chapter 10 - Chapter 10 – Sinusoidal Steady-State...

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Unformatted text preview: Chapter 10 – Sinusoidal Steady-State Analysis Exercises Ex. 10.3-1 (a) T = 2π / ω = 2π / 4 (b) v leads i by 30 − (−70) = 100° Ex. 10.3-2 v ( t ) = 3 cos 4 t + 4 sin 4 t = ( 3 )) = 5 cos(4 t −53° ) ( (3) 2 + (4) 2 cos 4 t − tan −1 4 Ex. 10.3-3 12 i ( t ) = −5 cos 5t + 12 sin 5t = (−5) 2 + (12) 2 cos 5 t − 180 + tan −1 = 13 cos (5 t −112.6°) −5 Ex. 10.4-1 KCL: is ( t ) = v( t ) d I v (t ) + = m cos ω t dt RC C v( t ) d + C v (t ) ⇒ R dt Try v f ( t ) = A cos ω t + B sin ω t & plug into above differential equation to get −ω A sin ω t + ω B cos ω t + 1 I ( A cos ω t + B sin ω t ) = m cos ω t RC C Equating sin ω t & cos ω t terms yields R Im ω R2 C Im and B = A= 1+ω 2 R 2 C 2 1+ω 2 R 2 C 2 Therefore v f (t ) = R Im ω R2 C Im cos ω t + sin ω t = 1+ω 2 R 2 C 2 1+ω 2 R 2 C 2 R Im 1+ ω R C 2 2 2 cos ω t − tan −1 (ω RC ) 10-1 Ex. 10.4-2 KVL : − 10 + j 3 I + 2 I = 0 ⇒ I = 10 = 2+ j 3 10∠0 13 ∠56.3 = 10 ∠ − 56.3 A 13 Therefore i (t ) = 10 cos (3 t − 56.3 ) A 13 Ex. 10.5-1 10 = 4.24 e− j 45 = 3− j 3 2.36 e j 45 Ex. 10.5-2 j 32 32e j 90 32 j (90 -111) = = = 3.75 e − j 21 e −3+ j 8 8.54 e j111 8.54 Ex. 10.6-1 (a) i = 4 cos(ω t − 80 ) = Re{4 e j ω t e − j 80° } ⇒ I = 4 e − j 80° = 4∠ − 80° A (b) i = 10 cos(ω t + 20° ) = Re{10 e j ω t e j 20° } ⇒ I = 10e j 20° = 10∠20° (c) i = 8 sin (ω t − 20 ) = 8 cos (ω t − 110 ) = 8 Re{e j ω t e − j110° } ⇒ I = 8e − j110° = 8∠ − 110° A Ex. 10.6-2 (a) V = 10∠ − 140° = 10 e − j140° V ⇒ v(t ) = Re{10 e − j140° e j ω t } = 10 cos (ω t − 140°) V (b) V = 80 + j 75 = 109.7∠43.2° = 109.7 e j 43.2° ⇒ v (t ) = Re{109.7 e j 43.2° e jω t } = 109.7 cos(ω t + 43.2°) V 10-2 Ex. 10.6-3 d v + v = 10 cos 100 t dt ( 0.01)( j 100 )V + V =10 0.01 10 =7.071 ∠− 45° 1+ j v = 7.071 cos 100 t V V= Ex. 10.6-4 { vs = 40 cos 100t = Re 4 e j100 t KVL: i (t ) + 10 × 10 −3 di (t ) 1 + dt 5×10 −3 ∫ t −∞ } i (t ) dt = vS Assume i (t ) = A e j100 t where A is complex number to be determined. Plugging into the differential equation yields Ae j100 t + j A e j100 t + (− j 2 A)e j100 t = 4 e j100 t ⇒ A= 4 = 2 2 e j 45° 1− j In the time domain: { } { } i (t ) = Re 2 2 e j100 t e j 45° = Re 2 2 e j (100 t -45°) = 2 2 cos (100 t + 45° ) A Ex. 10.7-1 (a) v = R i = 10 (5 cos 100 t ) = 50 cos 100 t (b) di v = L = 0.01[5(−100) sin 100 t ] = −5 sin 100 t = 5 cos (100 t + 90° ) V dt (c) 1 v = ∫ i dt = 103 ∫ 5 cos 100 t dt = 50 sin 100 t = 50 cos (100 t − 90°) V C Ex. 10.7-2 i=C dv = 10 × 10−6 [100(−500) sin (500 t + 30°)] dt = − 0.5 sin (500 t +30°) = 0.5 sin (500 t + 210°) = 0.5 cos(500t +120°) A 10-3 Ex. 10.7-3 From Figure E10.7-3 we get i (t ) = I m sin ω t = I m cos(ω t − 90°) A ⇔ I = I m ∠ − 90°A v(t ) = Vm cos ω t ⇔ V = Vm ∠0° V The voltage leads the current by 90° so the element is an inductor: Z eq = V ∠0° V V =m = m ∠90° Ω I m ∠−90° Im I Also Z eq = j ω L = ω L ∠90° ⇒ ω L = Vm Im ⇒ L= Vm ω Im Ex. 10.8-1 1 ZR = 8 Ω, ZC = = 2.4 j 2.4 = = − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω, j j× j 1 12 ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V. j5 Ex. 10.8-2 ZR = 8 Ω, ZC = 1 = j4 4 = = − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω, j j× j 1 12 ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° V. j3 10-4 Ex 10.9-1 V1 (ω ) = j10 5 e − j 90 = 3.9 e − j 51 8 + j10 V 2 (ω ) = j 20 5 e − j 90 = 5.68 e − j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 Ex 10.9-2 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 10-5 Ex. 10.10-1 KCL at Va: Va V −V + a b =1 4− j 2 − j10 (4 − j12) Va + (−4 + j 2) Vb = −20 − j 40 KCL at Vb: Vb − Va V + b + 0.5∠ − 90° = 0 ⇒ (−2 − j 4) Va + (2 − j 6) Vb = 10 + j 20 − j10 2+ j 4 Cramer’s rule yields: (−20− j 40) (−4+ j 2) (10+ j 20) (2 − j 6) −200+ j100 = = 5∠296.5° V V= a (4 − j12) (−4+ j 2) −80− j 60 (−2− j 4) (2- j 6) Therefore v (t ) = 5 cos (100 t + 296.5° ) = 5 cos (100 t − 63.5° ) V a Ex. 10.10-2 The mesh equations are: j15 I1 + 10 (I1 − I 2 ) = 20 ⇒ (10 + j15) I1 − 10 I 2 = 20 − j 5 I 2 +10(I 2 − I1 ) = −30∠−90° ⇒ −10I1 + (10− j 5) I 2 = j 30 Cramer’s rule yields: 20 −10 j 30 10− j 5 200+ j 200 = = 2.263∠ − 8.1° A I1 = 10 + j15 −10 75+ j100 10 − j 5 −10 Next VL = ( j15) I1 = (15∠90°) (2.263∠ −8.1°) = 24 2∠82° V Therefore vL (t ) = 24 2 cos (ω t + 82°) V 10-6 Ex. 10.10-3 The mesh equations are: (10+ j 50) I1 −10 I 2 = j 30 −10 I1 + (10− j 20) I 2 + j 20 I 3 = j 50 j 20 I 2 + (30− j10) I 3 = 0 Solving the mesh equations gives: I1 = − 0.87 − j 0.09 A, I 2 = −1.32+ j 1.27 A, I 3 = 0.5+ j 1.05 A Then Va = 10 (I1 − I 2 ) = 14.3∠ − 72° V and Vb = Va + j 50 = 36.6 ∠83° V Ex 10.11-1 V1 = V2 = j10 5 e − j 90 = 3.9 e− j 51 8 + j10 j 20 5 e − j 90 = 5.68 e− j 90 j 20 − j 2.4 Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e− j 90 = 3.58 e j 47 10-7 Zt = 8 ( j10 ) − j 2.4 ( j 20 ) + = 4.9 + j 1.2 8 + j10 − j 2.4 + j 20 Ex 10.11-2 V1 (ω ) = V 2 (ω ) = j10 5 e − j 90 = 3.9 e− j 51 8 + j10 j 20 5 e − j 90 = 5.68 e− j 90 j 20 − j 2.4 V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V1 (ω ) = V 2 (ω ) = 8 ( j6) 4 e j15 = 19.2 e j 68 8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75 j12 − j 4 V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t - 22° ) 10-8 Ex. 10.11-3 Use superposition. First, find the response to the voltage source acting alone: Z eq = − j 10⋅10 = 5(1 − j ) Ω 10− j10 Replacing the parallel elements by the equivalent impedance. The write a mesh equation : −10 + 5 I1 + j15 I1 + 5(1 − j) I1 = 0 ⇒ I1 = 10 = 0.707∠ − 45° A 10+ j10 Therefore: i1 (t ) = 0.707 cos(10 t − 45° ) A Next, find the response to the dc current source acting alone: Current division: Using superposition: I2 = − 10 × 3 = −2 A 15 i (t ) = 0.707 cos(10 t − 45°) − 2 A Ex. 10.12-1 ω2 = 1 1 = = 106 −3 −3 LC (1×10 )(1×10 ) ⇒ ω = 1000 rad sec Ex. 10.12-2 Diagram drawn with relative magnitudes arbitrarily chosen: 10-9 Ex. 10.12-3 Two possible phasor diagrams for currents: In both cases: I CL = I LC = ( 25 )−(15 ) 2 2 = 20 A In the first case: I LC = I L − I C ⇒ I C = 6 − 20 = −14 A That isn’t possible. Turning to the second case: I C L = I C − I L ⇒ I C = 20 + 6 = 2 6 A Ex. 10.14-1 Z1 = 1 1 R1 X 1 ( X 1 − jR1 ) and R1 = 1 kΩ , X 1 = = = 1 kΩ 2 2 ω C1 (1000 )(10−6 ) R1 + X 1 Z1 = (1)(1)(1− j1) 1 1 = − j kΩ and Z 2 = R 2 = 1 kΩ 1+1 2 2 Vo −1 Z =− 2 = = −1− j 1 1 Vs Z1 −j 2 2 10-10 Problems Section 10-3: Sinusoidal Sources P10.3-1 (a) i (t ) = 2 cos(6t + 120° ) + 4 sin(6t − 60° ) = 2 (cos 6t cos 120° −sin 6t sin 120° ) + 4 (sin 6t cos 60° − cos 6t sin 60° ) = 2.46 cos 6t + 0.27 sin 6t = 2.47 cos(6t − 6.26° ) (b) v(t ) = 5 2 cos 8t + 10 sin(8t + 45° ) = 5 2 cos 8t +10[sin 8t cos 45° + cos 8t sin 45° ] = 10 2 cos 8t + 5 2 sin 8t v(t ) = P10.3-2 250 cos(8t − 26.56° ) = 5 10 sin(8t + 63.4° ) V 2π 2π = = 6283 rad sec T 1×10−3 v(t ) = Vm sin(ω t + φ ) = 100 sin(6283 t + φ ) ω = 2π f = v(0) = 10 = 100 sin φ ⇒ φ = sin −1 (0.1) = 6° v(t ) = 100 sin(6283 t + 6°) V P10.3-3 ω 1200π = = 600 Hz 2π 2π i (2 × 10−3 ) = 300 cos(1200 π (2 × 10−3 ) + 55°) = 3 cos(2.4π + 55°) f= 180° −3 2.4π × = 432° ⇒ i (2 × 10 ) = 300 cos(432°+55°) = 300 cos(127°) = −180.5 mA π P10.3-4 10-11 P10.3-5 A = 18 V T = 18 − 2 = 16 ms ω= 2π 2π = = 393 rad/s T 0.016 16 = 18 cos (θ ) ⇒ θ = 27° v ( t ) = 18 cos ( 393 t + 27° ) V P10.3-6 A = 15 V T = 43 − 21 = 32 ms ω= 2π 2π = = 196 rad/s T 