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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

1. (a)

2π103 = 290.9t rad/s 21.6 ∴ f (t ) = 8.5sin (290.9t + Φ ) ∴ 0 = 8.5sin (290.9 × 2.1× 10−3 + Φ ) T = 4 (7.5 − 2.1)10−3 = 21.6 × 10−3 , ω =

∴Φ = −0.6109rad + 2π = 5.672rad or 325.0° ∴ f (t ) = 8.5sin (290.9t + 325.0°) (b)

8.5sin (290.9t + 325.0°) = 8.5 cos(290.9t + 235°) = 8.5 cos (290.9t − 125°)

(c)

8.5cos (−125°) cos ωt + 8.5sin125° sin ωt = −4.875+ cos 290.9t + 6.963sin 290.9t

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

2. (a)

−10 cos ωt + 4sin ωt + ACos ( wt + Φ ), A > 0, − 180° < Φ ≤ 180° A = 116 = 10.770, A cos Φ = −10, A sin Φ = −4 ∴ tan Φ = 0.4, 3d quad ∴Φ = 21.80° = 201.8°, too large ∴Φ = 201.8° − 360° = −158.20°

(b)

200 cos (5t + 130°) = Fcos 5t + G sin 5t ∴ F = 200cos130° = −128.6 G = −200sin130° = −153.2

(c)

(d)

i(t ) = 5cos10t − 3sin10t = 0, 0 ≤ t ≤ 1 s sin10t 5 ∴ = , 10t = 1.0304, cos10t 3 t = 0.10304 s; also, 10t = 1.0304 + π, t = 0.4172 s; 10t = 1.0304 + 2π, t = 0.7314 s 0 < t < 10ms, 10 cos100πt ≥ 12sin100πt ; let 10cos100πt =12sin100πt 10 ∴ tan100πt = , 100πt = 0.6947 ∴ t = 2.211 ms ∴ 0 < t < 2.211 ms 12

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

3. (a)

⎛ ⎛ −B ⎞ ⎞ Note that A cos x + B sin x = A2 + B 2 cos ⎜ x + tan −1 ⎜ ⎟ ⎟ . For f(t), the angle is in the ⎝ A ⎠⎠ ⎝ second quadrant; most calculators will return –30.96o, which is off by 180o. f (t ) = −50 cos ωt − 30sin ωt = 58.31cos (ωt + 149.04°) g (t ) = 55cos ωt − 15sin ωt = 57.01cos (ωt + 15.255°) ∴ ampl. of f (t ) = 58.31, ampl. of g (t ) = 57.01

(b)

f (t ) leads g (t ) by 149.04° − 15.255° = 133.8°

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Engineering Circuit Analysis, 7th Edition

4.

Chapter Ten Solutions

10 March 2006

i (t ) = A cos (ωt − θ), and L(di / dt ) + Ri = Vm cos ωt ∴ L[−ωA sin (ωt − θ)] + RA cos (ωt − θ) = Vm cos ωt −ωLA sin ωt cos θ + ωLA cos ωt sin θ + RA cos ωt cos θ + RA sin ωt sin θ = Vm cos ωt ∴ ωLA cos θ = RA sin θ and ωLA sin θ + RA cos θ = Vm ωL R ωL

Thus, tan θ = and ωLA

R 2 + ω2 L2

* + RA

R R 2 + ω2 L2

= Vm

⎛ ⎞ R2 ω2 L2 so that ⎜ + ⎟ A = Vm 2 2 2 R 2 + ω2 L2 ⎠ ⎝ R +ω L Thus,

(

)

R 2 + ω2 L2 A = ( Vm ) and therefore we may write A =

Vm R 2 + ω2 L2

*

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Engineering Circuit Analysis, 7th Edition

5.

Chapter Ten Solutions

10 March 2006

f = 13.56 MHz so ω = 2πf = 85.20 Mrad/s. Delivering 300 W (peak) to a 5-Ω load implies that

Vm2 = 300 so Vm = 38.73 V. 5

Finally, (85.2×106)(21.15×10–3) + φ = nπ, n = 1, 3, 5, … Since (85.2×106)(21.15×10–3) = 1801980, which is 573588π, we find that φ = 573589π - (85.2×106)(21.15×10–3) = 573589π - 573588π = π

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Engineering Circuit Analysis, 7th Edition

6.

(a)

Chapter Ten Solutions

10 March 2006

-33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o 12 cos (8t – 1o) → 12∠-1o 33∠81o

-33 sin(8t – 9o) leads 12 cos (8t – 1o) by 81 – (-1) = 82o.

12∠-1o

(b)

15 cos (1000t + 66o) -2 cos (1000t + 450o)

→ →

15 ∠ 66o -2 ∠ 450o = -2 ∠90o = 2 ∠ 270o 15 cos (1000t + 66o) leads -2 cos (1000t + 450o) by 66 – -90 = 156o.

15∠66o

2∠270o

(c)

sin (t – 13o) cos (t – 90o)

→ →

1∠-103o 1 ∠ -90o cos (t – 90o) leads sin (t – 13o) by 66 – -90 = 156o.

1∠-103o

(d)

1 ∠ -90o

sin t cos (t – 90o)

→ →

1 ∠ -90o 1 ∠ -90o

These two waveforms are in phase. Neither leads the other.

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Engineering Circuit Analysis, 7th Edition

7.

(a)

6 cos (2π60t – 9o) -6 cos (2π60t + 9o)

Chapter Ten Solutions

10 March 2006

→ 6∠-9o → 6∠189o -6 cos (2π60t + 9o) lags 6 cos (2π60t – 9o) by 360 – 9 – 189 = 162o.

6∠-9o

6∠189o

(b)

→ →

cos (t - 100o) -cos (t - 100o)

1 ∠ -100o -1 ∠ -100o = 1 ∠80o -cos (t - 100o) lags cos (t - 100o) by 180o.

1∠80o

1∠-100o (c)

-sin t sin t

→ →

-1∠-90o = 1∠90o 1∠ -90o -sin t lags sin t by 180o.

1∠90o

1 ∠ -90o

(d)

7000 cos (t – π) 9 cos (t – 3.14o)

→ →

7000 ∠ -π = 7000 ∠ -180o 9 ∠ -3.14o 7000 cos (t – π) lags 9 cos (t – 3.14o) by 180 – 3.14 = 176.9o.

7000 ∠ -180o 9 ∠ -3.14o

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Engineering Circuit Analysis, 7th Edition

8.

Chapter Ten Solutions

v(t) = V1 cos ωt - V2 sin ωt

10 March 2006

[1]

We assume this can be written as a single cosine such that v(t) = Vm cos (ωt + φ) = Vm cos ωt cos φ - Vm sin ωt sin φ [2] Equating terms on the right hand sides of Eqs. [1] and [2], V1 cos ωt – V2 sin ωt = (Vm cos φ) cos ωt – (Vm sin φ) sin ωt yields

Dividing, we find that

V1 = Vm cos φ and V2 = Vm sin φ V2 V sin φ = m = tan φ and φ = tan-1(V2/ V1) V1 Vm cos φ

V12 + V22

V2

φ

V1 Next, we see from the above sketch that we may write Vm = V1/ cos φ or Vm =

V1 V1

V +V 2 1

2 2

Thus, we can write v(t) = Vm cos (ωt + φ) =

=

V12 + V22 V12 + V22 cos [ ωt + tan-1(V2/ V1)].

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Engineering Circuit Analysis, 7th Edition

9.

Chapter Ten Solutions

10 March 2006

(a) In the range 0 ≤ t ≤ 0.5, v(t) = t/0.5 V. Thus, v(0.4) = 0.4/0.5 = 0.8 V. (b) Remembering to set the calculator to radians, 0.7709 V. (c) 0.8141 V. (d) 0.8046 V.

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Engineering Circuit Analysis, 7th Edition

10.

(a)

Vrms

⎡ V2 = ⎢ m ⎣T ⎡ V2 = ⎢ m ⎣T ⎡ V2 = ⎢ m ⎣ 2T ⎡ V2 = ⎢ m ⎣ 2T

∫

T

0

∫

T

∫

T

∫

T

0

0

0

Chapter Ten Solutions

⎤ cos 2 ωt dt ⎥ ⎦

1

2πt ⎤ cos dt ⎥ T ⎦

10 March 2006

2

1

2

2

4πt ⎞ ⎤ ⎛ ⎜1 + cos ⎟ dt ⎥ T ⎠ ⎦ ⎝ V2 dt + m 2T

∫

T

0

1

2

4πt ⎤ cos dt ⎥ T ⎦

⎡ V2 V2 4π ⎤ = ⎢ m T + m cos u 0 ⎥ 8π ⎣ 2T ⎦ V = m 2

1

1

2

2

*

(b)

Vm = 110 2 = 155.6 V , 115 2 = 162.6 V , 120 2 = 169.7 V

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Engineering Circuit Analysis, 7th Edition

11.

