Chapter 05

CHAPTER Exercises E5.1 (a) We are given v (t ) = 150 cos(200πt − 30 o ) The angular frequency is the coefficient of t so we have ω = 200π radian/s Then f = ω / 2π = 100 Hz T = / f = 10 ms Vrms = Vm / = 150 / = 106.1 V Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero Thus keeping in mind that ωt has units of radians, the positive peak occurs when ωtmax = 30 × π 180 ⇒ tmax = 0.8333 ms (b) Pavg = Vrms / R = 225 W (c) A plot of v(t) is shown in Figure 5.4 in the book E5.2 We use the trigonometric identity sin(z ) = cos(z − 90 o ) Thus 100 sin(300πt + 60 o ) = 100 cos(300πt − 30 o ) E5.3 ω = 2πf ≅ 377 radian/s T = / f ≅ 16.67 ms Vm = Vrms ≅ 155.6 V The period corresponds to 360 therefore ms corresponds to a phase angle of (5 / 16.67) × 360 o = 108o Thus the voltage is o v (t ) = 155.6 cos(377t − 108o ) E5.4 (a) V1 = 10∠0 o + 10∠ − 90 o = 10 − j 10 ≅ 14.14∠ − 45 o 10 cos(ωt ) + 10 sin(ωt ) = 14.14 cos(ωt − 45 o ) (b) I1 = 10∠30 o + 5∠ − 60 o ≅ 8.660 + j + 2.5 − j 4.330 ≅ 11.16 + j 0.670 ≅ 11.18∠3.44 o 10 cos(ωt + 30 o ) + sin(ωt + 30 o ) = 11.18 cos(ωt + 3.44 o ) (c) I2 = 20∠0 o + 15∠ − 60 o ≅ 20 + j + 7.5 − j 12.99 ≅ 27.5 − j 12.99 ≅ 30.41∠ − 25.28o 20 sin(ωt + 90 o ) + 15 cos(ωt − 60 o ) = 30.41 cos(ωt − 25.28o ) E5.5 The phasors are V1 = 10∠ − 30 o V2 = 10∠ + 30 o and V3 = 10∠ − 45 o v1 lags v2 by 60 o (or we could say v2 leads v1 by 60 o ) v1 leads v3 by 15 o (or we could say v3 lags v1 by 15 o ) v2 leads v3 by 75 o (or we could say v3 lags v2 by 75 o ) E5.6 (a) Z L = jωL = j 50 = 50∠90 o VL = 100∠0 o IL = VL / Z L = 100 / j 50 = 2∠ − 90 o (b) The phasor diagram is shown in Figure 5.11a in the book E5.7 (a) Z C = / jωC = − j 50 = 50∠ − 90 o VC = 100∠0 o IC = VC / Z C = 100 /( − j 50) = 2∠90 o (b) The phasor diagram is shown in Figure 5.11b in the book E5.8 (a) Z R = R = 50 = 50∠0 o VR = 100∠0 o IR = VR / R = 100 /(50) = 2∠0 o (b) The phasor diagram is shown in Figure 5.11c in the book E5.9 (a) The transformed network is: Vs 10∠ − 90 o = 28.28∠ − 135 o mA I= = Z 250 + j 250 i (t ) = 28.28 cos(500t − 135 o ) mA VR = RI = 7.07∠ − 135 o VL = jωLI = 7.07∠ − 45 o (b) The phasor diagram is shown in Figure 5.17b in the book (c) i(t) lags vs(t) by 45 o E5.10 The transformed network is: Z = = 55.47∠ − 56.31 o Ω / 100 + /( − j 50) + /( + j 200) V = ZI = 277.4∠ − 56.31 o V IC = V /( − j 50) = 5.547 ∠33.69 o A IL = V /( j 200) = 1.387∠ − 146.31 o A IR = V /(100) = 2.774∠ − 56.31 o A E5.11 The transformed network is: We write KVL equations for each of the meshes: j 100I1 + 100( I1 − I2 ) = 100 − j 200 I2 + j 100I2 + 100( I2 − I1 ) = Simplifying, we have (100 + j 100) I1 − 100I2 = 100 − 100 I1 + (100 − j 100) I2 = Solving