0.032 8 = 15 cos (θ ) ⇒ θ = 58° v ( t ) = 15 cos (196 t + 58° ) V 10-12 Section 10-4: Steady-State Response of an RL Circuit for a Sinusoidal Forcing Function P10.4-1 L di + R i = − vs dt Try i f = A cos 300 t + B sin 300 t then equating coefficients gives Then ⇒ di f dt di + 120 i = −400 cos 300 t dt = −300 A sin 300 t + 300 B cos 300 t . Substituting and −300 A+120 B = 0 A = − 0.46 B = −1.15 300 B +120 A = −400 i (t ) = −0.46 cos 300 t − 1.15 sin 300 t = 1.24 cos (300 t − 68°) A P10.4-2 v dv dv +C =0 ⇒ + 500 v = 500 cos 1000 t dt dt 2 dv f Try v f = A cos 1000 t + B sin 1000 t then = −1000 A cos 1000 t + 1000 B cos 1000 t . dt Substituting and equating coefficients gives −is + −1000 A+ 500 B =0 ⇒ 1000 B +500 A=500 Then A = 0.2 B = 0.4 v (t ) = 0.2 cos 1000 t + 0.4 sin 1000 t = 0.447 cos (1000t − 63°) V P10.4-3 ( j 4) (.05) = j (0.2) I (ω ) = 12 e j 45° ~ 12 e j 45° j 45° = ( 2 ⋅ 10 − 3 ) e ⇒ i (t ) = 2 cos (4 t + 45°) mA 6000 + j (0.2) 6000 10-13 Section 10.5: Complex Exponential Forcing Function P10.5-1 (5∠36.9° ) (10∠ −53.1° ) 50∠−16.2° 10∠ −16.2° = = = 2 5∠10.36° ° (4 + j3)(6− j8) 10− j5 5∠− 26.56 P10.5-2 3 2∠− 45° 3 5∠ + 81.87° 4− j 3+ = 5∠ + 81.87°[4 − j 3 + ∠ − 36.87°] 5 5 2∠−8.13° = 5∠+81.87° (4.48− j 3.36) = 5∠+81.87° (5.6∠−36.87°) = 28∠+ 45°= 14 2 + j14 2 P10.5-3 A*C* (3− j 7) 5e − j 2.3 = = 0.65 − j 6.31 B 6 e j15 P10.5-4 (6∠120° ) (−4 + j 3 + 2e j15 ) = −12.1 − j 21.3 ⇒ a =−12.1 and b =−21.3 P10.5-5 3− b (a) j tan −1 2 2 j120 −4 Ae = −4 + j (3 − b) = 4 + (3−b) e 3−b ° 120 = tan −1 ⇒ b = 3 + 4 + tan (120 ) = −3.93 −4 A = 42 + (3−b) 2 = 42 + (3− (−3.93)) 2 = 8.00 (b) −4 + 8 cos θ + j (b + 8 sin θ ) = 3e − j120 = − 1.5 − j 2.6 2.5 −4+8 cos θ = −1.5 ⇒ θ = cos −1 = 72° 8 b + 8 sin (72° ) = − 26 ⇒ b = −10.2 (c) −10 + j 2a = A e j 60 = A cos 60° − j A sin 60 A= −10 −20 sin 60° = −20 and a = = −8.66 cos 60° 2 10-14 P 10.5-6 d 5 0.1 v + v = cos 2 t ⇒ dt d v + 2 v = 2 cos 2 t dt Replace the real excitation by a complex exponential excitation to get d v + 2 v = 2 e j 2t dt Let ve = A e j 2t so d ve = j 2 A e j 2t and dt d ve + 2 ve = 2 e j 2t dt j 2 A e j 2 t + 2 A e j 2t = 2 e j 2 t ⇒ ( j 2 + 2 ) A e j 2t = 2 e j 2 t ⇒ A= 2 1 = ∠ − 45° 2 + j2 2 1 j ( 2t −45° ) 1 − j 45° j 2t ve = e e e = 2 2 so Finally v ( t ) = Re ( ve ) = 1 cos ( 2t − 45° ) V 2 P 10.5-7 d d2 0.45 v + v + 0.15 2 v = 4 cos 5 t ⇒ dt dt d2 d 20 80 v+3 v+ v= cos 5 t 2 dt dt 3 3 Replace the real excitation by a complex exponential excitation to get d2 dt 2 v+3 d 20 80 j 5t v+ v= e dt 3 3 d d2 ve = j 5 A e j 5 t , and 2 ve = −25 A e j 5 t dt dt 2 d d 20 80 j 5t 20 80 j 5t v+3 v+ v= e ⇒ − 25 A e j 5t + 3 j 5 A e j 5t + A e j 5t = e 2 dt 3 3 3 3 dt 80 20 80 j 5t 80 j 5t 3 e ⇒ A= = = 1.126∠ − 141 −25 + j15 + A e = 20 −55 + j 45 3 3 −25 + j15 + 3 Let ve = A e j 5 t so ( so ( ) ( ) ) j 5t −141° ) ve = 1.126 e− j141° e j 5t = 1.126 e ( Finally v ( t ) = Re ( ve ) = 1.126 cos ( 2t − 141° ) V 10-15 Section 10-6: The Phasor Concept P10.6-1 Apply KVL 6i+2 d i − 15 cos 4 t = 0 dt or 2 d i + 6 i = 15 cos 4 t dt j 4 t +θ ) Now use i = I m Re{e ( } and 15 cos 4 t = 15 Re{e 4 t } to write 2 ( )( ) d j 4 t +θ j 4 t +θ I m Re{e ( ) } + 6 I m Re{e ( ) } = 15 Re{e 4 t } dt d Re 2 ( I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ ) = Re{15 e 4 t } dt { } Re 2 ( j 4 I m e j 4 t e jθ ) + 6 ( I m e j 4 t e jθ ) = Re{15 e 4 t } j8 ( I m e jθ ) + 6 ( I m e jθ ) = 15 I m e jθ = 15 15 = = 1.5∠ − 53° 6 + j8 10∠53° i ( t ) = 1.5 cos ( 4 t − 53° ) A Finally v (t ) = 2 d d i ( t ) = 2 (1.5 cos ( 4 t − 53° ) ) = 3 ( −4 sin ( 4 t − 53° ) ) dt dt = −12 ( cos ( 4 t − 143° ) ) = 12 cos ( 4 t + 37° ) V 10-16 P10.6-2 Apply KCL at node a: v − 4 cos 2 t d + 0.25 v + i = 