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Engineering Circuit Analysis, 7th Edition

1.

i=C

Chapter Seven Solutions

10 March 2006

dv dt

(a) i = 0

(DC)

(

)(

)

(b) i = C

dv = − 10 × 10−6 115 2 (120π ) sin120π t = −613sin120π t mA dt

(c) i = C

dv = − (10 × 10−6 )( 4 ×10−3 ) e− t = −40e − t nA dt

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

2.

i=C

v=

Chapter Seven Solutions

10 March 2006

dv dt

6−0 dv t + 6 = 6 − t , therefore i = C = −4.7 × 10−6 μA 0−6 dt

i (μA)

t (s) –4.7

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

3.

i=C

(a)

Chapter Seven Solutions

10 March 2006

dv dt

dv = 30 ⎡⎣e − t − te− t ⎤⎦ dt

therefore i = 10−3

dv = 30 (1 − t ) e− t mA dt

(b) dv = 4 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦ dt dv therefore i = 10−3 = 4e −5t (100 cos100t − 5sin100t ) mA dt

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

4.

Chapter Seven Solutions

10 March 2006

1 W = CV 2 2

⎛1⎞ (a) ⎜ ⎟ 2000 × 10−6 1600 = 1.6 J ⎝2⎠

(

)

2 ⎛1⎞ (b) ⎜ ⎟ ( 25 ×10−3 ) ( 35 ) = 15.3 J ⎝2⎠ 2 ⎛1⎞ (c) ⎜ ⎟ 10−4 ( 63) = 198 mJ ⎝2⎠

(

)

⎛1⎞ (d) ⎜ ⎟ 2.2 × 10−3 ( 2500 ) = 2.75 J ⎝2⎠

(

)

2 ⎛1⎞ (e) ⎜ ⎟ ( 55 )( 2.5 ) = 171.9 J ⎝2⎠ 2 ⎛1⎞ (f) ⎜ ⎟ 4.8 × 10−3 ( 50 ) = 6 J ⎝2⎠

(

)

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

5.

(a) C =

εA d

=

Chapter Seven Solutions

10 March 2006

8.854 × 10 −12 (78.54 × 10 −6 ) = 6.954 pF 100 × 10 − 6

(

)

2 1×10−3 1 2E 2 (b) Energy, E = CV ∴V = = = 16.96 kV 2 6.954 × 10−12 C (c) E =

C=

εA d

1 2 E 2(2.5 × 10 −6 ) = 500 pF CV 2 ∴ C = 2 = 2 (100 2 ) V ∴ε =

Cd (500 × 10 −12 )(100 × 10 −6 ) = 636.62 pF .m −1 = −6 A (78.54 × 10 )

\Relative permittivity :

ε 636.62 × 10−12 = = 71.9 ε 0 8.854 × 10−12

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

6.

(a)

For VA = –1V, W =

Chapter Seven Solutions

(b)

) )

2K sε 0 × 10 −12 (Vbi − V A ) = 2(11.8) 8−.854 (0.57 + 1) qN 1.6 × 10 19 1 × 10 24

(

= 45.281 × 10 Cj =

(

10 March 2006

(

−9

m

)(

11.8 8.854 × 10 −12 1 × 10 −12 45.281 × 10

)(

−9

) = 2.307 fF (

) )

2K sε 0 2(11.8) 8.854 × 10 −12 (Vbi − V A ) = (0.57 + 5) For VA = –5V, W = qN 1.6 × 10 −19 1 × 10 24

(

)(

= 85.289 × 10 −9 m Cj =

(

)(

11.8 8.854 × 10 −12 1 × 10 −12 85.289 × 10

−9

) = 1.225 fF

(c)

For VA = –10V,

W=

2K sε 0 × 10 −12 (Vbi − V A ) = 2(11.8) 8−.854 (0.57 + 10) qN 1.6 × 10 19 1 × 10 24

(

(

) )

)(

= 117.491 × 10 −9 m Cj =

(

)(

11.8 8.854 × 10 −12 1 × 10 −12 117.491 × 10

−9

) = 889.239aF

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

7.