we find I1 = 1.414∠ − 45 o A and I2 = 1∠0 o A Thus we have i1 (t ) = 1.414 cos(1000t − 45 o ) A and i2 (t ) = cos(1000t ) E5.12 (a) For a power factor of 100%, we have cos(θ ) = 1, which implies that the current and voltage are in phase and θ = Thus, Q = P tan(θ ) = Also I rms = P /[Vrms cos(θ )] = 5000 /[500 cos(0)] = 10 A Thus we have I m = I rms = 14.14 and I = 14.14∠40 o (b) For a power factor of 20% lagging, we have cos(θ ) = 0.2, which implies that the current lags the voltage by θ = cos−1 (0.2) = 78.46o Thus, Q = P tan(θ ) = 24.49 kVAR Also, we have I rms = P /[Vrms cos(θ )] = 50.0 A Thus we have I m = I rms = 70.71 A and I = 70.71∠ − 38.46o (c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a) Wiring costs would be lower for the load of part (a) E5.13 The first load is a 10 µF capacitor for which we have Z C = /( jωC ) = 265.3∠ − 90 o Ω θ C = −90 o I Crms = Vrms / Z C = 3.770 A PC = Vrms I Crms cos(θ C ) = QC =Vrms I Crms sin(θ C ) = −3.770 kVAR The second load absorbs an apparent power of Vrms I rms = 10 kVA with a power factor of 80% lagging from which we have θ = cos −1 (0.8) = 36.87 o Notice that we select a positive angle for θ because the load has a lagging power factor Thus we have P2 = Vrms I 2rms cos(θ ) = 8.0 kW and Q2 = Vrms I 2rms sin(θ ) = kVAR Now for the source we have: Ps = PC + P2 = kW Qs = QC + Q2 = 2.23 kVAR Vrms I srms = Ps + Qs2 = 8.305 kVA I srms = Vrms I srms /Vrms = 8.305 A power factor = Ps /(Vrms I srms ) × 100% = 96.33% E5.14 First, we zero the source and combine impedances in series and parallel to determine the Thévenin impedance Zt = 50 − j 25 + = 50 − j 25 + 50 + j 50 / 100 + / j 100 = 100 + j25 = 103.1∠14.04 o Then we analyze the circuit to determine the open-circuit voltage Vt = Voc = 100 × 100 = 70.71∠ − 45 o 100 + j 100 In =Vt / Zt = 0.6858∠ − 59.04 o E5.15 (a) For a complex load, maximum power is transferred for Z L = Zt* = 100 − j 25 = RL + jX L The Thévenin equivalent with the load attached is: The current is given by 70.71∠ − 45 o I= = 0.3536∠ − 45 o 100 + j 25 + 100 − j 25 The load power is PL = RL I rms = 100(0.3536 / ) = 6.25 W (b) For a purely resistive load, maximum power is transferred for RL = Zt = 100 + 252 = 103.1 Ω The Thévenin equivalent with the load attached is: The current is given by 70.71∠ − 45 o I= = 0.3456∠ − 37.98 o 103.1 + 100 − j 25 The load power is PL = RL I rms = 103.1(0.3456 / ) = 6.157 W E5.16 The line-to-neutral voltage is 1000 / = 577.4 V No phase angle was specified in the problem statement, so we will assume that the phase of Van is zero Then we have Van = 577.4∠0 o Vbn = 577.4∠ − 120 o Vcn = 577.4∠120 o The circuit for the a phase is shown below (We can consider a neutral connection to exist in a balanced Y-Y connection even if one is not physically present.) The a-phase line current is V 577.4∠0 o IaA = an = = 4.610∠ − 37.02o Z L 100 + j 75.40 The currents for phases b and c are the same except for phase IbB = 4.610∠ − 157.02o IcC = 4.610∠82.98o V I 577.4 × 4.610 P = Y L cos(θ ) = cos(37.02 o ) = 3.188 kW 2 VI 577.4 × 4.610 Q = Y L sin(θ ) = sin(37.02o ) = 2.404 kVAR E5.17 The a-phase line-to-neutral voltage is Van = 1000 / ∠ o = 577 ∠ o The phase impedance of the equivalent Y is ZY = Z ∆ / = 50 / = 16.67 Ω Thus the line current is V 577.4∠0 o = 34.63∠0 o A IaA = an = ZY 16.67 Similarly, IbB = 34.63∠ − 120 o A and IcC = 34.63∠120 o A Finally, the power is P = 3(I aA / ) Ry = 30.00 kW E5.18 Writing KCL equations at nodes and we obtain V1 V − V2 + = 1∠60 o 100 + j 30 50 − j 80 V2 V − V1 + = 2∠30 o j 50 50 − j 80 In matrix form, these become   1  +  100 + 30 50 − 80 j j    −  50 − j 80      V1  = 1∠60°    V2  2∠30°   +  j 50 50 − j 80  − 50 − j 80 The MATLAB commands are Y = [(1/(100+j*30)+1/(50-j*80)) (-1/(50-j*80)); (-1/(50-j*80)) (1/(j*50)+1/(50-j*80))]; I = [pin(1,60); pin(2,30)]; V = inv(Y)*I; pout(V(1)) pout(V(2)) The results are V1 = 79.98∠106.21o and V2 = 124.13∠116.30 o Answers for Selected Problems P5.4* ω = 1000π rad/s f = 500 Hz phase angle = θ = −60 o = − π radians T = ms Vrms = 7.071 V P =1W t peak = 0.3333 ms P5.6* v (t ) = 28.28 cos(2π10 4t − 72o ) V P5.12* Vrms = 10.61 V P5.13* Vrms = 3.808 A P5.23* P5.24* cos(ωt + 75 o ) − cos(ωt − 75 o ) + sin(ωt ) = 3.763 cos(ωt + 82.09 o ) v s (t ) = 141.4 cos (ωt − 45 o ) V2 lags V1 by 90 o Vs lags V1 by 45o Vs leads V2 by 45o P5.25* v (t ) = 10 cos (400πt + 30 o ) v (t ) = cos( 400πt + 150 o ) v (t ) = 10 cos (400πt + 90 o ) v (t ) lags v (t ) by 120 o v (t ) lags v (t ) by 60 o v (t ) leads v (t ) by 60 o P5.35* Z L = 200 π∠90 o VL = 10∠0 o IL = (1 20 π )∠ − 90 o iL (t ) = (1 20 π) cos(2000 πt − 90 o ) = (1 20 π) sin(2000 πt ) iL (t ) lags v L (t ) by 90 o P5.37* Z C = 15.92∠ − 90 o Ω VC = 10∠0 o IC = VC Z C = 0.6283∠90 o iC (t ) = 0.6283 cos(2000πt + 90 o ) = −0.6283 sin(2000πt ) iC (t ) leads v C (t ) by 90 o P5.42* I = 70.71∠ − 45 o mA VR = 7.071∠ − 45 o V VL = 7.071∠45 o V I lags Vs by 45 o P5.44* I = 4.472∠63.43 o mA VR = 4.472∠63.43 o V VC = 8.944 ∠ − 26.57 o V I leads Vs by 63.43o P5.46* ω = 500 : Z = 158.1∠ − 71.57 o ω = 1000 : Z = 50∠0 o ω = 2000 : Z = 158.1∠71.57 o P5.49* IR = 10∠0 o mA IL = 50∠ − 90 o mA IC = 50∠90 o mA The peak value of iL (t ) is five times larger than the source current! 