Chapter Ten Solutions

10 March 2006

We begin by defining a clockwise current i. Then, KVL yields –2×10–3cos5t + 10i + vC = 0. Since i = iC = C

dvC , we may rewrite our KVL equation as dt dv 30 C + vC = 2 × 10−3 cos 5t dt

[1]

dvC = −5 A sin(5t + θ ) , dt we now may write Eq. [1] as –150Asin(5t + θ) + Acos(5t + θ) = 2×10–3 cos5t. Using a common trigonometric identity, we may combine the two terms on the left hand side into a single cosine function:

We anticipate a response of the form vC(t) = Acos(5t + θ). Since

(150 A)

2

150 A ⎞ ⎛ −3 + A2 cos ⎜ 5t + θ + tan −1 ⎟ = 2 × 10 cos 5t A ⎝ ⎠

Equating terms, we find that A = 13.33 μV and θ = –tan–1 150 = –89.62o. Thus, vC(t) = 13.33 cos (5t – 89.62o) μV.

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Engineering Circuit Analysis, 7th Edition

12.

Chapter Ten Solutions

10 March 2006

KVL yields –6cos400t + 100i + vL = 0. di di Since vL = L = 2 , we may rewrite our KVL equation as dt dt di 2 + 100i = 6 cos 400t dt We anticipate a response of the form i(t) = Acos(400t + θ). Since di = −400 A sin(400t + θ ) , dt we now may write Eq. [1] as

[1]

–800Asin(400t + θ) + 100Acos(400t + θ) = 6 cos400t. Using a common trigonometric identity, we may combine the two terms on the left hand side into a single cosine function:

(800 A) + (100 A) 2

2

800 A ⎞ ⎛ cos ⎜ 400t + θ + tan −1 ⎟ = 6 cos 400t 100 A ⎠ ⎝

Equating terms, we find that A = 7.442 mA and θ = –tan–1 8 = –82.88o. Thus, i(t) = 7.442 cos (400t – 82.88o) mV, so vL = L

di di = 2 = 5.954cos (400t + 7.12o) dt dt

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Engineering Circuit Analysis, 7th Edition

13.

Chapter Ten Solutions

10 March 2006

20cos500t V → 20∠0o V. 20 mH → j10 Ω. Performing a quick source transformation, we replace the voltage source/20-Ω resistor series combination with a 1∠0o A current source in parallel with a 20-Ω resistor. 20 || 60k = 19.99 Ω. By current division, then, IL =

19.99 = 0.7427∠-21.81o A. Thus, iL(t) = 742.7 cos (500t – 21.81o) mA. 19.99 + 5 + j 10

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

14. At x − x : R th = 80 20 = 16Ω 80 cos 500t 85 ∴ voc = 4.8cos 500t V voc = −0.4 (15 85)

10 ⎞ ⎛ cos ⎜ 500t − tan −1 ⎟ 15 ⎠ ⎝ 162 + 102 = 0.2544 cos (500t − 32.01°) A 4.8

(a)

iL =

(b)

vL = LiL′ = 0.02 × 0.02544 (−500) sin (500t − 32.01°) = − 2.544sin (500t − 32.01°) V ∴ vL = 2.544 cos (500t + 57.99°) V, ix = 31.80 cos (500t + 57.99°) mA

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

15.

800 ⎞ ⎛ 5 cos ⎜105 t − ⎟ = 0.10600 cos (10 t − 57.99°) A 2 2 500 ⎠ ⎝ 500 + 800 π 57.99° pR = 0 when i = 0 ∴105 t − π = , t = 25.83μs 180 2 100

(a)

i=

(b)

± vL = Li′ = 8 × 10−3 × 0.10600 (−105 ) sin (105 t − 57.99°) ∴ vL = −84.80sin (105 t − 57.99°) ∴ pL = vL i = −8.989sin (105 t − 57.99°) cos (105 t − 57.99°) = −4.494 sin (2 × 165 t − 115.989°) ∴ pL = 0 when 2 × 105 t − 115.989° = 0°, 180°, ∴ t = 10.121 or 25.83μs

(c)

ps = vs iL = 10.600 cos105 t cos (105 t − 57.99°) ∴ ps = 0 when 105 t =

π , t = 15.708μs and also t = 25.83μs 2

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Engineering Circuit Analysis, 7th Edition

16.

Chapter Ten Solutions

10 March 2006

vs = 3cos105 t V, is = 0.1cos105 t A vs in series with 30Ω → 0.1cos105 t A 30Ω Add, getting 0.2 cos105 t A 30 Ω change to 6 cos 105 t V in series with 30Ω; 30Ω + 20Ω = 50Ω 10 ⎞ ⎛ cos ⎜105 t − tan −1 ⎟ = 0.11767 cos (105 t − 11.310°) A 50 ⎠ ⎝ 502 + 102 At t = 10μs, 105 t = 1∴ iL = 0.1167 cos (1rad − 11.310°) = 81.76mA

∴ iL =

6

∴ vL = 0.11767 × 10 cos (1rad − 11.30° + 90°) = −0.8462V

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Engineering Circuit Analysis, 7th Edition

17.

Chapter Ten Solutions

10 March 2006

cos500t V → 1∠0o V. 0.3 mH → j0.15 Ω. Performing a quick source transformation, we replace the voltage source-resistor series combination with at 0.01∠0o A current source in parallel with a 100-Ω resistor. Current division then leads to 100 = IL ( 0.01 + 0.2I L ) 100 + j 0.15 1 + 20IL = (100 + j0.15) IL Solving, we find that IL = 0.0125∠-0.1074o A, so that iL(t) = 12.5cos(500t – 0.1074o) mA.

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Engineering Circuit Analysis, 7th Edition

18.

Chapter Ten Solutions

10 March 2006

vs1 = Vs 2 = 120 cos120 π t V 120 120 = 2A, = 1A, 2 + 1 = 3A, 60 120 = 40Ω 60 12 3 × 40 = 120 V, ωL = 12π = 37.70Ω 37.70 ⎞ ⎛ cos ⎜120 πt − tan −1 ⎟ 40 ⎠ ⎝ 402 + 37.702 = 2.183cos (120 π t − 43.30°) A

∴ iL =

(a)

(b)

120

1 ∴ ωL = × 0.1× 2.1832 cos 2 (120π t − 43.30°) 2 = 0.2383cos 2 (120π t − 43.30°) J ωL , av =

1 × 0.2383 = 0.11916 J 2

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

19. vs1 = 120 cos 400t V, vs 2 = 180 cos 200t V Performing two quick source transformations, 120 180 = 2 A, = 1.5 A, and noting that 60 120 = 40 Ω, 60 120 results in two current sources (with different frequencies) in parallel, and also in parallel with a 40 Ω resistor and the 100 mH inductor. Next we employ superposition. Open-circuiting the 200 rad/s source first, we perform a source transformation to obtain a voltage source having magnitude 2 × 40 = 80 V. Applying Eqn. 10.4, 80 400(0.1) cos (400t − tan −1 ) iL′ = 40 402 + 4002 (0.1) 2 Next , we open-circuit the 400 rad/s current source, and perform a source transformation to obtain a voltage source with magnitude 1.5 × 40 = 60 V. Its contribution to the inductor current is 60 200(0.1) cos (200t − tan −1 )A iL′′ = 40 402 + 2002 (0.1) 2 so that iL = 1.414 cos (400t − 45°) + 1.342 cos (200t − 26.57°) A

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Engineering Circuit Analysis, 7th Edition

20.

R i = ∞, R o = 0, A = ∞, ideal, R 1C1 = iupper = −

v Vm cos ω t , ilower = out R R1

∴ ic1 = iupper + ilower =

Chapter Ten Solutions

10 March 2006

L R

i ′ (vout − Vm cos ω t ) = −C1vout R1

L ′ vout R d ⎛v ⎞ For RL circuit, Vm cos ω t = vr + L ⎜ R ⎟ dt ⎝ R ⎠ L ∴ Vm cos ω t = vR + v′R R By comparison, vR = vout ′ = vout + ∴ Vm cos ω t = vout + R1C1vout

*

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

21. (a)

1 idt (ignore I.C) C∫ 1 ∴−ω Vm sin ω t = Ri′ + i C

(b)

Assume i = A cos (ω t + Φ )

10 March 2006

Vm cos ω t = Ri +

∴−ω Vm sin ω t = −Rω A sin (ω t + Φ ) +

A cos (ω t + Φ ) C

∴−ω Vm sin ω t = − Rω A cos Φ sin ω t-Rω A sin Φ cos ω t +

A A cos ω t cos Φ − sin ω t sin Φ C C

Equating terms on the left and right side, A 1 [1] Rω A sin Φ = cos Φ∴ tan Φ = so Φ = tan −1 (1 ω CR ) , and ω CR C

ω CR

[2] −ω Vm = −Rω A ∴ω Vm = ∴i =

1+ ω C R 2

2

A ⎡ R ω 2C2 + 1 ⎢ C ⎣ 1 + ω 2 C2 R 2 2

2

−

A 1 C 1 + ω 2C2 R 2

⎤ A ω CVm 1 + ω 2 C2 R 2 ∴ A = ⎥= 1 + ω 2C2 R 2 ⎦ C

ω CVm

⎛ 1 ⎞ cos ⎜ ω t + tan −1 ⎟ ω CR ⎠ 1 + ω 2 C2 R 2 ⎝

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Engineering Circuit Analysis, 7th Edition

22.