Chapter Seven Solutions

10 March 2006

We require a capacitor that may be manually varied between 100 and 1000 pF by rotation of a knob. Let’s choose an air dielectric for simplicity of construction, and a series of 11 half-plates: fixed

Top view

Side view with no overlap between plates

Side view with a small overlap between plates.

Constructed as shown, the half-plates are in parallel, so that each of the 10 pairs must have a capacitance of 1000/ 10 = 100 pF when rotated such that they overlap completely. If we arbitrarily select an area of 1 cm2 for each half-plate, then the gap spacing between each plate is d = εA/C = (8.854×10-14 F/cm)(1 cm2)/ (100×10-12 F) = 0.8854 mm. This is tight, but not impossible to achieve. The final step is to determine the amount of overlap which corresponds to 100 pF for the total capacitor structure. A capacitance of 100 pF is equal to 10% of the capacitance when all of the plate areas are aligned, so we need a pieshaped wedge having an area of 0.1 cm2. If the middle figure above corresponds to an angle of 0o and the case of perfect alignment (maximum capacitance) corresponds to an angle of 180o, we need to set out minimum angle to be 18o.

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

8.

t

(a)

Energy stored = ∫ v ⋅ C

(b)

Vmax = 3 V

t0

Chapter Seven Solutions

10 March 2006

t − 2×10−3 ⎛ 3 −t ⎞ dv = C∫ 3e 5 ⋅ ⎜ − e 5 ⎟ dt = −1.080 μ J 0 dt ⎝ 5 ⎠

1 CV 2 = 1.35mJ ∴ 37% E max = 499.5μJ 2 V at 37% Emax = 1.825 V Max. energy at t=0, =

v (t ) = 1.825 = 3e

−

t 5

∴ t = 2.486s ⇒≈ 2 s

(c)

1.2 ⎞ ⎛ − dv −6 ⎜ 3 = 300 × 10 − e 5 ⎟ = −141.593μA i =C ⎜ 5 ⎟ dt ⎝ ⎠

(d)

P = vi = 2.011 − 120.658 × 10 −6 = −242.6μW

(

)

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

9.

(

)2 =

(a)

v=

1 π . 1 × 10 −3 C 2

(b)

v=

1 ⎛π .⎜ 1 × 10 −3 C ⎝2

(c)

v=

1 ⎛π .⎜ 1 × 10 −3 C ⎝2

Chapter Seven Solutions

1 47 × 10 −6

.

10 March 2006

(3.14159) (1 × 10 −3 )2 = 33.421mV 2

(

)2 + 0 ⎞⎟⎠ =

(

)2 + π4 (1 × 10 −3 )2 ⎞⎟⎠ =

1 47 × 10 −6

.

(3.14159 ) (1 × 10 −3 )2 = 33.421mV 2

1 47 × 10 −6

(

⎛ 3π 1 × 10 −3 .⎜ ⎝ 4

)2 ⎞⎟⎠ = 50.132mV

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

10.

1 V= C

E=

200ms

∫0

Chapter Seven Solutions

⎞⎤ 1 ⎡⎛ 7 × 10 −3 cos πt ⎟⎥ idt = ⎢⎜ − ⎟ π C ⎣⎢⎜⎝ ⎠⎦⎥

10 March 2006

200ms

= 0

0.426 C

1 181.086 × 10 −9 181.086 × 10 −9 CV 2 = 3 × 10 −6 = ∴C = = 30181μF 2 2C 2 3 × 10 −6

(

)

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

Chapter Seven Solutions

10 March 2006

11. (a)

c = 0.2 μ F, vc = 5 + 3cos 2 200tV; ∴ ic = 0.2 × 10−6 (3) (−2) 200 sin 200t cos 200t

∴ ic = −0.12sin 400tmA (b)

1 1 wc = cvc2 = × 2 × 10−7 (5 + 3cos 2 200t ) 2 ∴ wc max = 10−7 × 64 = 6.4 μ J 2 2

(c)

vc =

(d)

vc = 500 − 400e −100t V

t 1 × 106 ∫ 8e −100t ×10−3 dt = 103 × 40(−0.01) (e−100t − 1) = 400(1 − e100t )V 0 0.2

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

12.

