Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
9th Edition
ISBN: 9781259989452
Author: Hayt
Publisher: Mcgraw Hill Publishers
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Textbook Question
Chapter 12, Problem 11E

The single-phase three-wire system of Fig. 12.31 has three separate load impedances. If the source is balanced and Van = 110 +j0 V rms, (a) express Van and Vbn in phasor notation. (b) Determine the phasor voltage which appears across the impedance Z3. (c) Determine the average power delivered by the two sources if Z1 = 50 + j0 Ω, Z2 = 100 + j45 Ω, and Z3 = 100 – j90 Ω. (d) Represent load Z3 by a series connection of two elements, and state their respective values if the sources operate at 60 Hz.

Chapter 12, Problem 11E, The single-phase three-wire system of Fig. 12.31 has three separate load impedances. If the source

(a)

Expert Solution
Check Mark
To determine

The expression of phase to neutral voltage of phase a and phase b in phasor notations.

Answer to Problem 11E

The expression of phase to neutral voltage of phase a in phasor notation is 1100°Vrms and of phase b is 110180°Vrms.

Explanation of Solution

Given data:

The phase to neutral voltage of phase a Van is (110+j0)Vrms.

Calculation:

The given diagram is shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 12, Problem 11E , additional homework tip  1

The general expression for the phasor notation is given by,

F=|F|θ        (1)

Here,

|F| is the magnitude of the F.

θ is the angle measured from the reference.

The magnitude of F is given by,

|F|=x2+y2        (2)

Here,

x is the real part.

y is the imaginary part.

The angle measured from the reference θ is,

θ=tan1(yx)        (3)

Substitute 110Vrms for x, 0Vrms for y and Van for F in equation (2).

|Van|=(110Vrms)2+(0Vrms)2=110Vrms

Substitute 110Vrms for x and 0Vrms for y in equation (3).

θ=tan1(0Vrms110Vrms)=0°

Substitute 110Vrms for |F|, 0° for θ and Van for F in equation (1).

Van=1100°Vrms

The voltage Vnb is equals to voltage Van since the system is a single phase system.

The voltage Vbn using double subscript notation is,

Vbn=Vnb

Substitute Van for Vnb in the above equation.

Vbn=Van

Substitute 1100°Vrms for Van in the above equation.

Vbn=1100°Vrms=110180°Vrms

Conclusion:

Therefore, the expression of phase to neutral voltage of phase a in phasor notation is 1100°Vrms and of phase b is 110180°Vrms.

(b)

Expert Solution
Check Mark
To determine

The phasor voltage which appears across the impedance Z3.

Answer to Problem 11E

The phasor voltage which appears across the impedance Z3 is 2200°Vrms.

Explanation of Solution

Calculation:

The voltage across the impedance Z3 is the sum of the voltages across the impedances Z1 and Z2.

The voltage across the impedance Z1 as per KVL is Van.

The voltage across the impedance Z2 as per KVL is Vnb.

The voltage across the impedance Z3 VZ3 is

VZ3=Van+Vnb

Substitute 1100°Vrms for Van and 1100°Vrms for Vnb in the above equation.

VZ3=1100°Vrms+1100°Vrms=2200°Vrms

Conclusion:

Therefore, the phasor voltage which appears across the impedance Z3 is 2200°Vrms.

(c)

Expert Solution
Check Mark
To determine

The average power delivered by the two sources.

Answer to Problem 11E

The average power delivered by source Van is 375.16W and by source Vnb is 234.07W.

Explanation of Solution

Given data:

The value of the impedance Z1 is (50+j0)Ω.

The value of the impedance Z2 is (100+j45)Ω.

The value of the impedance Z3 is (100j90)Ω.

Calculation:

The required diagram is shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 12, Problem 11E , additional homework tip  2

The formula to find current I3 as per KVL in mesh 3 is given by,

I3=VZ3Z3

Substitute 2200°Vrms for VZ3 and (100j90)Ω for Z3 in the above equation.

I3=2200°Vrms(100j90)Ω=2200°(100j90)×100+j90100+j90A=22000+j198001002+902A=1.215+j1.093A

Substitute 1.215A for x, 1.093A for y and I3 for F in equation (2).

|I3|=(1.215A)2+(1.093A)2=1.63A

Substitute 1.215A for x and 1.093A for y in equation (3).

θ=tan1(1.093A1.215A)=41.97°

Substitute 1.63A for |F|, 41.97° for θ and I3 for F in equation (1).

I3=1.6341.97°A

The formula to find current I1 using KCL and KVL in Figure 2 is given by,

I1=VanZ1+I3

Substitute (50+j0)Ω for Z1, 1100°Vrms for Van and 1.6341.97°A for I3 in the above equation.

I1=1100°Vrms(50+j0)Ω+1.6341.97°A=11050+1.215+j1.093A=3.415+j1.093A

Substitute 3.415A for x, 1.093A for y and I1 for F in equation (2).

|I1|=(3.415A)2+(1.093A)2=3.58A

Substitute 3.415A for x and 1.093A for y in equation (3).

θ=tan1(3.415A1.215A)=17.7°

Substitute 3.58A for |F|, 17.7° for θ and I1 for F in equation (1).

I1=3.5817.7°A

The formula to find current I2 using KVL and KVL in Figure 2 is given by,

I2=VnbZ2+I3

Substitute (100+j45)Ω for Z2, 1100°Vrms for Vnb and 1.6341.97°A for I3 in the above equation.

I2=1100°Vrms(100+j45)Ω+1.6341.97°A=1100°109.6524.22°+1.6341.97°A=1.00324.22°+1.6341.97°A=2.23317.646°A

The power P1 delivered by the source Van is given by,

P1=VanI1cos(θvθi)

Substitute 110V for Van, 3.58A for I1, 0° for θv and 17.7° for θi in the above equation.

P1=(110V)(3.58A)cos(0°(17.7°))=375.16W

The power P2 delivered by the source Vnb is given by,

P2=VnbI2cos(θvθi)

Substitute 110V for Vnb, 2.233A for I1, 0° for θv and 17.646° for θi in the above equation.

P2=(110V)(2.233A)cos(0°(17.646°))=234.07W

Conclusion:

Therefore, the average power delivered by source Van is 375.16W and by source Vnb is 234.07W.

(d)

Expert Solution
Check Mark
To determine

The representation of load Z3 for the given condition.

Answer to Problem 11E

The impedance Z3 is the combination of resistance 100Ω and the capacitance 29.5×106F in series.

Explanation of Solution

Given data:

The operating frequency is 60Hz.

Calculation:

The value of the impedance Z3 is (100j90)Ω.

The real part of Z3 is the resistance.

Hence, the resistance R is 100Ω.

The imaginary part of Z3 should be capacitive reactance since series only series capacitive reactance can have negative sign.

Hence, The series capacitive reactance Xc is given by,

Xc=j2πfC

Here,

f is the operating frequency.

C is the capacitance,

Substitute j90Ω for Xc and 60Hz for f in the above equation.

j90Ω=j2π(60Hz)C90=1120πCC=29.5×106F

Conclusion:

Therefore, the impedance Z3 is the combination of resistance 100Ω and the capacitance 29.5×106F in series.

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Chapter 12 Solutions

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf

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