Chapter Ten Solutions

10 March 2006

(a) 7 ∠ -90o = -j 7 (b) 3 + j + 7 ∠ -17o = 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047 o

(c) 14ej15 = 14 ∠ 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3. 623 (d) 1 ∠ 0o = 1 (e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34o (f) 3 = 3 ∠ 0o

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Engineering Circuit Analysis, 7th Edition

23.

Chapter Ten Solutions

10 March 2006

(a) 3 + 15 ∠ -23o = 3 + 13.81 – j 5.861 = 16.81 – j 5.861 (b) (j 12)(17 ∠ 180o) = (12 ∠ 90o)(17 ∠ 180o) = 204 ∠ 270o = –j 204 (c) 5 – 16(9 – j 5)/ (33 ∠ -9o) = 5 – (164 ∠ -29.05o)/ (33 ∠ -9o) = 5 – 4.992 ∠ -20.05o = 5 – 4.689 – j 1.712 = 0.3109 + j 1.712

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Engineering Circuit Analysis, 7th Edition

24.

Chapter Ten Solutions

10 March 2006

(a) 5 ∠ 9o – 9 ∠ -17o = 4.938 + j 0.7822 – 8.607 + j 2.631 = -3.668 + j 3.414 = 5.011∠ 137.1o (b) (8 – j 15)(4 + j 16) – j = 272 + j 68 – j = 272 + j 67 = 280.1 ∠ 13.84o (c) (14 – j 9)/ (2 – j 8) + 5 ∠ -30o = (16.64 ∠-32.74o)/ (8.246 ∠ - 75.96o) + 4.330 – j 2.5 = 1.471 + j 1.382 + 4.330 – j 2.5 = 5.801 – j 1.118 = 5.908 ∠ -10.91o (d) 17 ∠ -33o + 6 ∠-21o + j 3 = 14.26 – j 9.259 + 5.601 – j 2.150 + j 3 = 19.86 – j 8.409 = 21.57 ∠ -22.95o

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Engineering Circuit Analysis, 7th Edition

25.

Chapter Ten Solutions

10 March 2006

o

(a) ej14 + 9 ∠ 3o – (8 – j 6)/ j2 = 1 ∠ 14o + 9 ∠ 3o – (8 – j 6)/ (-1) = 0.9703 + j 0.2419 + 8.988 + j 0.4710 + 8 – j 6 = 17.96 – j 5.287 = 18.72 ∠ -16.40o o

(b) (5 ∠ 30o)/ (2 ∠ -15o) + 2 e j5 / (2 – j 2) = 2.5 ∠ 45o + (2 ∠ 5o)/ (2.828 ∠ -45o) = 1.768 + j 1.768 + 0.7072 ∠ 50o = 1.768 + j 1.768 + 0.4546 + j 0.5418 = 2.224 + j 2.310 = 3.207 ∠ 46.09o

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

26. (a)

5∠ − 110° = −1.7101 − j 4.698

(b)

6e j160° = −5.638 + j 2.052

(c)

(3 + j 6) (2∠50°) = −5.336 + j12.310

(d)

−100 − j 40 = 107.70∠ − 158.20°

(e)

2∠50° + 3∠ − 120° = 1.0873∠ − 101.37°

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

27. (a)

40∠ − 50° − 18∠25° = 39.39∠ − 76.20°

(b)

3+

(c)

(2.1∠25°)3 = 9.261∠75° = 2.397 + j8.945+

(d)

0.7e j 0.3 = 0.7∠0.3rad = 0.6687 + j 0.2069

2 2 − j5 + = 4.050− ∠ − 69.78° j 1+ j2

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

28. ic = 20e(40t +30°) A ∴ vc = 100∫ 20e j (40t +30°) dt vc = − j 50e j (40t +30°) , iR = − j10e j (40t + 30°) A ∴ iL = (20 − j10) e j (40t +30°) , vL = j 40 × 0.08(20 − j10) e j (40t + 30°) ∴ vL = (32 + j 64) e j (40t +30°) V ∴ vs = (32 + j 64 − j 50) e j (40t + 30°) ∴ vs = 34.93e j (40t −53.63° ) V

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

29. iL = 20e j (10t + 25°) A vL = 0.2

d [20e j (10t + 25°) ] = j 40e(10t = 25° ) dt

vR = 80e j (10t + 25°) vs = (80 + j 40) e j (10t + 25°) , ic = 0.08(80 + j 40) j10e j (10t + 25°) ∴ ic = (−32 + j 64) e j (10t + 25°) ∴ is = (−12 + j 64) e j (10t + 25°) ∴ is = 65.12e j (10t +125.62°) A

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

30.

80 cos(500t − 20°) V → 5 cos (500t + 12°) A

(a)

vs = 40 cos (500t + 10°) ∴ iout = 2.5cos (500t + 42°) A

(b)

vs = 40sin (500t + 10°) = 40 cos (500t − 80°)

10 March 2006

∴ iout = 2.5cos (500t − 48°) A (c)

vs = 40e j (500t +10°) = 40 cos (500t + 10°) + j 40sin (500t + 10°) ∴ iout = 2.5e j (500t + 42°) A

(d)

vs = (50 + j 20) e j 500t = 53.85+ e j 21.80°+ j 500t ∴ iout = 3.366e j (500t +53.80° ) A

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

31. (a)

12 sin (400t + 110°) A → 12∠20°A

(b)

−7 sin 800t − 3cos 800t → j 7 − 3 = −3 + j 7 = 7.616∠113.20° A

(c)

4 cos (200t − 30°) − 5cos (200t + 20°) → 4∠ − 30° − 5∠20° = 3.910∠ − 108.40° A

(d)

ω = 600, t = 5ms : 70∠30° V → 70 cos (600 × 5 × 10−3rad + 30°) = −64.95 V

(e)

ω = 600, t = 5ms : 60 + j 40 V = 72.11∠146.3° → 72.11cos (3rad + 146.31°) = 53.75 V

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Engineering Circuit Analysis, 7th Edition

32.

ω = 4000, t = 1ms

(a)

I x = 5∠ − 80° A

Chapter Ten Solutions

10 March 2006

∴ ix = 5cos (4rad − 80°) = −4.294 A (b)

I x = −4 + j1.5 = 4.272∠159.44° A ∴ ix = 4.272 cos (4rad + 159.44°) = 3.750− A

(c)

vx (t ) = 50sin (250t − 40°) = 50 cos (250t − 130°) → Vx = 50∠ − 130° V

(d)

vx = 20 cos108t − 30sin108t → 20 + j 30 = 36.06∠56.31° V

(e)

vx = 33cos (80t − 50°) + 41cos (80t − 75°) → 33∠ − 50° + 41∠ − 75° = 72.27∠ − 63.87° V

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

33. V1 = 10∠90° mV, ω = 500; V2 = 8∠90° mV, ω = 1200, M by − 5, t = 0.5ms vout = (−5) [10 cos (500 × 0.5 ×10−3rad + 90°) + 8cos (1.2 × 0.5 + 90°)] = 50sin 0.25rad + 40sin 0.6rad = 34.96mV

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Engineering Circuit Analysis, 7th Edition

34.

Chapter Ten Solutions

10 March 2006

Begin with the inductor: (2.5 ∠40o) (j500) (20×10-3) = 25∠130o V across the inductor and the 25-Ω resistor. The current through the 25-Ω resistor is then (25∠130o) / 25 = 1∠130o A. The current through the unknown element is therefore 2.5∠40o + 1∠130o = 2.693∠61.80o A; this is the same current through the 10-Ω resistor as well. Armed with this information, KVL provides that Vs = 10(26.93∠61.8o) + (25∠ -30o) + (25∠130o) = 35.47 ∠58.93o

and so vs(t) = 35.47 cos (500t + 58.93o) V.

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

35.

ω = 5000 rad/s.

(a)

The inductor voltage = 48∠ 30o = jωL IL = j(5000)(1.2×10-3) IL So IL = 8∠-60o and the total current flowing through the capacitor is 10 ∠ 0o - IL = 9.165∠49.11o A and the voltage V1 across the capacitor is V1 = (1/jωC)(9.165∠49.11o) = -j2 (9.165∠49.11o) = 18.33∠-40.89o V.