Chapter Seven Solutions

10 March 2006

0.1

vc (0) = 250V, c = 2mF (a) vc (0.1) = 250 + 500∫ 5dt 0

0.2

∴ vc (0.1) = 500V; vc (0.2) = 500∫ 10dt = 1000V 0.1

∴ vc (0.6) = 1750V, vc (0.9) = 2000V t

∴ 0.9 < t < 1: vc = 2000 + 500∫ 10dt = 2000 + 5000(t − 0.9) 0.9

∴ vc = 2100 = 2000 + 5000(t2 − 0.9) ∴ t2 = 0.92 ∴ 0.9 < t < 0.92s

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

Chapter Seven Solutions

10 March 2006

13. (a)

1 1 wc = Cv 2 = × 10−6 v 2 = 2 × 10−2 e −1000t ∴ v = ±200e −500t V 2 2 −6 i = Cv′ = 10 (±200) (−500)e−500t = m0.1e−500t −v 200 ∴R = = = 2k Ω 0.1 i

(b)

PR = i 2 R = 0.01× 2000e −1000t = 20e −1000t W ∞

∴ WR = ∫ 20e −1000t dt = −0.02e −1000t 0

∞ 0

= 0.02J

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

14.

Chapter Seven Solutions

10 March 2006

(a) Left circuit: By Voltage division, VC =

1k (5) = 0.877V 4.7k + 1k

Right circuit: 2 V1 = 1(1 // 2 ) = V 3 1 1 By Voltage Division, V2 = V ∴VC = − V 3 3 (b)

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

15.

v=L

Chapter Seven Solutions

10 March 2006

di dt

(a) v = 0 since i = constant (DC)

(

)

(

)

(b) v = −10−8 115 2 (120π ) sin120π t = −613sin120π t μ V (c) v = −10−8 115 2 ( 24 ×10−3 ) e−6t = − 240e −6t pV

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

16.

v=L

Chapter Seven Solutions

10 March 2006

di dt

⎡ ( 6 − 0 ) × 10−9 ⎤ i=⎢ t + 6 × 10−9 = 6 × 10−9 − 10−6 t , therefore −3 ⎥ ⎢⎣ ( 0 − 6 ) × 10 ⎥⎦ di v = L = − (10−12 )(10−6 ) = −10−18 V = − 1 aV dt v (aV)

t (s) –1

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Engineering Circuit Analysis, 7th Edition

17.

Chapter Seven Solutions

v=L

di dt

(a) L

di = 5 ×10−6 30 ×10−9 ⎣⎡e− t − te− t ⎦⎤ = 150 (1 − t ) e− t fV dt

(

10 March 2006

)

(b)

(

)(

)

di = 5 ×10−6 4 × 10−3 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦ dt therefore v = 100e −5t ( 20 cos100t − sin100t ) pV L

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Engineering Circuit Analysis, 7th Edition

18.

W=

Chapter Seven Solutions

10 March 2006

1 2 LI . Maximum energy corresponds to maximum current flow, so 2

Wmax =

(

)

1 2 5 × 10−3 (1.5 ) = 5.625 mJ 2

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

Chapter Seven Solutions

10 March 2006

19.

(a)

(b)

PL = vLiL ∴ PL max = (−100) (−5) = 500W at t = 40− ms

(c)

PL min = 100(−5) = −500W at t = 20+ and 40+ ms

(d)

WL =

1 2 1 Li L ∴ WL (40ms) = × 0.2(−5) 2 = 2.5J 2 2

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

Chapter Seven Solutions

10 March 2006

20. L = 50 × 10−3 , t < 0 : i = 0; t > 0 i = 80te −100t mA = 0.08te −100t A ∴ i′= 0.08e-100t − 8te −100t ∴ 0.08 = 8t , tm , = 0.01s, i max = 0.08 × 0.01e −1 ∴ i max = 0.2943mA; v = 0.05i′ = e −100t (0.004 − 0.4t ) 0.8 = 0.02 s 40 = 0.004V at t=0