0 1 dt Apply KVL to the right mesh: 4i + 4 d d i − v = 0 ⇒ v = 4 i + 4 iL dt dt After some algebra: d2 d i + 5 i + 5 i = 4 cos 2 t 2 dt dt j 2 t +θ ) } and 4 cos 2 t = 4 Re{e 2 t } to write Now use i = I m Re{e ( d2 d j 2 t +θ j 2 t +θ j 2 t +θ I Re{e ( ) } + 5 I m Re{e ( ) } + 5 I m Re{e ( ) } = 4 Re{e 2 t } 2m dt dt d2 d j 2 t +θ j 2 t +θ j 2 t +θ Re 2 I m e ( ) + 5 I m e ( ) + 5 I m e ( ) = Re{4 e 2 t } dt dt { } Re −4 e jθ I m e j 2 t + 5 ( j 2 e jθ I m e j 2 t ) + 5 e jθ I m e j 2 t = Re{4 e2 t } −4 e j θ I m + 5 ( j 2 e j θ I m ) + 5 e j θ I m = 4 I m e jθ = 4 4 4 = = = 0.398∠ − 84° −4 + 5 ( j 2 ) + 5 1 + j 10 10.05∠84 i ( t ) = 0.398 cos ( 2 t − 85° ) A 10-17 P10.6-3 VS = 2∠ − 90° V Z R = R; Z C = −j −j = = − j 16000 Ω ω C (500)(0.125×10−6 ) − j 16000 (16000∠−90° )( 2∠−90° ) = 1.25∠ − 141° V V (ω ) = ( 2∠−90° ) = 25612∠−39° 20000 − j 16000 therefore v(t) = 1.25 cos (500t −141° ) V Section 10-7: Phasor Relationships for R, L, and C Elements P10.7-1 P10.7-2 10-18 P10.7.3 P10.7-4 10-19 P10.7-5 (a) v = 15 cos (400 t + 30°) V i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V v leads i by 90° ⇒ element is an inductor v 15 Z L = peak = = 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH 3 ipeak (b) i leads v by 90° ⇒ the element is a capacitor (c) 8 1 1 =4= = ⇒ C = 277.77 µ F ω C 900 C ipeak 2 v = 20 cos (250 t + 60°) V Zc = vpeak = i = 5 sin (250 t +150°) =5 cos (250 t + 60°) A Since v & i are in phase ⇒ element is a resistor v 20 ∴ R = peak = =4Ω 5 ipeak P10.7-6 V1 = 150 cos(−30°) + j150 sin(−30°) = 130 − j 75 V V2 = 200 cos 60°+ j 200 sin 60° = 100+ j173 V V = V1 + V2 = 230+ j 98 = 250∠23.1° V Thus v(t ) = v1 (t ) + v2 (t ) = 250 cos (377 t + 23.1°) V 10-20 Section 10-8: Impedance and Admittance P10.8-1 ω = 2π f = 2π (10 ×103 ) = 62830 rad sec Z R = R = 36 Ω ⇔ YR = 1 1 = 0.0278 S = Z R 36 Z L = jω L = j (62830)(160×10−6 ) = j10 Ω ⇔ YL = ZC = 1 = − 0.1 j S ZL −j −j 1 = = − j 16 Ω ⇔ YC = = 0.0625 j S −6 ZC ω C (62830)(1×10 ) Yeq = YR + YL + YC = 0.0278 − j0.00375 = 0.027 ∠9° S Z eq = P10.8-2 Z= 1 = 36.5∠ 9° = 36 − j5.86 Ω Yeq V −10 ∠40° = = − 5000∠ − 155°Ω = 4532 + 2113 j = R + j ω L −I 2×10−3 ∠195° so R =4532 Ω and L = 2113 ω = 2113 = 1.06 m H 2×106 P10.8-3 j L R ( R + jω L) −j C ωC = Z(ω ) = ω C j 1 − + ( R + jω L) R + j ω L − ωC ωC − 1 R L −j R − j ω L − ωC C ω C = 2 1 R 2 + ω L − ωC = Z(ω ) will be purely resistive when 1 R2 L 1 RL R − + ω L− ω L− − j ωC ωC C ωC C ωC 1 R 2 + ω L − ωC 2 10-21 R2 L 1 1 R 2 + ω L− − = 0 ⇒ ω = CL L ωC C ωC 2 when R =6 Ω , C = 22 µ F, and L = 27 mH, then ω = 1278 rad/s. P10.8-4 R Zc R R + j (ω L −ω R 2 C +ω 3 R 2 L C 2 ) jω C = jω L + = 1 R +Z c 1+ (ω R C ) 2 R+ jω C Set real part equal to 100 Ω to get C Z = ZL + R = 100 ⇒ C = 0.158 µF 1+ (ω R C ) 2 Set imaginary part of numerator equal to 0 to get L ( ω = 2π f = 6283 rad sec ) L − R 2C + ω 2 R 2 LC 2 = 0 ⇒ L = 0.1587 H P10.8-5 Z L = j ω L = j (6.28×106 ) (47×10−6 ) = j 300 Ω 1 ( 300 + j 300 ) jω C Z eq = Z c ||(Z R +Z L ) = = 590.7 Ω 1 + 300+ j 300 jω C 300+300 j 590.7 = ⇒ 590.7 −(590.7)(300 ω C ) + j (590.7)(300ω C ) = 300 + j 300 1+300 j ω C −300 ω C ( Equating imaginary terms ω =2π f = 6.28×106 rad sec ) (590.7) (300ω C ) = 300 ⇒ C = 0.27 nF 10-22 Section 10-9: Kirchhoff’s Laws Using Phasors P10.9-1 (a) (b) (c) P10.9-2 Z1 =3+ j 4 = 5∠53.1° Ω and Z 2 =8− j8 = 8 2 ∠− 45° Ω Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω I= 100∠0° 100 100 = = ∠20.0° ⇒ Z1 +Z 2 11.7 ∠− 20° 11.7 i (t ) = 8.55 cos (1250 t + 20.0°) A V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125° = ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 ) = ( 5.23 + 0.91) + j ( 5.62 − 1.30 ) = 6.14 + j 4.32 = 7.51∠35° v1 ( t ) = 