Section 8-2 and 8-3: Average and Complex Power

Section 8-2 and 8-3: Average and Complex Power Problem 8.9 Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by: (a) v(t) = 100 cos(377t − 30◦ ) V, i(t) = 2.5 cos(377t − 60◦ ) A. (b) v(t) = 25 cos(2π × 103 t + 40◦ ) V, i(t) = 0.2 cos(2π × 103 t − 10◦ ) A. (c) Vrms = 110∠60◦ V, Irms = 3∠45◦ A. (d) Vrms = 440∠0◦ V, Irms = 0.5∠75◦ A. (e) Vrms = 12∠60◦ V, Irms = 2∠−30◦ A. Solution: (a) 100 ◦ Vrms = √ e− j30 V, 2 2.5 − j60◦ Irms = √ e V. 2 100 ◦ 2.5 ◦ ◦ S = Vrms I∗rms = √ e− j30 × √ e j60 = 125e j30 2 2 S = |S| = 125 VA

(VA)

Pav = Re[S] = 125 cos 30◦ = 108.25 W Q = Im[S] = 125 sin 30◦ = 62.5 VAR

φs = φv − φi = −30◦ + 60◦ = 30◦

(hence pf is lagging)

◦

pf = cos 30 = 0.866. (b)

25 ◦ Vrms = √ e j40 V, 2 0.2 ◦ Irms = √ e− j10 A, 2 ◦

S = Vrms I∗rms = 2.5e j50

(VA),

φs = 50◦ (lagging)

S = 2.5 VA,

Pav = 2.5 cos 50◦ = 1.61 W, Q = 2.5 sin 50◦ = 1.92 VAR, pf = cos 50◦ = 0.64 lagging. (c) ◦

◦

◦

S = Vrms I∗rms = 110e j60 × 3e− j45 = 330e j15 VA. S = 330 VA,

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Pav = 330 cos 15◦ = 318.76 W, Q = 330 sin 15◦ = 85.41 VAR, pf = cos 15◦ = 0.97 (lagging). (d) ◦

◦

S = Vrms I∗rms = 440 × 0.5e− j75 = 220e− j75 VA. S = 220 VA,

Pav = 220 cos(−75◦ ) = 56.94 W, Q = 220 sin(−75◦ ) = −212.50 VAR,

pf = cos(−75◦ ) = 0.26 (leading). (e) ◦

◦

◦

S = Vrms I∗rms = 12e j60 × 2e j30 = 24e j90 VA. S = 24 VA,

Pav = 24 cos 90◦ = 0, Q = 24 sin 90◦ = 24 VAR, pf = cos 90◦ = 0

(purely inductive with I lagging V by 90◦ )



Problem 8.11 In the circuit of Fig. P8.11, is (t) = 0.2 sin 105 t A, R = 20 Ω, L = 0.1 mH, and C = 2 µ F. Show that the sum of the complex powers for the three passive elements is equal to the complex power of the source. + is(t) _

R

L

C

Figure P8.11: Circuit for Problem 8.11.

Solution: V

Is

IR

IL

20 Ω

j10 Ω

IC

+ _ −j5 Ω

Is = 0.2∠0◦ A ZL = jω L = j105 × 10−4 = j10 Ω

−j −j = 5 = − j5 Ω. ωC 10 × 2 × 10−6

ZC =

V V V + + = Is = 0.2 20 j10 − j5 V = 1.79e− j63.4◦ V. ◦

V 1.79e− j63.4 IR = = A. 20 20 1 (1.79)2 = 0.08 VA. SR = VI∗R = 2 2 × 20 V 1.79 − j153.4◦ IL = = e j10 10 1 1.79 − j63.4◦ 1.79 j153.4◦ ◦ SL = VI∗L = e × e = 0.16e j90 = 0 + j0.16 VA. 2 2 10 V 1.79 j26.6◦ IC = = e A. − j5 5 1 1 ◦ 1.79 − j26.6◦ ◦ SC = VI∗C = 1.79e− j63.4 × e = 0.32e− j90 = 0 − j0.32 VA. 2 2 5 ST = SR + SL + SC = 0.08 + j0.16 − j0.32 = 0.08 − j0.16 VA. For the source, ◦ ◦ 1 1 Ss = VI∗s = 1.79e− j63.4 × 0.2 = 0.179e− j63.4 2 2



= 0.179 cos 63.4◦ − j0.179 sin 63.4◦ = 0.08 − j0.16 VA. Hence ST = Ss .