∴ v′ = e −100t (−0.4) − 100e −100t (0.004 − 0.4t ) ∴−0.4 = 0.4 − 40t , t = v = e −2 (0.004 − 0.008) = −0.5413mV this is minimum∴ v max

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

21. (a)

t > 0 : is = 0.4t 2 A ∴ vin = 10is + 5is′ = 4t 2 + 4tV

(b)

iin′ = 0.1vs +

1 t 40tdt + 5 = 4t + 4t 2 + 5A 5 ∫0

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

Chapter Seven Solutions

22.

vL = 20 cos1000tV, L = 25mH, iL (0) = 0

(a)

iL = 40 ∫ 20 cos 1000tdt = 0.8sin 1000tA ∴ p = 8sin 2000t W

(b)

1 w = × 25 ×10−3 × 0.64sin 2 1000t = 8sin 2 1000t mJ 2

10 March 2006

t

0

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

Chapter Seven Solutions

10 March 2006

23.

t

(a)

0 < t < 10 ms: iL = −2 + 5∫ 100dt = −2 + 500t ∴ iL (10ms) = 3A, iL (8ms) = 2A

(b)

iL (0) = 0 ∴ iL (10ms) = 500 × 0.01 = 5A ∴ iL (20ms) = 5 + 5∫

0

0.02

0.01

104 (0.02 − t )dt

4 ∴ iL (20ms) = 5 + 5 × 104 (0.02t − 0.5t )0.02 0.01 = 5 + 5 × 10 (0.0002 − 0.00015) = 7.5A

1 ∴ wL = × 0.2 × 7.52 = 5.625J 2 (c)

If the circuit has been connected for a long time, L appears like short circuit. V8Ω =

8 (100V ) = 80V 2+8

I 2Ω =

20V = 10 A 2Ω

∴i x =

80V = 1A 80Ω

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

24.

Chapter Seven Solutions

10 March 2006

After a very long time connected only to DC sources, the inductors act as short circuits. The circuit may thus be redrawn as

⎛ 80 ⎞ ⎜ ⎟ ⎛ 100 ⎞ And we find that ix = ⎜ 9 ⎟ ⎜ ⎟= 1A ⎜ 80 + 80 ⎟ ⎝ 2 + 8 ⎠ 9 ⎠ ⎝

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

Chapter Seven Solutions

25.

L = 5H, VL = 10(e −t − e −2t )V, iL (0) = 0.08A

(a)

vL (1) = 10(e −1 − e−2 ) = 2.325+ V

(b)

iL = 0.08 + 0.2 ∫ 10(e − t − e −2t ) dt = 0.08 + 2(−e − t + 0.5e −2t )t0

10 March 2006

t

0

iL = 0.08 + 2(−e − t + 0.5e −2t + 1 − 0.5) = 1.08 + e −2t − 2e − t ∴ iL (1) = 0.4796A

(c)

iL (∞) = 1.08A

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

Chapter Seven Solutions

10 March 2006

26. (a)

(b)

40 12 + 40 × 5 × 12 + 20 + 40 12 + 20 + 40 200 100 = + = 100V 3 3

vx = 120 ×

vx =

15 12 120 15 × × 40 + 40 × 5 12 + 15 60 15 + 60 15 12 + 60

120 1 6.667 × × 40 + 200 12 + 12 5 66.667 = 40 + 20 = 60V =

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

Chapter Seven Solutions

10 March 2006

27. (a)

1 wL = × 5 × 1.62 = 6.4J 2

(b)

1 wc = × 20 × 10−6 ×1002 = 0.1J 2

(c)

Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V)

(d)

Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A)

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

28. (a)

Chapter Seven Solutions

10 March 2006

vs = 400t 2 V, t > 0; iL (0) = 0.5A; t = 0.4s 1 vc = 400 × 0.16 = 64V, wc × 10−5 × 642 = 20.48mJ 2

(b)

(c)

0.4 1 iL = 0.5 + 0.1∫ 400t 2 dt = 0.5 + 40 × × 0.43 = 1.3533A 0 3 1 ∴ wL = × 10 × 1.35332 = 9.1581J 2

0.4

iR = 4t 2 , PR = 100 × 16t 4 ∴ wR = ∫ 1600t 4 dt = 320 × 0.45 = 3.277J 0

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

29.