7.51 cos ( 2 t + 35° ) V P10.9-3 I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠196° i ( t ) = 460 cos (2 t + 196°) mA P10.9-4 Vs = 2 ∠30° V and I = 2 ∠30° = 0.185 ∠ − 26.3° A 6+ j12+ 3 / j i (t ) = 0.185 cos (4 t − 26.3°) A 10-23 P10.9-5 j 15 = j (2π ⋅ 796) (3 ⋅10−3 ) 12 = 0.48 ∠ − 37° A 20 + j15 i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A I= P10.9-6 Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A 8 I B ∠−51.87° Z1 = = = = 2 ∠−15° Is Z1 + Z 2 8+ j 3L 8 ∠0° 3L 82 + (3L) 2 ∠ tan −1 8 Equate the magnitudes and the angle. 3L angles: + 36.87 = + tan −1 ⇒ L = 2 H 8 8 B = ⇒ B =1.6 magnitudes: 64+ 9 L2 2 P10.9-7 The voltage V can be calculated using Ohm's Law. V = (1.72 ∠ - 69°) (4.24∠45°) = 7.29∠ - 24° V The current I can be calculated using KCL. 10-24 I = (3.05 ∠ - 77°) - (1.72∠ - 69°) = 1.34∠ - 87° A Using KVL to calculate the voltage across the inductor and then Ohm's Law gives: j 2L = 24 - 4(1.34∠-87°) ⇒ L=4 H 3.05∠-77° P10.9-8 10 10 V10 = Vs = 20∠0° 10 2∠− 45° 10 − j10 = 0 2∠45° v10 (t ) = 10 2 cos (100 t + 45°) V P10.9-9 (a) (b) 160 ∠0° 160 ∠0° = (−1326) (300 + j 37.7) 303 ∠−5.9° − j1326 + 300 + j 37.7 = 0.53 ∠5.9° A i(t ) = 0.53 cos (120π t +5.9°) A I= I= 160∠0° 160∠0° = (− j199)(300 + j 251) 256∠−59.9° − j199+ 300+ j 251 = 0.625∠59.9° A i(t ) =0.625 cos (800π t +59.9°) A 10-25 Section 10-10: Node Voltage and Mesh Current Analysis Using Phasors P10.10-1 Draw frequency domain circuit and write node equations: VA VA − VC + = 0 ⇒ (2 + j )VA − 2VC = j 20 10 j5 VC − VA VC KCL at C: + − (1+ j ) = 0 ⇒ 4VA + VC = 20− j 20 j5 − j4 KCL at A − 2 + Solve using Cramers rule: (2 + j ) j 20 4 20 − j 20 60 − j100 116.6 ∠−59° Vc = = = = 11.6 ∠ − 64.7° V (2 + j ) −2 10 + j 101 ∠5.7° 4 1 P10.10-2 KCL: V V V (V −100) + + = 0 ⇒ V = 57.6 ∠22.9° V + − j125 j80 250 150 IS = 100− V = 0.667 − 0.384 ∠22.9° = 0.347 ∠ − 25.5° A 150 IC = V = 0.461 ∠112.9° A 125 ∠−90° 10-26 IL = V = 0.720 ∠− 67.1° A 80∠90° IR = V = 0.230∠22.9° A 250 P10.10-3 KCL at node A: Va Va − Vb + =0 200 j 100 (1) KCL at node B: Vb − Va V V −1.2 + b+b =0 j 100 − j 50 j 80 1 3 ⇒ Va = Vb − 4 2 (2) Substitute Eqn (2) into Eqn (1) to get Vb = 2.21 ∠ − 144° V Then Eqn (2) gives Va = ( 0.55∠−144° ) − 1.5 = 1.97∠ − 171° V Finally va (t ) = 1.97 cos (4000 t − 171°) V and vb (t ) = 2.21 cos (4000 t − 144°) V 10-27 P10.10-4 ω = 104 rad s I s = 20∠53° A The node equations are: 1 1 j 1 KCL at a: + + Va + − Vb = 20∠53.13° 20 40 60 40 1 1 j j KCL at b: − Va + − + Vb − j Vc = 0 80 40 40 40 80 −j 1 j KCL at c: Vb + + Vc = 0 80 40 80 Solving thes equation yields Va = 2 ⋅ 240∠45° V ⇒ P10.10-5 va (t ) = 339.4 cos (ω t + 45°) V vs = sin ( 2π ⋅ 400 t ) V R = 100 Ω LR = 40 mH 40 mH LS = 60 mH door opened door closed With the door open VA − VB = 0 since the bridge circuit is balanced. With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω. The node equations are: KCL at node B: VB − VC VB j100.5 + = 0 ⇒ VB = VC R j100.5+100 Z LR KCL at node A : VA − VC VA + =0 Z LS R 10-28 Since VC = Vs =1 V VB =0.709∠44.86° V and VA = 0.833∠33.55 V Therefore VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040 = 0.195∠ − 11.83° V P10.10-6 The node equations are: V1 −(−1+ j ) V1 V1 − V2 ++ =0 j2 2 −j2 V2 − V1 V2 + −I C = 0 − j2 − j2 Also, expressing the controlling signal of the dependent source in terms of the node voltages yields −1+ j −1 + j Ix = ⇒ IC = 2 I x = 2 = −1 − j A -2 j -2 j Solving these equations yields V2 = −3− j = 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V 1+ j 2 10-29 P10.10-7 V2 = 0.7571∠66.7° V V3 = 0.6064∠ − 69.8° V I1 = I 2 + I 3 V3 − V2 I2 = yields j 10 V I3 = 3 −j2 I 3 = 0.3032 ∠20.2° A I 2 = 0.1267∠ −184° A I =0.195∠36° A 1 