. 7. 9. The following information is given. in connection with Fig. 15-4: E = 120,0°, Z. = 12/-90°, Z, = 12/90°, Z. = 12, ZL = 8.5/45°. Calculate the load er- rent Is and the load voltage EL, using Thévenin's theorem. 20. Solve Prob. 9 using Norton's theorem. 11. Referring to Prob. 9, assume the same values for E, Z,, Z, and Z. If three losd impedances Zı = S.5/45°, ZL: = 8.5/-45°, and Zz: = 8.5/0° are connected to the output terminsls, calculate the values of I Lı, ILa, and IL:. A8. The current I = 15/-30° when E = 120/10° in Fig. 15.8a. Whst will be the current I in Fig. 15-85 if E = 110/40°? 18. In Fig. 15.8a the current I = 4.33 - j2.5 when E 110/0°. For what voltage in Fig. 15.8b will the current I be 6/30°? A. When switch S: is open and switch S, is closed in Fig. 15-9, the currents are I = (8 + j5) and I: (4 -j3). Calculate the current I when switch S: is open and (a) switch S: is closed to the right, and (b) switch S; is closed to the left. 16. A 48-volt battery has an internal resistance of 0.22 ohm and-is connected to a variable-resistance load through a line resistance of 1.28 ohms. For what value of load resistance will the load power be a maximum, snd what will be the loid current, load power, and power loss under this condition? 16. A 228-volt constant-potential generator delivers load to a variable impedance whose values ofR and X are readily adjustable. Assuming a "looking-bsek" impedance of (6 + j4.5) ohms calculate, for maximum load power, (a) the losd impedance, (b) the current, (c) the load power and power factor. 17. If the load in Prob. 16 is a variable resistor calculate, for maximum load power, the load resistance and power. 18. If the load in Prob. 16 contains a constant inductive reactance of 3.5 ohms and a variable resistance, calculate, for maximum load power, (a) the load resistance and impedance, (b) the load current and power. 19. If the load in Prob. 16 contains a constant resistance of 4.8 ohms and a variavie inductive reactance, calculate, for maximum load power, (e) the load reaetance and impedance, (b) the lond current and power. 20. Referring to Fig. 15-13a, transfora the upper delta into an equivalent star determine the total current I as in Example 13.
NETWORK LAWS, THEOREMS, AND PRINCIPLES
e point in a circuit to a second
ing from the generator into the circuit, is
In any network of bilateral-linear impedances, a shift of a source of voltage
405
accompanied by a corresponding
B from one
Zo
Zb
Z.
Zc
Zc
(o) Original circuit
(b) Circuit showing shifted generator
FIG. 15-8 Diagrams illustrating the reciprocity theorem.
Referring to Fig. 15-8a, the equivalent impedance of the network, look-
ing from the generator into the circuit, is
Z.(Z +ZL)
Z + Z. + ZL
Z.Z, + Z,Z. + Z.Z. + ZL(Z. + Z)
Z, + Z. + ZL
Z = Za +
%3D
%3D
and the current is
EZ.
E
Z.
%3D
Zea^ Z + Z. + ZL
ZaZb + ZZ. + Z.Z. + Z1(Z. + Z.)
ZaZb + Z.Z. + ZZ. + Z1(Z. + Z.)
Za+ Z.
Z= Z, + ZL+
Z.Z.
%3D
%3D
Z. + Z.
And the current is
EZ.
17 + Zz(Z. + Ze)
72
Transcribed Image Text:NETWORK LAWS, THEOREMS, AND PRINCIPLES e point in a circuit to a second ing from the generator into the circuit, is In any network of bilateral-linear impedances, a shift of a source of voltage 405 accompanied by a corresponding B from one Zo Zb Z. Zc Zc (o) Original circuit (b) Circuit showing shifted generator FIG. 15-8 Diagrams illustrating the reciprocity theorem. Referring to Fig. 15-8a, the equivalent impedance of the network, look- ing from the generator into the circuit, is Z.(Z +ZL) Z + Z. + ZL Z.Z, + Z,Z. + Z.Z. + ZL(Z. + Z) Z, + Z. + ZL Z = Za + %3D %3D and the current is EZ. E Z. %3D Zea^ Z + Z. + ZL ZaZb + ZZ. + Z.Z. + Z1(Z. + Z.) ZaZb + Z.Z. + ZZ. + Z1(Z. + Z.) Za+ Z. Z= Z, + ZL+ Z.Z. %3D %3D Z. + Z. And the current is EZ. 17 + Zz(Z. + Ze) 72
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