Problem 8.14 Determine S for the RL load in the circuit of Fig. P8.14, given that Is = 4∠0◦ A, R1 = 10 Ω, R2 = 5 Ω, ZC = − j20 Ω, R = 10 Ω, and ZL = j20 Ω. R2

a R

R1

ZC

Is ZL b Load


Solution: V I

R2

a R

R1

ZC

Is ZL b

1 1 1 V + + = Is ZC R1 R2 + R + ZL 1 1 1 + + =4 V − j20 10 5 + 10 + j20 Solution gives

◦

V = 31.6 − j4.6 = 31.93e− j8.28 V. For RL load: 1 Vab I∗ 2 ∗ 1 (10 + j20) V = V× 2 15 + j20 15 + j20

S=

= (5 + j10)

|V|2 (31.93)2 ◦ j63.4◦ = 11.18e × = 18.24e j63.4 VA. 2 |15 + j20| 625



Problem 8.20 The apparent power entering a certain load Z is 250 VA at a power factor of 0.8 leading. If the rms phasor voltage of the source is 125 V at 1 MHz: (a) Determine Irms going into the load (b) Determine S into the load (c) Determine Z (d) The equivalent impedance of the load circuit should be of the form Z = R + jω L or Z = R − j/ωC. Determine the value of L or C, whichever is applicable. Solution: (a) From S = Vrms Irms , Irms =

S Vrms

=

250 = 2 A. 125

(b) pf = 0.8 leading means φz is negative. Hence,

φz = − cos−1 0.8 = −36.87◦ . S = S cos φz + jS sin φz

= 250[cos(−36.87◦ ) + j sin(−36.87◦ )] = (200 − j150) VA. (c) 2 Pav = Irms R

R=

Pav 200 = = 50 Ω. 2 Irms 4

Also, 2 Q = Irms X

X= Hence,

(d)

−j = − j37.5, or ωC C=

Q −150 = = −37.5 Ω. 2 Irms 4

Z = R + jX = (50 − j37.5) Ω.

1 1 = = 4.24 nF. 37.5ω 37.5 × 2π × 106



Section 8-4: Power Factor Problem 8.22 The RL load in Fig. P8.22 is compensated by adding the shunt capacitance C so that the power factor of the combined (compensated) circuit is exactly unity. How is C related to R, L, and ω in that case?

R C L


Solution: For the combined load, the impedance is 1 Z = (R + jω L) k jω C (R + jω L) ω−Cj = R + j ω L − ω1C =

= =

ω L − jR ω RC − j(1 − ω 2 LC)

(ω L − jR)[ω RC + j(1 − ω 2 LC)] [ω RC − j(1 − ω 2 LC)][ω RC + j(1 − ω 2 LC)]

[ω 2 RLC + R(1 − ω 2LC)] j[ω L(1 − ω 2 LC) − ω R2C] + ω 2 R2C2 + (1 − ω 2 LC)2 ω 2 R2C2 + (1 − ω 2 LC)2

For the pf to be unity, the imaginary component of Z has to be zero, which is realized if ω L(1 − ω 2 LC) − ω R2C = 0. or C=

L . R2 + ω 2 L2



Problem 8.23 The generator circuit shown in Fig. P8.23 is connected to a distant load via a long coaxial transmission line. The overall circuit can be modeled as in Fig. P8.23(b), in which the transmission line is represented by an equivalent impedance Zline = (5 + j2) Ω. a

10 Ω

Vs

c

+ _

ZL = (50 + j40) Ω Transmission line

b

d

(a) Transmission-line circuit

a

10 Ω

5Ω

j2 Ω

c

Transmission line Vs

+ _

ZL = (50 + j40) Ω

C

(b) Equivalent circuit

b

d Figure P8.23: Circuit for Problem 8.23.

(a) Determine the power factor of voltage source Vs . (b) Specify the capacitance of a shunt capacitor C that would raise the power factor of the source to unity when connected between terminals (a, b). The source frequency is 1.5 kHz. Solution: (a) The power factor of the source is the same as the phase of the impedance representing the entire circuit connected to Vs . Thus, ZT = 10 + (5 + j2) + (50 + j40) = (65 + j42) Ω, 42 = 32.87◦ , 65 pf = cos φZ = 0.84, lagging.