therefore i1 (t ) =0.195 cos (2 t + 36°) A P10.10-8 The mesh equations are (4 + j 6) I1 − j 6 I 2 = 12 + j12 3 - j 6 I1 + (8 + j 2) I 2 = 0 Using Cramer’s rule yields I1 = (12+ j 12 3 ) (8+ j 2) = 2.5∠29° = 2.2 + j 1.2 A (4+ j 6) (8+ j 2) − (− j 6) (− j 6) I2 = j6 6∠90° (2.5∠29°) = (2.5∠29°) = 1.82∠105° A 8+ j 2 68∠14° Then and VL = j 6(I1 − I 2 ) = (6∠90°) (2.5∠29° − 1.82∠105°) = (6∠90°) (2.71∠ − 11.3°) = 16.3∠78.7° V Finally V = − j 4I 2 = ( 4∠ − 90°)(1.82∠105°) = 7.28∠15° V c 10-30 P10.10-9 The mesh equations are: (10 − j ) I1 + ( j ) I 2 + 0 I 3 = 10 j I1 − j I 2 + j I 3 = 0 0 I1 + j I 2 + (1− j ) I 3 = j10 Solving these mesh equations using Cramer’s rule yields: (10− j ) j 0 I2 = (10− j ) j 0 10 0 j 0 j 10 (1− j ) 90 − j 20 = = 8.38∠77.5° A ⇒ i (t ) = 8.38 cos (103 t + 77.5° ) A j 0 −11 j −j j j (1− j ) (checked using LNAPAC on 7/3/03) P10.10-10 The mesh equations are: −1 − j 4 I1 10∠30° (2 + j 4) −1 (2 +1 / j 4) −1 I 2 = 0 − j4 (3+ j 4) I 3 0 −1 Using Cramer’s rule yields I3 = Then 2 + j8 (10∠30° ) = 3.225∠44° A 12+ j 22.5 V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45 cos (105 t + 44° ) V 10-31 P10.10-11 Mesh Equations: j 75 I1 − j 100 I 2 = 375 − j 100 I1 + (100+ j 100) I 2 = 0 Solving for I 2 yields I 2 = 4.5 + j 1.5 A ⇒ i 2 (t ) = 4.74 ∠18.4° A 10-32 Section 10-11: Superposition, Thèvenin and Norton Equivalents and Source Transformations P10.11-1 Use superposition I1 = 12∠45° = 3.3∠11.3° mA 3000 + j 2000 I2 = −5∠0° = 1.5∠153° mA 3000+ j1500 i (t ) = 3.3 cos (4000 t + 11.3°) + 1.5 cos (3000 t + 153°) mA P10.11-2 Use superposition I1 = 3 = 0.5 mA 6000 I 2 (ω ) = −1∠45° = −0.166 ×10−3 ∠45° A 6000+ j 0.2 i (t ) = i 2 (t ) + i1 (t ) = − 0.166 cos (4 t + 45°) + 0.5 mA = 0.166 cos (4 t − 135°) + 0.5 mA P10.11-3 Use superposition 12∠ 45° = 2∠45° mA I1 (ω ) = 6000 + j 0.2 5∠−90° = 0.833∠ − 90° mA I 2 (ω ) = 6000 + j 0.15 i (t ) = i1 (t ) − i 2 (t ) = 2 cos (4 t + 45°) − 0.833 cos (3 t − 90°) mA 10-33 P10.11-4 Find Voc : 80 + j80 Voc = ( 5 ∠−30° ) 80 + j80 − j 20 80 2∠− 21.9° = ( 5 ∠−30° ) 100∠36.90° = 4 2∠ − 21.9° V Find Z t : Zt = ( − j 20 ) ( 80 + j80 ) = 23 ∠ − 81.9° − j 20 + 80 + j80 Ω The Thevenin equivalent is 10-34 P10.11-5 First, determine Voc : The mesh equations are 600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0° −2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒ j 3 I1 + (1 − j 3) I 2 = 0 Using Cramer’s rule: I 2 = 0.0124∠ − 16° A Then Voc = 300 I 2 = 3.71∠ − 16° V Next, determine I sc : −2 V − V = 0 ⇒ V = 0 ⇒ I sc = The Thevenin impedance is ZT = 9∠0° = 0.015∠0° A 600 Voc 3.71∠−16° = = 247∠ − 16° Ω 0.015∠0° I sc The Thevenin equivalent is 10-35 P10.11-6 First, determine Voc : The node equation is: Voc Voc − (6 + j8) 3 Voc − (6+ j8) + − =0 − j4 j2 j2 2 Voc =3+ j 4=5∠53.1° V Vs = 10∠53° = 6 + j 8 V Next, determine I sc : The node equation is: V V V − (6 + j8) 3 V − (6 + j8) + + − =0 2 − j4 j2 2 j2 V= I sc = 3 + j4 1− j V 3+ j 4 = 2 2− j 2 Vs = 10∠53° = 6 + j 8 V The Thevenin impedance is ZT = 2− j 2 Voc = 3 + j4 = 2 − j2 Ω I sc 3+ j 4 The Thevenin equivalent is 10-36 P10.11-7 Y = G + YL + YC 1 Y = G when YL + YC = 0 or + jω C = 0 jω L 1 1 1 , fO = ωO = = 2π LC 2π 39.6×10−15 LC = 0.07998×107 Hz =800 KHz (80 on the dial of the radio) P10.11-8 In general: I= Voc ZL and V = Voc Z t +Z L Z t +Z L In the three given cases, we have Z1 = 50 Ω ⇒ Z2 = I1 = V1 25 = = 0.5 A Z1 50 1 1 = = − j 200 Ω ⇒ jω C j (2000)(2.5×10 −6 ) Z 3 = jω L = j ( 2000)(50 × 10−3 ) = j100 Ω ⇒ I2 = I3 = V2 100 = = 0.5 A Z 2 200 V3 50 = = 0.5 A Z 3 100 Since |I| is the same in all three cases, Z t +Z1 = Z t +Z 2 = Z t +Z3 . Let Z t = R + j X . Then ( R + 50) 2 + X 2 = R 2 + ( X − 200) 2 = R 2 + ( X + 