Thus, v1(t) = 18.33 cos (5000t – 40.89o) V. (b)

V2 = V1 + 5(9.165∠49.11o) + 60∠120o = 75.88∠79.48o V ∴ v2 (t ) = 75.88cos (5000t + 79.48°) V

(c)

V3 = V2 – 48∠30o = 75.88 ∠79.48o – 48∠30o = 57.70∠118.7o V ∴ v3 (t ) = 57.70 cos (5000t + 118.70°) V

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Engineering Circuit Analysis, 7th Edition

36.

Chapter Ten Solutions

10 March 2006

VR = 1∠0o V, Vseries = (1 + jω –j/ω)(1∠0o)

VR = 1 and Vseries = 1 + (ω - 1/ω )

2

We desire the frequency w at which Vseries = 2VR or Vseries = 2 2 Thus, we need to solve the equation 1 + (ω - 1/ω ) = 4 or ω 2 - 3 ω - 1 = 0 Solving, we find that ω = 2.189 rad/s.

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Engineering Circuit Analysis, 7th Edition

37.

Chapter Ten Solutions

10 March 2006

With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is jωL = j4 Ω, and the impedance of the 1-mF capacitor is –j/ωC = -j2.5 Ω. ∴ Vc = 2∠40° (− j 2.5) = 5∠ − 50° A ∴ I L = 3 − 2∠40° = 1.9513∠ − 41.211° A ∴ VL = 4 × 1.9513∠90° − 4.211° = 7.805+ ∠48.79° V ∴ Vx = VL − Vc = 7.805+ ∠48.79° − 5∠ − 50° ∴ Vx = 9.892∠78.76° V, vx = 9.892 cos (400t + 78.76°) V

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

38. If I si = 2∠20° A, I s 2 = 3∠ − 30° A → Vout = 80∠10° V I s1 = I s 2 = 4∠40° A → Vout = 90 − j 30 V Now let I s1 = 2.5∠ − 60° A and I s 2 = 2.5∠60° A Let Vout = AI s1 + BI s 2 ∴ 80∠10° = A(2∠20°) + B(3∠ − 30°) and 90 − j 30 = (A + B) (4∠40°) ∴ A + B =

90 − j 30 = 12.415+ − j 20.21 4∠40°

80∠10° 3∠ − 30° =A+B ∴ A = 40∠ − 10° − B(1.5∠ − 50°) 2∠20° 2∠20 ∴12.415+ − j 20.21 − B = 40∠ − 10° − B(1.5∠ − 50°) ∴

∴12.415+ − j 20.21 − 40∠ − 10° = B (1 − 1.5∠ − 50°) = B (1.1496∠ + 88.21°) 30.06∠ − 153.82° = 26.148∠117.97° 1.1496∠ + 88.21° ∴ A = 12.415+ − j 20.21 − 10.800 + j 23.81 ∴B =

= 49.842∠ − 60.32° Vout = (49.842∠ − 60.32°) (2.5∠ − 60°) + (26.15∠117.97°) (2.5∠60°) = 165.90∠ − 140.63°V

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Engineering Circuit Analysis, 7th Edition

39.

Chapter Ten Solutions

10 March 2006

We begin by noting that the series connection of capacitors can be replaced by a single 1 equivalent capacitance of value C = = 545.5 μ F . Noting ω = 2πf, 1+ 1 + 1 2 3 (a) ω = 2π rad/s, therefore ZC = –j/ωC =

− j106 = − j 291.8 Ω . 2π ( 545.5 )

(b) ω = 200π rad/s, therefore ZC = –j/ωC =

− j106 = − j 2.918 Ω . 200π ( 545.5 )

− j106 (c) ω = 2000π rad/s, therefore ZC = –j/ωC = = − j 291.8 mΩ . 2000π ( 545.5 )

(d) ω = 2×109π rad/s, therefore ZC = –j/ωC =

− j106 = − j 291.8 nΩ . 2 × 109 π ( 545.5 )

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Engineering Circuit Analysis, 7th Edition

40.

Chapter Ten Solutions

10 March 2006

We begin by noting that the parallel connection of inductors can be replaced by a single 1 5 equivalent inductance of value L = = nH . In terms of impedance, then, we have 1+ 5 6

Noting ω = 2πf,

5 ⎛ ⎞ 5 ⎜ jω × 10−9 ⎟ 6 ⎠ Z= ⎝ 5 −9 5 + jω ×10 6

(a) ω = 2π rad/s, therefore Z = j5.236×10–9 Ω (the real part is essentially zero). (b) ω = 2×103π rad/s, therefore Z = 5.483×10–12 + j5.236×10–6 Ω. (c) ω = 2×106π rad/s, therefore Z = 5.483×10–6 + j5.236×10–6 Ω. (d) ω = 2×109π rad/s, therefore Z = 2.615 + j2.497 Ω. (e) ω = 2×1012π rad/s, therefore Z = 5 + j4.775×10–3 Ω .

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Engineering Circuit Analysis, 7th Edition

41. (a)

Chapter Ten Solutions

10 March 2006

ω = 800 : 2μF → − j 625, 0.6H → j 480 300(− j 625) 600( j 480) + 300 − j 625 600 + j 480 = 478.0 + j175.65Ω

∴ Zin =

(b)

ω = 1600 : Zin = +

300(− j 312.5) 300 − j 312.5

600( j 960) = 587.6 + j119.79Ω 600 + j 960

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

42.

(a)

At ω = 100 rad/s, 2 mF → - j 5 Ω; 0.1 H → j10 Ω. 50 − j 50 10 − j10 2 − j1 (10 + j10) (− j 5) = = 10 + j 5 2 + j1 2 − j1 = 2 − j 6 Ω∴ Zin = 20 + 2 − j 6 = 22 − j 6 Ω

(b)

SCa, b : 20 10 = 6.667, (6.667 − j 5) j10 50 + j 66.67 150 + j 200 30 + j 40 4 − j 3 = = × 6.667 + j 5 20 + j15 4 + j3 4 − j3 = Z in ∴ Z in (1.2 + j1.6) (4 − j 3) = 9.6 + j 2.8 Ω =

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

43. ω = 800 : 2μF → − j 625, 0.6H → j 480 300(− j 625) 600( j 480) + 300 − j 625 600 + j 480 = 478.0 + j175.65Ω

∴ Zin =

− j 625 120 × 478.0 + j175.65 300 − j 625 or I = 0.2124∠ − 45.82° A ∴I =

Thus, i(t) = 212.4 cos (800t – 45.82o) mA.

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

44. (a)

3 Ω + 2mH : V = (3∠ − 20°) (3 + j 4) = 15∠33.13° V

(b)

3 Ω + 125μF : V = (3∠ − 20°) (3 − j 4) = 15∠ − 73.3° V

(c)

3 Ω 2mH 125μF : V = (3∠ − 20°) 3 = 9∠ − 20° V

(d)

same: ω = 4000 ∴ V = (3∠ − 20°) (3 + j8 − j 2) ∴ V = (3∠ − 20°) (3 + j 6) = 20.12∠43.43° V

10 March 2006

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

45. (a)

C = 20μF, ω = 100 Zin =

1

ω = 100 rad/s ∴ Z in =

1 0.005 − j 0.001 + j100C 1

=

1 0.005 − j 0.01 + j 0.002

1 1 + + j1000 × 20 × 10−6 200 j1000 1 ∴ Zin = = 196.12∠ − 11.310°Ω 0.005 + j 0.001

(b)

Z in = 125 =

(c)

0.0052 + (100C − 0.001)

2

or

64 × 10-6 = 0.0052 + (100C − 0.001)

so or

6.245 ×10-3 = 39 × 10-6 = 100C − 0.001 C = 72.45 μF

C = 20μF ∴ Zin =

2

1 1 = 100∠ = −5 0.0005 − j 0.1/ ω + j 2 ×10 ω 0.01∠ 2

2

0.1 ⎞ 0.1 ⎞ ⎛ ⎛ −5 −5 ∴ 0.0052 + ⎜ 2 ×10−5 ω − ⎟ = 0.0001, ⎜ 2 × 10 − ⎟ = 7.5 × 10 ω ⎠ ω ⎠ ⎝ ⎝ 0.01 ∴ 2 ×10−5 − ∓ 866.0 × 10−5 = 0 ∴ 2 × 10−5 ω2 ∓ 866.0 × 10−5 ω − 0.1 = 0 ω use − sign: ω =

866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6 = 444.3 and < 0 4 × 10−5

−866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6 = 11.254 and 0) = 1200C

X=

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

49.

At ω = 4 rad/s, the 1/8-F capacitor has an impedance of –j/ωC = -j2 Ω, and the 4-H inductor has an impedance of jωL = j16 Ω.

