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

1 Section 8-2 and 8-3: Average and Complex Power Problem 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) = cos(377t 60 ) A. (b) v(t) = 25 cos(2 103 t + 40 ) V, i(t) = cos(2 103 t 10 ) A. (c) Vrms = 110 60 V, Irms = 3 45 A. (d) Vrms = 440 0 V, Irms = 75 A. (e) Vrms = 12 60 V, Irms = 2 30 A. Solution: (a). 100 . Vrms = e j30 V, 2. j60 . Irms = e V. 2. 100 . S = Vrms I rms = e j30 e j60 = 125e j30 (VA). 2 2. S = |S| = 125 VA. Pav = Re[S] = 125 cos 30 = W. Q = Im[S] = 125 sin 30 = VAR. s = v i = 30 + 60 = 30 (hence pf is lagging).. pf = cos 30 = (b). 25 . Vrms = e j40 V, 2.. Irms = e j10 A, 2.

2 S = Vrms I rms = j50 (VA), S = VA, s = 50 (lagging). Pav = cos 50 = W, Q = sin 50 = VAR, pf = cos 50 = lagging. (c).. S = Vrms I rms = 110e j60 3e j45 = 330e j15 VA. S = 330 VA, All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Pav = 330 cos 15 = W, Q = 330 sin 15 = VAR, pf = cos 15 = (lagging). (d).. S = Vrms I rms = 440 j75 = 220e j75 VA. S = 220 VA, Pav = 220 cos( 75 ) = W, Q = 220 sin( 75 ) = VAR, pf = cos( 75 ) = (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 ). All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Problem In the circuit of Fig. , is (t) = sin 105 t A, R = 20 , L = mH, and C = 2 F.

3 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 : Circuit for Problem Solution: V. IR IL IC. +. Is _. 20 j10 j5 . Is = 0 A. ZL = j L = j105 10 4 = j10 . j j ZC = = 5 = j5 . C 10 2 10 6. V V V. + + = Is = 20 j10 j5. V = V.. V IR = = A. 20 20. 1 ( )2. SR = VI R = = VA. 2 2 20. V . IL = = e j10 10. 1 . SL = VI L = e e = j90 = 0 + VA. 2 2 10. V . IC = = e A. j5 5. 1 1 . SC = VI C = e = j90 = 0 VA. 2 2 5. ST = SR + SL + SC = + = VA. For the source, 1 1 . Ss = VI s = = 2 2. All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press = cos sin . = VA. Hence ST = Ss . All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Problem Determine S for the RL load in the circuit of Fig.

4 , given that Is = 4 0 A, R1 = 10 , R2 = 5 , ZC = j20 , R = 10 , and ZL = j20 . R2 a R. ZC R1 Is ZL. b Load Figure : Circuit for Problem Solution: V I R2 a R. ZC R1 Is ZL. b . 1 1 1. V + + = Is ZC R1 R2 + R + ZL.. 1 1 1. V + + =4. j20 10 5 + 10 + j20. Solution gives . V = = V. For RL load: 1. S= Vab I . 2.. 1 (10 + j20) V. = V . 2 15 + j20 15 + j20. |V|2 ( )2 . = (5 + j10) 2. = = VA. |15 + j20| 625. All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Problem The apparent Power entering a certain load Z is 250 VA at a Power factor of 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 + j L or Z = R j/ C.

5 Determine the value of L or C, whichever is applicable. Solution: (a) From S = Vrms Irms , S 250. Irms = = = 2 A. Vrms 125. (b) pf = leading means z is negative. Hence, z = cos 1 = . S = S cos z + jS sin z = 250[cos( ) + j sin( )] = (200 j150) VA. (c). 2. Pav = Irms R. Pav 200. R= 2. = = 50 . Irms 4. Also, 2. Q = Irms X. Q 150. X= 2. = = . Irms 4. Hence, Z = R + jX = (50 ) . j (d) = , or C. 1 1. C= = = nF. 2 106. All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Section 8-4: Power Factor Problem The RL load in Fig. 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 in that case? R. C. L. Figure : Circuit for Problem Solution: For the combined load, the impedance is.