100) 2 This requires ( X − 200) 2 = ( X + 100) 2 ⇒ X = 50 Ω Then ( R + 50) 2 + (50) 2 = R 2 + (−150) 2 ⇒ R = 175 Ω so Z t =175+ j 50 Ω and Voc = I1 Z t + R1 =(0.5) (175+ 50) 2 + (50) 2 =115.25 V 10-37 P10.11-9 Z1 = (− j 3)(4) = 2.4∠ − 53.1° Ω − j 3+ 4 =1.44 − j1.92 Ω Z 2 = Z1 + j 4 = 1.44 + j 2.08 = 2.53∠55.3° Ω Z3 = 3.51∠ − 37.9° Ω = 2.77 − j 2.16 Ω 3.51∠−37.9° ( 3.51∠−37.9° ) = 1.9∠ − 92° A I = ( 2.85∠ − 78.4° ) = ( 2.85∠ − 78.4° ) ( 5.24∠− 24.4° ) 2.77 − j 2.16 + 2 10-38 P10.11-10 Z2 = I= (200)(− j 4) = 4∠ − 88.8° Ω 200− j 4 0.4∠− 44° = 4∠ − 44° mA −4 j +100+ j 4 i (t ) = 4 cos (25000 t − 44°) mA 10-39 Section 10-12: Phasor Diagrams P10-12-1 V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( 3+ j 2 ) = −4 + j 3 * * P10.12-2 I= 10∠0° = 0.74∠42° A 10+ j1− j10 VR = R I = 7.4∠42° V VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V VS = 10∠0° V P10.12-3 I = 72 3 + 36 3∠(140° − 90°) + 144∠210° + 25∠φ = 40.08 − j 24.23 + 25∠φ = 46.83∠ − 31.15° + 25∠φ To maximize I , require that the 2 terms on the right side have the same angle ⇒ φ = −31.15°. 10-40 Section 10-14: Phasor Circuits and the Operational Amplifier P10.14-1 Vo (ω ) 104 || − j104 10 − j 225 −j e = − = = −10 Vs (ω ) 1− j 2 1000 10 − j 225 − j 225 Vs (ω ) = 2 ⇒ Vo (ω ) = e 2 = 10e 2 vo ( t ) = 10 cos (1000t − 225°) V H (ω ) = P10.14-2 Node equations: V1 − VS VS + j ω C 1 V1 = 0 ⇒ V1 = 1 + j ω C1 R1 R1 R3 V1 V1 − V0 V + = 0 ⇒ V0 = 1 + R2 1 R2 R3 Solving: 1+ R3 R2 V0 = VS 1 + j ω C 1 R1 P10.14-3 Node equations: j ω C 1 VS V1 + j ω C 1 ( V1 − VS ) = 0 ⇒ V1 = R1 1 + j ω C 1 R1 R3 V1 V1 − V0 + = 0 ⇒ V0 = 1 + V R2 1 R2 R3 Solving: R j ω C 1 1 + 3 R2 V0 = 1 + j ω C 1 R1 VS 10-41 P10.14-4 Node equations: V1 − VS V VS + 1 = 0 ⇒ V1 = 175 − j1.6 1 + j 109 V − V0 V1 +1 = 0 ⇒ V0 = 11 V1 1000 10000 Solving: V0 = 11 11 VS = ( 0.005∠0° ) 1 + j 109 110∠89.5° = 0.5∠89.5° mV Therefore v0 (t ) = 0.5 cos (ω t − 89.5°) mV 10-42 PSpice Problems SP10-1 10-43 SP10-2 10-44 SP 10-3 10-45 SP 10-4 The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of Vm and Im are Vm = 12.5 V and Im = -1.667 A. 10-46 Verification Problems VP 10-1 Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the division in rectangular form. Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40 KCL at node 1: 2− V1 10 − V1 − V 2 j 10 = 2− − j 20 − j 20 − ( 20 − j 40 ) − = 2+ j2−2− j2 = 0 10 j 10 KCL at node 2: V1 − V 2 j 10 − V1 − j 20 − ( 20 − j 40 ) 20 − j 40 − j 20 + 3 = − + 3 = ( 2 + j 2) − ( 2 − j4) − j 6 = 0 10 j 10 10 10 10 V2 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. VP 10-2 I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180° Generally, it is more convenient to multiply complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the multiplication in rectangular form. Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284 KVL for mesh 1: 8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10 Since KVL is not satisfied for mesh 1, the mesh currents are not correct. Here is a MATLAB file for this problem: 10-47 % Impedance and phasors for Figure VP 10-2 Vs = -j*5; Z1 = 8; Z2 = j*10; Z3 = -j*2.4; Z4 = j*20; % Mesh equations in matrix form Z = [ Z1+Z2 0; 0 Z3+Z4 ]; V = [ Vs; -Vs ]; I = Z\V abs(I) angle(I)*180/3.14159 % Verify solution by obtaining the algebraic sum of voltages for % each mesh. KVL requires that both M1 and M2 be zero. M1 = -Vs + Z1*I(1) +Z2*I(1) M2 = Vs + Z3*I(2) + Z4*I(2) VP 10-3 V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V KCL at node 1 : 19.2 ∠ 68° 19.2 ∠ 68° + − 4∠15 = 0 2 j6 KCL at node 2: 24 ∠105° 24 ∠105° + + 4∠15 = 0 − j4 j 12 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. Here is a MATLAB file for this problem: % Impedance