(a)

Terminals ab open circuited: Zin = 8 + j16 || (2 – j2) = 10.56 – j1.92 Ω

(b)

Terminals ab short-circuited: Zin = 8 + j16 || 2 = 9.969 + j0.2462 Ω

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Engineering Circuit Analysis, 7th Edition

50.

f = 1 MHz, ω = 2πf = 6.283 Mrad/s 2 μF → -j0.07958 Ω 3.2 μH → j20.11 Ω 1 μF → -j0.1592 Ω 1 μH → j6.283 Ω 20 μH → j125.7 Ω 200 pF → -j795.8 Ω

Chapter Ten Solutions

10 March 2006

= Z1 = Z2 = Z3 = Z4 = Z5 = Z6

The three impedances at the upper right, Z3, 700 kΩ, and Z3 reduce to –j0.01592 Ω Then we form Z2 in series with Zeq: Z2 + Zeq = j20.09 Ω. Next we see 106 || (Z2 + Zeq) = j20.09 Ω. Finally, Zin = Z1 + Z4 + j20.09 = j26.29 Ω.

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Engineering Circuit Analysis, 7th Edition

51.

Chapter Ten Solutions

10 March 2006

As in any true design problem, there is more than one possible solution. Model answers follow: (a) Using at least 1 inductor, ω = 1 rad/s. Z = 1 + j4 Ω. Construct this using a single 1 Ω resistor in series with a 4 H inductor. (b) Force jL = j/C, so that C = 1/L. Then we construct the network using a single 5 Ω resistor, a 2 H inductor, and a 0.5 F capacitor, all in series (any values for these last two will suffice, provided they satisfy the C = 1/L requirement). (c) Z = 7∠80o Ω. R = Re{Z} = 7cos80o= 1.216 Ω, and X = Im{Z} = 7sin80o = 6.894 Ω. We can obtain this impedance at 100 rad/s by placing a resistor of value 1.216 Ω in series with an inductor having a value of L = 6.894/ω = 68.94 mH. (d) A single resistor having value R = 5 Ω is the simplest solution.

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Engineering Circuit Analysis, 7th Edition

52.

Chapter Ten Solutions

10 March 2006

As in any true design problem, there is more than one possible solution. Model answers follow: (a) 1 + j4 kΩ at ω = 230 rad/s may be constructed using a 1 kΩ resistor in series with an inductor L and a capacitor C such that j230L – j/(230C) = 4000. Selecting arbitrarily C = 1 F yields a required inductance value of L = 17.39 H. Thus, one design is a 1 kΩ resistor in series with 17.39 H in series with 1 F. (b) To obtain a purely real impedance, the reactance of the inductor must cancel the reactance of the capacitor, In a series string, this is obtained by meeting the criterion ωL = 1/ωC, or L = 1/ω2C = 1/100C. Select a 5 MΩ resistor in series with 1 F in series with 100 mH. (c) If Z = 80∠–22o Ω is constructed using a series combination of a single resistor R and single capacitor C, R = Re{Z} = 80cos(–22o) = 74.17 Ω. X = –1/ωC = Im{Z} = 80sin(–22o) = –29.97 Ω. Thus, C = 667.3 μF. (d) The simplest solution, independent of frequency, is a single 300 Ω resistor.

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Engineering Circuit Analysis, 7th Edition

53.

Chapter Ten Solutions

10 March 2006

Note that we may replace the three capacitors in parallel with a single capacitor having value 10 −3 + 2 × 10−3 + 4 × 10−3 = 7 mF . (a) ω = 4π rad/s.

Y = j4πC = j87.96 mS

(b) ω = 400π rad/s.

Y = j400πC = j8.796 S

(c) ω = 4π×103 rad/s. Y = j4π×103C = j879.6 S (d) ω = 4π×1011 rad/s. Y = j4π×1011C = j8.796×109 S

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Engineering Circuit Analysis, 7th Edition

54.

Chapter Ten Solutions

10 March 2006

(a) Susceptance is 0 (b) B = ωC = 100 S (c) Z = 1 + j100 Ω, so Y =

1 1 − j100 =G + jB , where B = –9.999 mS. = 1 + j100 1+1002

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

55. 2 H → j 2, 1F → − j1 Let I∈ = 1∠0° A ∴ VL = j 2V ∴ I c = I in + 0.5 VL = 1 + j1 ∴ Vin = j 2 + (1 + j1) (− j1) = 1 + j1 ∴ Vin =

1∠0° 1 1 − j1 = = 0.5 − j 0.5 1 + j1 1 − j1 Vin

Now 0.5 S → 2 Ω, − j 0.5 S =

1 →2H j2

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Engineering Circuit Analysis, 7th Edition

56. (a)

Chapter Ten Solutions

10 March 2006

ω = 500, ZinRLC = 5 + j10 − j1 = 5 + j 9 1 5 − j9 9 = ∴ Yc = = 500C 5 + j9 106 106 9 ∴C = = 169.8 μF 53, 000 ∴ YinRLC =

106 = 21.2 Ω 5

(b)

R in , ab =

(c)

ω = 1000 rad/s ∴ Z S = 5 + j 2 − j 5 = 5 − j 3 = 5.831∠ − 30.96o Ω and Z C = − j 58.89 Ω. Thus, Yin ,ab =

1 1 + = 0.1808∠35.58o S Z S ZC

= 147.1 + j105.2 mS

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Engineering Circuit Analysis, 7th Edition

57. (a)

Chapter Ten Solutions

10 March 2006

j 0.1ω 100 + j 0.001ω 50, 000 + j 0.6ω 100 − j 0.001ω ∴ Zin = × 100 + j 0.001ω 100 − j 0.001ω R in = 550 Ω : Zin = 500 +

5 × 106 + 0.0006ω2 + j (60ω − 50ω) 104 + 10−6 ω2 5 × 106 + 0.006ω2 ∴ R in = = 550 ∴ 5.5 ×106 4 −6 2 10 + 10 ω −4 2 + 5.5 × 10 ω = 5 ×106 × 10−4 ω2 ∴ Zin =

∴ 0.5 × 10−4 ω2 = 0.5 ×106 , ω2 = 1010 , ω = 105 rad/s (b)

10ω = 0.5 × 106 + 0.5 × 10−4 ω2 − 10ω −6 2 10 + 10 ω 2 = 0, ω − 2 × 105 ω + 1010 = 0 X in = 50 Ω =

∴ω =

(c)

4

2 × 105 ± 4 × 1010 − 4 × 1010 = 105 ∴ω = 105 rad/s 2

G in = 1.8 × 10−3 : Yin =

100 + j 0.001ω 50, 000 − j 0.6ω × 50, 000 + j 0.6ω 50, 000 − j 0.6ω

5 × 106 + 6 × 10−4 ω2 + j (50ω − 6ω) 25 × 108 + 0.36ω2 5 × 106 + 6 × 10−4 ω2 ∴1.8 × 103 = 25 × 108 + 0.36ω2 ∴ 5 × 106 + 6 ×10−4 ω2 = 4.5 × 106 + 648 × 10−6 ω2 =

∴ 0.5 ×106 = 48 × 10−6 ω2 ∴ω = 102.06 krad/s (d)

−10ω 25 ×108 + 0.36ω2 ∴10ω = 37.5 × 104 + 54 × 10−6 ω2 Bin = 1.5 × 10−4 =

∴ 54 × 10−6 ω2 − 10ω + 37.5 × 104 = 0, ω = 10 ±

100 − 81 = 52.23 and 133.95 krad/s 108 × 10−6

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

58. I1 0.1∠30° = = 20∠ − 23.13°∴ V1 = 20 V Y1 (3 + j 4)10−3

(a)

V1 =

(b)

V2 = V1 ∴ V2 = 20V

(c)

I 2 = Y2 V2 = (5 + j 2)10−3 × 20∠ − 23.13° = 0.10770∠ − 1.3286° A ∴ I3 = I1 + I 2 = 0.1∠30° + 0.10770∠ − 1.3286° = 0.2∠13.740° A ∴ V3 =

(d)

I3 0.2∠13.740° = = 44.72∠77.18° V ∴ V3 = 44.72V Y3 (2 − j 4)10−3

Vin = V1 + V3 + 20∠ − 23.13° + 44.72∠77.18° = 45.60∠51.62° ∴ Vin = 45.60V

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

59. (a)

50 μ F → − j 20 Ω∴ Yin = 0.1 + j 0.05 1 1000 1 ∴ R1 − j = = 8 − j4 1000 C 0.1 + j 0.05 R1 − j C 1 ∴ R 1 = 8 Ω and C1 = = 250 μ F 4ω Yin =

(b)

ω = 2000 : 50μ F → − j10 Ω ∴ Yin = 0.1 + j 0.1 =

∴ R1 − j

1 R1 − j

500 C1

500 = 5 − j 5 ∴ R1 = 5 Ω, C1 = 100μ F C1

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

60. (a)

10 10 + jω = jω jω jω 10 − jω ∴ Yin × 10 + jω 10 − jω Zin = 1 +

ω 2 + j10ω ω 2 + 100 ω2 10ω G in = 2 , Bin = 2 ω + 100 ω + 100

∴ Yin =

ω

Gin

Bin

0 1 2 5 10 20 ∞

0 0.0099 0.0385 0.2 0.5 0.8 1

0 0.0099 0.1923 0.4 0.5 0.4 0

B

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Engineering Circuit Analysis, 7th Edition

61.