10 P5.52* V = 8.944 ∠ − 26.56o V IR = 89.44 ∠ − 26.56o mA IC = 44.72∠63.44 o mA V lags Is by 26.56o P5.67* I = 15.11∠20.66 o P = 10 kW Q = −3.770 kVAR Apparent power = 10.68 kVA Power factor = 93.57% leading P5.69* This is a capacitive load P = 22.5 kW Q = −11.25 kVAR power factor = 89.44% apparent power = P + Q = 25.16 KVA Ps = 22 kW Qs = 13.84 kVAR P5.78* Apparent power = 26 kVA Power factor = 84.62% lagging P5.83* (a) I = 400 2∠ − 75.52 o (b) C = 1027 µF The capacitor must be rated for at least 387.3 kVAR I = 100∠0 o (c) The line current is smaller by a factor of with the capacitor in place, reducing I 2R losses in the line by a factor of 16 11 P5.87* (a) In = 1.789∠ − 26.57 o (b) Pload = 50 W (c) Pload = 47.21 W P5.91* Rload = 12.5 Ω C load = 106.1 µF P5.95* Z ∆ = 70.29∠ − 62.05 o Ω P5.96* VL = 762.1 V rms I L = 14.67 A rms P = 19.36 kW P5.99* IaA = 59.87 ∠0 o VAn = 322.44∠ − 21.80 o VAB = 558∠8.20 o IAB = 34.56∠30 o Pload = 26.89 kW Pline = 5.38 kW P5.105* V1 = 9.402∠29.58o V2 = 4.986∠111.45 o P5.107* I1 = 1.372∠120.96o I2 = 1.955∠136.22o 12 Practice Test T5.1 I rms = T T ∫ i (t )dt = 2 (3t ) dt = t = = 2.828 A ∫ 30 P = I rms R = 8(50) = 400 W T5.2 V = 5∠ − 45° + 5∠ − 30° = 3.5355 − j 3.5355 + 4.3301 − j 2.5000 V = 7.8657 − j 6.0355 = 9.9144∠ − 37.50° v (t ) = 9.914 cos(ωt − 37.50°) T5.3 (a) V1rms = 15 = 10.61 V (b) f = 200 Hz (c) ω = 400π radians/s (d) T = / f = ms (e) V1 = 15∠ − 45° and V2 = 5∠ − 30° V1 lags V2 by 15° or V2 leads V1 by 15° T5.4 I= V1 Vs 10∠0° 10∠0° = = = 0.7071∠ − 45° A R + jωL − j / ωC 10 + j 15 − j 14.14∠45° VR = 10 I = 7.071∠ − 45° V VC = − j 5I = 5.303∠ − 135° V T5.5 V2 VL = j 15I = 10.606∠45° V S = 21 VI* = 21 (440∠30°)(25∠10°) = 5500∠40° = 4213 + j 3535 VA P = Re(S) = 4213 W Q = Im(S) = 3535 VAR Apparent power = S = 5500 VA Power factor = cos(θv − θ I ) = cos(40°) = 76.6% lagging T5.6 We convert the delta to a wye and connect the neutral points with an ideal conductor ZY = Z ∆ / = + j / Ztotal = Z line + ZY = 0.3 + j 0.4 + + j 2.667 = 2.3 + j 3.067 13 Ztotal = 3.833∠53.13° IaA = T5.7 Van Ztotal = 208∠30° = 54.26∠ − 23.13° A 3.833∠53.13° The mesh equations are: j 10I1 + 15( I1 − I2 ) = 10∠45° − j 5I2 + 15( I2 − I1 ) = −15 In matrix form these become − 15   I1  10∠45° (15 + j 10) =  − 15 (15 − j 5)  I2   − 15   The commands are: Z = [(15+j*10) -15; -15 (15-j*5)] V = [pin(10,45); -15] I = inv(Z)*V pout(I(1)) pout(I(2)) 14 ... have: Ps = PC + P2 = kW Qs = QC + Q2 = 2.23 kVAR Vrms I srms = Ps + Qs2 = 8. 305 kVA I srms = Vrms I srms /Vrms = 8. 305 A power factor = Ps /(Vrms I srms ) × 100% = 96.33% E5.14 First, we zero... − 62 .05 o Ω P5.96* VL = 762.1 V rms I L = 14.67 A rms P = 19.36 kW P5.99* IaA = 59.87 ∠0 o VAn = 322.44∠ − 21.80 o VAB = 558∠8.20 o IAB = 34.56∠30 o Pload = 26.89 kW Pline = 5.38 kW P5. 105* V1