and phasors for Figure VP 10-3 Is = 4*exp(j*15*3.14159/180); Z1 = 8; Z2 = j*6; Z3 = -j*4; 10-48 Z4 = j*12; % Mesh equations in matrix form Y = [ 1/Z1 + 1/Z2 0; 0 1/Z3 + 1/Z4 ]; I = [ Is; -Is ]; V = Y\I abs(V) angle(V)*180/3.14159 % Verify solution by obtaining the algebraic sum of currents for % each node. KCL requires that both M1 and M2 be zero. M1 = -Is + V(1)/Z1 + V(1)/Z2 M2 = Is + V(2)/Z3 + V(2)/Z4 VP 10-4 First, replace the parallel resistor and capacitor by an equivalent impedance ZP = (3000)(− j 1000) = 949 ∠ − 72° = 300 − j 900 Ω 3000− j 1000 The current is given by I= VS 100 ∠0° = = 0.2∠53° A j 500+ Z P j 500+ 300− j 900 Current division yields − j 1000 I1 = ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA 3000 − j 1000 3000 I2 = ( 0.2 ∠53° ) = 190∠71.4° mA 3000 − j 1000 The reported value of I1 is off by an order of magnitude. 10-49 Design Problems DP 10-1 R2 1 jω C = R2 1 + jω CR 2 R2 R2 1 + jω CR 2 R1 Vo (ω ) =− =− Vi (ω ) R1 1 + jω CR 2 R2 R1 Vo (ω ) j (180 − tan −1 ω CR 2 ) = e 2 Vi (ω ) 1 + (ω CR 2 ) In this case the angle of Vo (ω ) tan (180 − 76 ) is specified to be 104° so CR 2 = = 0.004 and the 1000 Vi (ω ) Vo (ω ) 8 so is specified to be 2.5 Vi (ω ) R2 R1 = R2 8 ⇒ 2.5 = 132 . One set of values R1 1 + 16 that satisfies these two equations is C = 0.2 µ F, R1 = 1515 Ω, R 2 = 20 kΩ . magnitude of DP 10-2 R2 Vo (ω ) = Vi (ω ) where K = 1 jω C = R2 1 + jω CR 2 R2 1 + jω CR 2 K = R2 1 + jω CR p R1 + 1 + jω CR 2 R1 R1 + R 2 and R p = R1 R 2 R1 + R 2 Vo (ω ) K − j tan −1 ω CR p = e 2 Vi (ω ) 1 + (ω CR p ) 10-50 In this case the angle of C Rp = C R1 R 2 R1 + R 2 =− Vo (ω ) is specified to be -76° so Vi (ω ) tan ( −76 ) V (ω ) 2.5 = 0.004 and the magnitude of o so is specified to be 12 1000 Vi (ω ) R2 2.5 K = ⇒ 0.859 = K = . One set of values that satisfies these two equations is R1 + R 2 1 + 16 12 C = 0.2 µ F, R1 = 23.3 kΩ, R 2 = 142 kΩ . DP 10-3 jω L R 2 L R 2 + jω L R1 Vo (ω ) = =− jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = jω R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) ω = L R1 L 1+ ω Rp 2 e L j 90 − tan −1 ω Rp Vo (ω ) L L ( R1 + R 2 ) tan ( 90 − 14 ) is specified to be 14° so = = = 0.1 Rp R1 R 2 Vi (ω ) 40 L 40 R1 V (ω ) 2.5 L 2.5 = ⇒ = 0.0322 . One so is specified to be and the magnitude of o 8 R1 Vi (ω ) 8 1 + 16 In this case the angle of set of values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω . 10-51 DP 10-4 jω L R 2 L R 2 + jω L R1 Vo (ω ) = =− jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = R1 R 2 R1 + R 2 Vo (ω ) Vi (ω ) In this case the angle of jω ω = L R1 L 1+ ω Rp 2 e L j 90 − tan −1 ω Rp Vo (ω ) is specified to be -14°. This requires Vi (ω ) L L ( R1 + R 2 ) tan ( 90 + 14 ) = = = −0.1 Rp R1 R 2 40 This condition cannot be satisfied with positive DP 10-5 Z1 =10 Ω 1 Z2 = jω C Z3 = R + jω L 1 S 10 Y2 = jω C 1 Y3 = R + jω L Y1 = v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 8∠ − θ V iS (t ) = 10 cos 100 t A ⇒ I s =10∠0° A so 1 1 + + jω C = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L 10 R + jω L (80∠−θ ) Equate real part: 40 − 40ω 2 LC = R where ω = 1000 rad sec Equate imaginary part: 40 RC = L Solving yields R = 40(1− 4×107 RC 2 ) 10-52 Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 ) which yields C = 2.5×10−5 F= 25 µ F so L = 40 RC = 0.02 H= 20 mH Now check the angle of the voltage. First Y1 = 1 / 10 = 0.1 S Y2 = j 0.25 S Y3 = 1 /(20+ j 20) = .025− j.025 S then Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V So the angle of the voltage is θ =0° , which satisfies the specifications. 10-53 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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