φZ = tan−1

(b) For the circuit to the right of terminals (a, b), the impedance—with a shunt capacitance C—is: −j Zab = k (55 + j42) ωC =

− jXC (55 + j42) 55 + j(42 − XC )



where XC = ω1C . Simplifying, Zab = =

42XC − j55XC · [55 − j(42 − XC)] 552 + (42 − XC)2

[42 × 55XC − 55XC(42 − XC )] − j[552 XC + 42XC(42 − XC )] 3025 + (42 − XC )2

For a pf of 1, the imaginary part of Zab should be zero, 552 XC + 42XC (42 − XC ) = 0. Solution gives XC = 114.02, or C=

1 1 = = 930.5 nF. ω XC 2π × 1.5 × 103 × 114.02



Problem 8.25 Use the power information given for the circuit in Fig. P8.25 to determine: (a) Z1 and Z2 (b) the rms value of Vs . 0.6 Ω

j0.4 Ω

1.2 Ω

+

+ _ Vs

Z2

Z1

Vrms = 440

0o V

_

Load Z1 : 24 kW @ pf = 0.66 leading Load Z2 : 18 kW @ pf = 0.82 lagging Figure P8.25: Circuit for Problem 8.25.

Solution: 0.6 Ω

I1

Is Vs

j0.4 Ω

1.2 Ω

V1

+ _

I2

Z1

Z2

+ Vrms = 440

_ (a) For load Z2 :

φZ2 = cos−1 0.82 = 34.9◦ .

Since φZ2 = φv2 − φi2 , and φv2 = 0,

=⇒

φi2 = −34.9◦ .

Pav2 = Vrms I2rms cos φZ2 18 × 103 = 440I2rms cos 34.9◦ , or I2rms = 49.88 A, and S2 = Vrms I2rms = 440 × 49.88 = 21.95 kVA.

Also, I2rms = 49.88∠−34.9◦ A. If Z2 = R2 + jX2 , Pav2 = I22rms R2

=⇒

R2 =

18 × 103 = 7.23 Ω (49.88)2

Q2 = S2 sin φZ2 = 21.95 × 103 sin 34.9◦ = 12.56 kVAR But Q2 = I22rms X2

=⇒

X2 =

12.56 × 103 = 5.05 Ω. (49.88)2

0o V

Hence,

Z2 = (7.23 + j5.05) Ω.

To determine Z1 , we first determine the voltage across it, V1rms : V1rms = (1.2 + j0.4 + Z2 )I2rms = (1.2 + j0.4 + 7.23 + j5.05)49.88e− j34.9

◦

= 500.8∠−2◦ V. For load Z1 :

φZ1 = − cos−1 0.66 = −48.7◦ S1 = I1rms =

Pav1 24 = = 36.36 kVA cos φZ1 0.66

36.36 × 103 S1 = 72.6 A = V1rms 500.8

φZ1 = φv1 − φi1

−48.7◦ = −2 − φi1

=⇒

φi1 = 46.7◦ .

I1rms = 72.6∠46.7◦ A. Pav1 = I21rms R1

=⇒

R1 =

24 × 103 = 4.55 Ω (72.6)2

Q1 = S1 sin φZ1 = 36.36 × 103 sin(−48.7◦ ) = −27.32 kVAR X1 =

−27.32 × 103 Q1 = = −5.18 Ω (72.6)2 I12rms

Hence,

Z1 = (4.55 − j5.18) Ω.

(b) Given I1 and I2 , we can now determine Is : Isrms = I1rms + I2rms ◦

= 72.6e j46.7 + 49.88e− j34.9

◦

= (90.7 + j24.3) A Vsrms = V1rms + 0.6Isrms ◦

= 500.8e− j2 + 0.6(90.7 + j24.3) = (554.9 − j2.9) V.

Problem 8.27 For the circuit in Fig. P8.27, choose the load impedance ZL so that the power dissipated in it is a maximum. How much power will that be? −j2 Ω

V1

1Ω

6

o

0

j2 Ω

V2

a

+ V _

ZL

2Ω

b


Solution: To determine Vs of the equivalent source circuit, we remove ZL and calculate Voc at terminals (a, b). −j2 Ω

1Ω

6

0o

I 1 V1

j2 Ω

V2

+

+ V _

Voc

_

2Ω

At node V2 : V2 − 6 V2 V2 − 6 + + =0 =⇒ V2 = 0.6(3 + j) V. − j2 2 1 + j2 V2 − 6 V1 − 6 I1 = = 1 + j2 1 Hence, Vs = V1 =

V2 − 6 0.6(3 + j) − 6 +6 = + 6 = 5.7∠18.4◦ V. 1 + j2 1 + j2

To determine ZTh at terminals (a, b), we suppress the 6-V source and simplify the circuit. The process leads to: Zs = ZTh = (0.6 + j0.2) Ω For maximum power transfer: ZL = Z∗s = (0.6 − j0.2) Ω, and Pav (max) =

1 |Vs |2 1 (4.9)2 = × = 6.78 W. 8 RL 8 0.6



Section 8-2 and 8-3: Average and Complex Power

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