Chapter Ten Solutions

10 March 2006

As in any true design problem, there is more than one possible solution. Model answers follow: (a) Y = 1 – j4 S at ω = 1 rad/s. Construct this using a 1 S conductance in parallel with an inductance L such that 1/ωL = 4, or L = 250 mH. (b) Y = 200 mS (purely real at ω = 1 rad/s). This can be constructed using a 200 mS conductance (R = 5 Ω), in parallel with an inductor L and capacitor C such that ωC – 1/ωL = 0. Arbitrarily selecting L = 1 H, we find that C = 1 F. One solution therefore is a 5 Ω resistor in parallel with a 1 F capacitor in parallel with a 1 H inductor. (c) Y = 7∠80o μS = G + jB at ω = 100 rad/s. G = Re{Y} = 7cos80o = 1.216 S (an 822.7 mΩ resistor). B = Im{Y} = 7sin80o = 6.894 S. We may realize this susceptance by placing a capacitor C in parallel with the resistor such that jωC = j6.894, or C = 68.94 mF. One solution therefore is an 822.7 mΩ resistor in parallel with a 68.94 mF. (d) The simplest solution is a single conductance G = 200 mS (a 5 Ω resistor).

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Engineering Circuit Analysis, 7th Edition

62.

Chapter Ten Solutions

10 March 2006

As in any true design problem, there is more than one possible solution. Model answers follow: (a) Y = 1 – j4 pS at ω = 30 rad/s. Construct this using a 1 pS conductance (a 1 TΩ resistor) in parallel with an inductor L such that –j4×10–12 = –j/ωL, or L = 8.333 GH. (b) We may realise a purely real admittance of 5 μS by placing a 5 μS conductance (a 200 kΩ resistor) in parallel with a capacitor C and inductance L such that ωC – 1/ωL = 0. Arbitrarily selecting a value of L = 2 H, we find a value of C = 1.594 μF. One possible solution, then, is a 200 kΩ resistor in parallel with a 2 H inductor and a 1.594 μF capacitor. (c) Y = 4∠–10o nS = G + jB at ω = 50 rad/s. G = Re{Y} = 4×10–9cos(–10o) = 3.939 nS (an 253.9 MΩ resistor). B = Im{Y} = 4×10–9sin(–10o) = –6.946×10–10 S. We may realize this susceptance by placing an inductor L in parallel with the resistor such that –j/ωL = –j6.946×10–10, or L = 28.78 μH. One possible solution, then, is a 253.9 MΩ resistor in parallel with a 28.78 μH inductor. (d) The simplest possible solution is a 60 nS resistor (a 16.67 MΩ resistor).

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Engineering Circuit Analysis, 7th Edition

63.

Chapter Ten Solutions

10 March 2006

v1 V1 − V2 v1 − V2 , − j 75 = 5V1 + j 3V1 − j 3V2 − j 5V1 + j 5V2 + + 3 − j5 j3 ∴ (5 − j 2) V1 + j 2V2 = − j 75 (1)

− j5 =

v2 − V1 V2 − V1 V2 + + = 10 6 − j5 j3 − j10V2 + j10V1 + j 6V2 − j 6V1 + 5V2 = 300 ∴ j 4V1 + (5 − j 4) V2 = 300

(2)

5 − j 2 − j 75 j4 300 1500 − j 600 − 300 1200 − j 600 = 34.36∠23.63° V = ∴ V2 = = j2 5 − j2 25 − j 30 17 − j30 + 8 j4 5 − j4

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

64. j 3I B − j 5(I B − I D ) = 0 ∴−2I B + j 5I D = 0 3(I D + j 5) − j 5(I D − I B ) + 6 (I D + 10) = 0 ∴ j 5I B + (9 − j 5) I D = −60 − j15 0 j5 −60 − j15 9 − j 5 −75 + j 300 = IB = − j2 j5 15 − j18 j5 9 − j5 = 13.198∠154.23° A

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Engineering Circuit Analysis, 7th Edition

65.

Chapter Ten Solutions

10 March 2006

vs1 = 20 cos1000t V, vs 2 = 20sin1000t V ∴ Vs1 = 20∠0° V, Vs 2 = − j 20V

0.01H → j10 Ω, 0.1mF → − j10 Ω vx − 20 vx vx + j 20 + + = 0, 0.04vx + j 2 − 2 = 0, − j10 j10 25 Vx = 25(2 − j 2) = 70.71∠ − 45° V ∴

∴ vx (t ) = 70.71cos(1000t − 45°) V

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66. (a)

Chapter Ten Solutions

10 March 2006

Assume V3 = 1V ∴ V2 = 1 − j 0.5V, I 2 = 1 − j 0.5 mA ∴ V1 = 1 − j 0.5 + (2 − j 0.5) (− j 0.5) = 0.75 − j1.5V ∴ I1 = 0.75 − j1.5 mA, ∴ Iin = 0.75 − j1.5 + 2 − j 0.5 = 2.75 − j 2 mA ∴ Vin = 0.75 − j1.5 − j1.5 + (2.75 − j 2) (− j 0.5) = −0.25 − j 2.875 V ∴ V3 =

(b)

100 = 34.65+ ∠94.97°V − j 0.25 − j 2.875

− j 0.5 → − jx Assume V3 = 1V ∴ I 3 = 1A, V2 = 1 − jX, I 2 = 1 − jX, → I12 = 2 − jX ∴ V1 = 1 − jX + (2 − jX) (− jX) = 1 − X 2 − j 3X, I1 = 1 − X 2 − j 3X, I in = 3 − X 2 − j 4 X ∴ Vin = 1 − X 2 − j 3X − 4X 2 + jX 3 − j3X = 1 − 5X 2 + j (X 3 − 6X) ∴ X 3 − 6X = 0 ∴ X 2 = 6, X = 6, Z c = − j 2.449 kΩ

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Engineering Circuit Analysis, 7th Edition

67.

Chapter Ten Solutions

10 March 2006

Define three clockwise mesh currents i1, i2, i3 with i1 in the left mesh, i2 in the top right mesh, and i3 in the bottom right mesh. Mesh 1: -10∠0o + (1 + 1 – j0.25)I1 – I2 – (-j0.25)I3 = 0 Mesh 2: – I1 + (1 + 1 + j4)I2 – I3 = 0 Mesh 3: (-j0.25 + 1 + 1)I3 – I2 – (-j0.25I1) = 0 −1 10 2 − j 0.25 −1 2 + j4 0 −1 j 0.25 0 Ix = 2 − j 0.25 −1 j 0.25 −1

2 + j4

−1

j 0.25

−1

2 − j 0.25

10 (1 + 1 − j 0.5) j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1) 20 − j 5 = ∴ I x = 1.217∠ − 75.96° A, ix (t ) = 1.2127 cos (100t − 75.96°) A 8 + j15

∴Ix =

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Engineering Circuit Analysis, 7th Edition

68.

Chapter Ten Solutions

10 March 2006

V1 − 10 − j 0.25V1 + j 0.25Vx + V1 − V2 = 0 ∴ (2 − j 0.25) V1 − V2 + j 0.25 Vx = 10 V2 − V1 + V2 − Vx + j 4V2 = 0 − V1 + (2 + j 4) V2 − Vx = 0 − j 0.25Vx + j 0.25V1 + Vx + Vx − V2 ∴ j 0.25V1 − V2 + (2 − j 0.25) Vx = 0 −1 10 2 − j 0.25 −1 2 + j4 0 −1 j 0.25 0 Vx = j 0.25 − j 0.25 −1 −1 −1 2 + j4 j 0.25 −1 2 − j 0.25 10 (1 + 1 − j 0.5) j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1) 20 − j 5 = = 1.2127∠ − 75.96° V 8 + j15 ∴ vx = 1.2127 cos(100t − 75.96°) V =

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69. (a)

Chapter Ten Solutions

10 March 2006

R1 = ∞, R o = 0, A = −Vo / Vi >> 0 I=

V1 + AVi = jω C1 (Vs − Vi ) Rf

∴ Vi (1 + A + jω C1R f ) = jω C1R f Vs Vo (1 + A + jω C1R f ) = jω C1R f Vs A jω C1R f A V V ∴ o =− As A → ∞, o → − jω C1R f Vs 1 + A + jω C1R f Vs Vo = − AVi ∴−

(b)

R f Cf =

I=

1 jω C f +

1 Rf

=

Rf 1 + jω C f R f

(V1 + AVi ) (1 + jω C f R f ) = (Vs − Vi ) jω C1 , Vo = − AVi Rf

∴ Vi (1 + A) (1 + jω C f R f ) = Vs jω C1R f − jω C1R f Vi , Vi [(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs Vo [(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs A − jω C1R f − jω C1R f A V V ∴ o = As A → ∞, o → Vs 1 + jω C f R f Vs (1 + A) (1 + jω C f R f ) + jω C1R f ∴−

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70.