6 1. Z = (R + j L) k j C.. (R + j L) Cj =. R + j L 1C.. L jR. =. RC j(1 2 LC). ( L jR)[ RC + j(1 2 LC)]. =. [ RC j(1 2 LC)][ RC + j(1 2 LC)]. [ 2 RLC + R(1 2LC)] j[ L(1 2 LC) R2C]. = +. 2 R2C2 + (1 2 LC)2 2 R2C2 + (1 2 LC)2. For the pf to be unity, the imaginary component of Z has to be zero, which is realized if L(1 2 LC) R2C = 0. or L. C= . R2 + 2 L2. All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Problem The generator circuit shown in Fig. is connected to a distant load via a long coaxial transmission line. The overall circuit can be modeled as in Fig. (b), in which the transmission line is represented by an equivalent impedance Zline = (5 + j2) . 10 a c +. Vs _ ZL = (50 + j40) . Transmission line b d (a) Transmission-line circuit 10 a 5 j2 c Transmission line +.

7 Vs _ C ZL = (50 + j40) . b d (b) Equivalent circuit Figure : Circuit for Problem (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 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) , 42. Z = tan 1 = , 65. pf = cos Z = , lagging. (b) For the circuit to the right of terminals (a, b), the impedance with a shunt capacitance C is: . j Zab = k (55 + j42). C. jXC (55 + j42). =. 55 + j(42 XC ). All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press where XC = 1C.

8 Simplifying, 42XC j55XC. Zab = [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 = , or 1 1. C= = = nF. XC 2 103 All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press Problem Use the Power information given for the circuit in Fig. to determine: (a) Z1 and Z2. (b) the rms value of Vs .. +. +. _ Vs Z1 Z2 Vrms = 440 0o V. _. Load Z1 : 24 kW @ pf = leading Load Z2 : 18 kW @ pf = lagging Figure : Circuit for Problem Solution: V1 . Is I1 I2 +. +. Vs _ Vrms = 440 0o V. Z1 Z2. _. (a) For load Z2 : Z2 = cos 1 = . Since Z2 = v2 i2 , and v2 = 0, = i2 = . Pav2 = Vrms I2rms cos Z2. 18 103 = 440I2rms cos , or I2rms = A, and S2 = Vrms I2rms = 440 = kVA.

9 Also, I2rms = A. If Z2 = R2 + jX2 , 18 103. Pav2 = I22rms R2 = R2 = = . ( )2. Q2 = S2 sin Z2 = 103 sin = kVAR. But 103. Q2 = I22rms X2 = X2 = = . ( )2. Hence, Z2 = ( + ) . To determine Z1 , we first determine the voltage across it, V1rms : V1rms = ( + + Z2 )I2rms . = ( + + + ) = 2 V. For load Z1 : Z1 = cos 1 = . Pav1 24. S1 = = = kVA. cos Z1 S1 103. I1rms = = = A. V1rms Z1 = v1 i1. = 2 i1 = i1 = . I1rms = A. 24 103. Pav1 = I21rms R1 = R1 = = . ( )2. Q1 = S1 sin Z1 = 103 sin( ) = kVAR. Q1 103. X1 = = = . I12rms ( )2. Hence, Z1 = ( ) . (b) Given I1 and I2 , we can now determine Is : Isrms = I1rms + I2rms . = + = ( + ) A. Vsrms = V1rms + . = j2 + ( + ) = ( ) V. Problem For the circuit in Fig. , choose the load impedance ZL so that the Power dissipated in it is a maximum. How much Power will that be? j2.

10 1 V1 j2 . V2. a +. 6 0o V _ ZL 2 . b Figure : Circuit for Problem Solution: To determine Vs of the equivalent source circuit, we remove ZL and calculate Voc at terminals (a, b). j2 . 1 I 1 V1 j2 . V2. + +. 6 0o V _ Voc 2 . _. At node V2 : V2 6 V2 V2 6. + + =0 = V2 = (3 + j) V. j2 2 1 + j2. V2 6 V1 6. I1 = =. 1 + j2 1. Hence, V2 6 (3 + j) 6. Vs = V1 = +6 = + 6 = 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 = ( + ) . For maximum Power transfer: ZL = Z s = ( ) , and 1 |Vs |2 1 ( )2. Pav (max) = = = W. 8 RL 8 All rights reserved. Do not reproduce or distribute. 2009 National Technology and Science Press