Chapter Ten Solutions

10 March 2006

Define the nodal voltage v1(t) at the junction between the two dependent sources. The voltage source may be replaced by a 3∠-3o V source, the 600-μF capacitor by a –j/ 0.6 Ω impedance, the 500-μF capacitor by a –j2 Ω impedance, and the inductor by a j2 Ω impedance. V1 - 3∠ - 3o (V - V ) 5V2 + 3V2 = + 1 2 100 − j / 0.6 - j2

-5V2 =

( V2 − V1 ) − j2

+

V2 j2

[1]

[2]

Solving, we find that V2 = 9.81 ∠ – 13.36o mV. Converting back to the time domain, v2(t) = 9.81 cos (103t – 13.36o) mV

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Engineering Circuit Analysis, 7th Edition

71.

Chapter Ten Solutions

10 March 2006

Define three clockwise mesh currents: i1(t) in the left-most mesh, i2(t) in the bottom right mesh, and i3(t) in the top right mesh. The 15-μF capacitor is replaced with a –j/ 0.15 Ω impedance, the inductor is replaced by a j20 Ω impedance, the 74 μF capacitor is replaced by a –j1.351 Ω impedance, the current source is replaced by a 2∠0o mA source, and the voltage source is replaced with a 5∠0o V source. Around the 1, 2 supermesh: (1 + j20) I1 + (13 – j1.351) I2 – 5 I3 = 0 and –I1 + I2 = 2×10–3 Mesh 3:

5∠0o – 5 I2 + (5 – j6.667) I3 = 0

Solving, we find that I1 = 148.0∠179.6o mA. Converting to the time domain, i1(t) = 148.0cos (104t + 179.6o) μA Thus, P1Ω = [i1(1 ms)]2 • 1 = (16.15×10–3)(1) W = 16.15 mW.

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Engineering Circuit Analysis, 7th Edition

72.

Chapter Ten Solutions

10 March 2006

We define an additional clockwise mesh current i4(t) flowing in the upper right-hand mesh. The inductor is replaced by a j0.004 Ω impedance, the 750 μF capacitor is replaced by a –j/ 0.0015 Ω impedance, and the 1000 μF capacitor is replaced by a –j/ 2 Ω impedance. We replace the left voltage source with a a 6 ∠ -13o V source, and the right voltage source with a 6 ∠ 0o V source. = 6 ∠ –13o

(1 – j/ 0.0015) I1 + j/0.0015I2 – I3 (0.005 + j/ 0.0015) I1 + (j0.004 – j/0.0015) I2 –I1

+ (1 – j500) I3

– j0.004 I4 = 0 +

j500 I4 = –6 ∠ 0o

–j0.004 I2 + j500I3 + (j0.004 – j500) I4 = 0

[1] [2] [3] [4]

Solving, we find that I1 = 2.002∠ –6.613o mA, I2 = 2.038 ∠ –6.500o mA, and I3 = 5.998 ∠ 179.8o A.

Converting to the time domain, i1(t) = 1.44 cos (2t – 6.613o) mA i2(t) = 2.038 cos (2t – 6.500o) mA i3(t) = 5.998 cos (2t + 179.8o) A

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Engineering Circuit Analysis, 7th Edition

73.

Chapter Ten Solutions

10 March 2006

We replace the voltage source with a 115 2 ∠0o V source, the capacitor with a –j/ 2πC1 Ω impedance, and the inductor with a j0.03142 Ω impedance. Define Z such that Z-1 = 2πC1 - j/0.03142 + 1/20 By voltage division, we can write that 6.014 ∠85.76o = 115 2

Z Z + 20

Thus, Z = 0.7411 ∠ 87.88o Ω. This allows us to solve for C1: 2πC1 – 1/0.03142 = -1.348 so that C1 = 4.85 F.

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Engineering Circuit Analysis, 7th Edition

74.

Chapter Ten Solutions

10 March 2006

Defining a clockwise mesh current i1(t), we replace the voltage source with a 115 2 ∠0o V source, the inductor with a j2πL Ω impedance, and the capacitor with a –j1.592 Ω impedance. Ohm’s law then yields I1 = Thus, 20 =

115 2 = 8.132∠0o 20 + j (2πL − 1.592 )

202 + (2πL − 1.592)

2

and we find that L = 253.4 mH.

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Engineering Circuit Analysis, 7th Edition

75.

Chapter Ten Solutions

10 March 2006

(a) By nodal analysis: 0 = (Vπ – 1)/ Rs + Vπ / RB + Vπ / rπ + jωCπ Vπ + (Vπ – Vout) jωCμ [1] -gmVπ = (Vout – Vπ) jωCμ + Vout / RC + Vout / RL

[2]

Simplify and collect terms: ⎡⎛ 1 ⎤ 1 1⎞ 1 [1] + + ⎟⎟ + jω (Cπ + C μ )⎥ Vπ - jωC μ Vout = ⎢⎜⎜ RS ⎣⎢⎝ R S R B rπ ⎠ ⎦⎥

(-gm + jωCμ) Vπ - (jωCμ + 1/RC + 1/RL) Vout = 0 Define

1

′ RS

Then Δ =

And Vout =

=

1 1 1 + + R S R B rπ

[2]

′ and R L = RC || RL

⎛ C + Cπ C μ ⎞⎟ -1 + ω 2 (2C2μ + C μ Cπ ) - jω ⎜ g mC μ + μ + ′ ′ ′ ′ ⎜ RS R L R R S ⎟⎠ L ⎝ g m R S − jω C μ R S

⎛ C μ + Cπ C μ ⎞⎟ -1 + ω 2 2C2μ + C μ Cπ - jω ⎜ g m C μ + + ′ ′ ′ ′ ⎜ RS R L RL R S ⎟⎠ ⎝ ⎛ ⎛ ⎞⎞ ⎜ − ω ⎜ g C + C μ + Cπ + C μ ⎟ ⎟ ⎜ ′ ′ ⎟ ⎜ m μ − jωCμ ⎞ RL R S ⎟⎠ ⎟ −1 ⎛ ⎝ −1 ⎜ ⎟ Therefore, ang(Vout) = tan ⎜⎜ tan 2 ⎟ ⎜ -1 ⎟ 2 2 ⎝ g m RS ⎠ ⎜ ′ ′ + ω (2C μ + C μ Cπ ) ⎟ ⎜ RS R L ⎟ ⎝ ⎠

(

)

(b)

(c) The output is ~180o out of phase with the input for f < 105 Hz; only for f = 0 is it exactly 180o out of phase with the input.

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Engineering Circuit Analysis, 7th Edition

76. OC : −

Chapter Ten Solutions

10 March 2006

Vx 100 − Vx + − 0.02Vx = 0 − j10 20

j10 = (0.05 + j 0.1 + 0.02) Vx , Vx =

j10 0.07 + j 0.1

∴ Vx = 67.11 + j 46.98 ∴ Vab ,oc = 100 − Vx = 32.89 − j 46.98 = 57.35∠ − 55.01° V SC :Vx = 100 ∴↓ I SC = 0.02 ×100 + ∴ Zth =

100 = 7A 20

57.35∠ − 55.01° = 4.698 − j 6.711Ω 7

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Engineering Circuit Analysis, 7th Edition

77.

Chapter Ten Solutions

10 March 2006

Let I in = 1∠0. Then VL = j 2ω I in = j 2ω ∴ 0.5VL = jω ∴ Vin = (1 + jω )

1 + j 2ω jω

1 + j 2ω jω ω V 1 ∴ Zin = in = 1 + + j 2ω so Yin = 1 jω ω + j (2ω 2 − 1) At ω = 1, Zin = 1 − j1 + j 2 = 1 + j = 1+

∴ Yin =

1 = 0.5 − j 0.5 1 + j1

R = 1/0.5 = 2 Ω

and

L = 1/0.5 = 2 H.

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Chapter Ten Solutions

10 March 2006

78. (a)

(b)

−15 (1 − j1)1 2 + j1 3 − j1 × = ∴ V1 = × 0.6 − j 0.2 2 − j1 2 + j1 5 j 2 + 0.6 − j 0.2 ∴ V1 = 5∠90°∴ v1 (t ) = 5cos (1000t + 90°) V Vs :

Is:

j2 1− j2 = 0.8 + j 0.4 ∴ V1 1+ j2 1− j2 −10 + j 20 0.8 + j 0.4 = j 25 = = 11.785+ ∠135° V 1 − j1 + 0.8 + j 0.4 1.8 − j 0.6 j2 1=

so v1(t) = 11.79 cos (1000t + 135o) V.

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Chapter Ten Solutions

10 March 2006

79. OC :VL = 0 ∴ Vab ,oc = 1∠0° V SC : ↓ I N ∴ VL = j 2I N ∴1∠0° = − j1[0.25( j 2I N ) + I N ] + j 2I N ∴1 = (0.5 − j + j 2) I N = (0.5 + j1) I N I 1 = 0.4 − j 0.8 ∴ YN = N = 0.4 − j 0.8 0.5 + j1 1∠0° 1 1 1 1 ∴RN = = 2.5 Ω, = = − j 0.8, L N = = 1.25H 0.4 0.8 jω L N jL N ∴ IN =

I N = 0.4 − j 0.8 = 0.8944∠ − 63.43° A

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Engineering Circuit Analysis, 7th Edition

80.

Chapter Ten Solutions

10 March 2006

To solve this problem, we employ superposition in order to separate sources having different frequencies. First considering the sources operating at w = 200 rad/s, we opencircuit the 100 rad/s current source. This leads to VL′ = (j)(2∠0) = j2 V. Therefore, vL′ ( t ) = 2cos(200t + 90o) V. For the 100 rad/s source, we find j VL′′ = (1∠0 ) , vL′′ = 0.5cos (100t + 90°) V 2 ∴ vL ( t ) = 2 cos (200t + 90°) + 0.5cos (100t + 90°) V

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Chapter Ten Solutions

10 March 2006

81. j100 j100 − j 300 − j 300 = −50∠0° V Right: Vab = j100 = j150 V − j 300 + j100 ∴ Vth = −50 + j150 = 158.11∠108.43° V Use superposition. Left: Vab = 100

Zth = j100 − j 300 =

30, 000 = j150 Ω − j 200

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Engineering Circuit Analysis, 7th Edition

82.

Chapter Ten Solutions

10 March 2006

This problem is easily solved if we first perform two source transformations to yield a circuit containing only voltage sources and impedances:

5∠17o + 0.240∠-90o − 2.920∠-45o 73 + 10 + j13 − j 4 = (4.264∠ 50.42o)/ (83.49 ∠ 6.189o) = 51.07 ∠ 44.23 mA

Then I =

Converting back to the time domain, we find that i(t) = 51.07 cos (103t + 43.23o) mA

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Engineering Circuit Analysis, 7th Edition

Chapter Ten Solutions

10 March 2006

83.

(a) There are a number of possible approaches: Thévenizing everything to the left of the capacitor is one of them. VTH = 6(j2)/ (5 + j2) = 2.228 ∠ 68.2o V ZTH = 5 || j2 = j10/ (5 + j2) = 1.857 ∠ 68.2o Ω

Then, by simple voltage division, we find that − j /3 VC = (2.228 ∠ 68.2o) 1.857∠68.2o - j / 3 + j 7 = 88.21 ∠-107.1o mV Converting back to the time domain, vC(t) = 88.21 cos (t – 107.1o) mV. (b) PSpice verification. Running an ac sweep at the frequency f = 1/2π = 0.1592 Hz, we obtain a phasor magnitude of 88.23 mV, and a phasor angle of –107.1o, in agreement with our calculated result (the slight disagreement is a combination of round-off error in the hand calculations and the rounding due to expressing 1 rad/s in Hz.

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Engineering Circuit Analysis, 7th Edition

84.

Chapter Ten Solutions

10 March 2006

(a) Performing nodal analysis on the circuit, Node 1:

1 = V1/ 5 + V1/ (-j10) + (V1 – V2)/ (-j5) + (V1 – V2)/ j10

[1]

Node 2:

j0.5 = V2/ 10 + (V2 – V1)/ (-j5) + (V2 – V1)/ j10

[2]

Simplifying and collecting terms, (0.2 + j0.2) V1 – j0.1 V2 = 1

[1]

-j V1 + (1 + j) V2 = j5

[2]

Solving, we find that V2 = VTH = 5.423 ∠ 40.60o V ZTH = 10 || [(j10 || -j5) + (5 || -j10)] = 10 || (-j10 + 4 – j2) = 5.882 – j3.529 Ω.

(b)

FREQ

VM($N 0002,0)

VP($N 0002,0)

1.592E+01

4.474E+00

1.165E+02

FREQ

VM($N_0005,0)

VP($N_0005,0)

1.592E+01

4.473E+00

1.165E+02

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Engineering Circuit Analysis, 7th Edition

85.

Chapter Ten Solutions

10 March 2006

Consider the circuit below: Vout Vin

1 jω C

Using voltage division, we may write: Vout = Vin

V out 1 / jωC , or R + 1 / jωC V in

=

1 1 + j ω RC

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is 1 Vout = 2 Vin 1 + (ωRC ) As ω → 0, this magnitude → 1, its maximum value. As ω → ∞, this magnitude → 0; the capacitor is acting as a short circuit to the ac signal. Thus, low frequency signals are transferred from the input to the output relatively unaffected by this circuit, but high frequency signals are attenuated, or “filtered out.” This is readily apparent if we plot the magnitude as a function of frequency (assuming R = 1 Ω and C = 1 F for convenience):

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7th Edition

86.

Chapter Ten Solutions

10 March 2006

Consider the circuit below:

1/jωC

Vin

Vout R

Using voltage division, we may write:

Vout = Vin

R jωRC V , or out = R + 1 / jωC Vin 1 + jωRC

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is ωRC Vout = 2 Vin 1 + (ωRC ) As ω → ∞, this magnitude → 1, its maximum value. As ω → 0, this magnitude → 0; the capacitor is acting as an open circuit to the ac signal. Thus, high frequency signals are transferred from the input to the output relatively unaffected by this circuit, but low frequency signals are attenuated, or “filtered out.” This is readily apparent if we plot the magnitude as a function of frequency (assuming R = 1 Ω and C = 1 F for convenience):

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7th Edition

87.

Chapter Ten Solutions

10 March 2006

(a) Removing the capacitor temporarily, we easily find the Thévenin equivalent:

Vth = (405/505) VS and Rth = 100 || (330 + 75) = 80.2 Ω

80.2 Ω 405 VS 505

31.57 fF

(b) Vout =

405 1/jωC VS 505 80.2 + 1 / jωC

and hence

Vout = VS

so

+ Vout -

Vout 1 ⎛ 405 ⎞ = ⎜ ⎟ −12 VS ⎝ 505 ⎠ 1 + j 2.532 × 10 ω

0.802 1 + 6.411 × 10− 24 ω 2

(c)

Both the MATLAB plot of the frequency response and the PSpice simulation show essentially the same behavior; at a frequency of approximately 20 MHz, there is a sharp roll-off in the transfer function magnitude.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Engineering Circuit Analysis, 7th Edition

88.

Chapter Ten Solutions

10 March 2006

From the derivation, we see that

- g m (R C || R L ) + jω (R C || R L )C μ Vout = Vin 1 + jω (R C || R L )Cμ so that 1

2 ⎡ 2 ⎛ R R ⎞2 ⎤ 2 2 2 ⎛ R CR L ⎞ C L ⎟ Cμ ⎥ ⎟ + ω ⎜⎜ ⎢ g m ⎜⎜ R C + R L ⎟⎠ R C + R L ⎟⎠ Vout ⎢ ⎥ ⎝ ⎝ = ⎢ 2 ⎥ Vin 2 2 ⎛ R CR L ⎞ ⎢ ⎥ ⎟⎟ C μ 1 + ω ⎜⎜ ⎢ ⎥ ⎝ RC + RL ⎠ ⎣ ⎦ This function has a maximum value of gm (RC || RL) at ω = 0. Thus, the capacitors reduce the gain at high frequencies; this is the frequency regime at which they begin to act as short circuits. Therefore, the maximum gain is obtained at frequencies at which the capacitors may be treated as open circuits. If we do this, we may analyze the circuit of Fig. 10.25b without the capacitors, which leads to

Vout VS

low frequency

⎛ R R ⎞ (rπ || R B ) ⎛ R R ⎞ rπ R B = - g m ⎜⎜ C L ⎟⎟ = - g m ⎜⎜ C L ⎟⎟ ⎝ R C + R L ⎠ R S + rπ || R B ⎝ R C + R L ⎠ R S (rπ + R B ) + rπ R B

The resistor network comprised of rπ, RS, and RB acts as a voltage divider, leading to a reduction in the gain of the amplifier. In the situation where rπ || RB >> RS, then it has minimal effect and the gain will equal its “maximum” value of –gm (RC || RL). (b) If we set RS = 100 Ω, RL = 8 Ω, RC | max = 10 kΩ and rπgm = 300, then we find that

Vou t rπ || R B = - g m (7.994) VS 100 + rπ || R B We seek to maximize this term within the stated constraints. This requires a large value of gm, but also a large value of rπ || RB. This parallel combination will be less than the smaller of the two terms, so even if we allow RB → ∞, we are left with Vou t - 2398 g r ≈ - (7.994) m π = VS 100 + rπ 100 + rπ Considering this simpler expression, it is clear that if we select rπ to be small, (i.e. rπ