(a)

P7Ω = 0W ; P10Ω =

Chapter Seven Solutions

10 March 2006

V 2 (2 )2 = = 0.4W 10 R

(b) PSpice verification We see from the PSpice simulation that the voltage across the 10-Ω resistor is –2 V, so that it is dissipating 4/10 = 400 mW. The 7-Ω resistor has zero volts across its terminals, and hence dissipates zero power. Both results agree with the hand calculations.

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Engineering Circuit Analysis, 7th Edition

30.

Chapter Seven Solutions

10 March 2006

(a) We find RTH by first short-circuiting the voltage source, removing the inductor, and looking into the open terminals.

Simplifying the network from the right, 3 || 6 + 4 = 6 Ω, which is in parallel with 7 Ω. 6 || 7 + 5 = 8.23 Ω. Thus, RTH = 8.23 || 8 = 4.06 Ω. To find VTH, we remove the inductor: + VTH – V1

V2

V3

REF Writing the nodal equations required: (V1 – 9)/3 + V1/6 + (V1 – V2)/4 = 0 (V2 – V1)/4 + V2/7 + (V2 – V3)/5 = 0 V3/8 + (V3 – V2)/5 = 0 Solving, V3 = 1.592 V, therefore VTH = 9 – V3 = 7.408 V. (b) iL = 7.408/4.06 = 1.825 A (inductor acts like a short circuit to DC). (c)

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

Chapter Seven Solutions

10 March 2006

31. C equiv

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ in series with 10μ in series with 10μ + ≡ 10 μ + ⎜ 1 ⎜ 1 1 ⎟ 1 ⎟ ⎜ 10μ + 10μ ⎟ ⎜ 10μ + 10μ ⎟ ⎝ ⎠ ⎝ ⎠

≡ 4.286μF

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

32.

Chapter Seven Solutions

10 March 2006

Lequiv ≡ (77 p // (77 p + 77 p )) + 77 p + (77 p // (77 p + 77 p )) = 179.6& pH

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

33.

Chapter Seven Solutions

10 March 2006

(a)

Assuming all resistors have value R, all inductors have value L, and all capacitors have value C,

(b)

At dc, 20μF is open circuit; 500μH is short circuit. 10k (9 ) = 3.6V Using voltage division, V x = 10k + 15k

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

34.

Chapter Seven Solutions

10 March 2006

(a) As all resistors have value R, all inductors value L, and all capacitors value C,

+ Vx -

(b)

Vx = 0 V as L is short circuit at dc.

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

35.

Chapter Seven Solutions

10 March 2006

Cequiv = { [(100 n + 40 n) || 12 n] + 75 n} || {7 μ + (2 μ || 12 μ)} C equiv ≡ 85.211nF

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

36.

Chapter Seven Solutions

10 March 2006

Lequiv = {[ (17 p || 4 n) + 77 p] || 12 n} + {1 n || (72 p + 14 p)} Lequiv ≡ 172.388 pH

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

37.

Chapter Seven Solutions

10 March 2006

C T − C x = (7 + 47 + 1 + 16 + 100) = 171μF E CT −C x =

1 (CT − C x )V 2 = 1 (171μ )(2.5)2 = 534.375μJ 2 2

E C x = E CT − E CT −C x = (534.8 − 534.375)μJ = 425nJ

1 425n (2 ) ∴ E C x = 425n = C xV 2 ⇒ C x = = 136nF 2 (2.5)2

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

Chapter Seven Solutions

10 March 2006

38. ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ = 2.75H = 1.5 + + ⎜ 1 + 1 ⎟ ⎜ 1 + 1 + 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1.5 1.5 ⎠ ⎝ 1.5 1.5 1.5 ⎠

(a)

For all L = 1.5H, Lequiv

(b)

For a general network of this type, having N stages (and all L values equiv), Lequiv =

n

LN

∑ NLN −1

N =1

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

39. ⎞ ⎛ ⎞ ⎛ ⎟ ⎜ 1 ⎟ ⎜ 1 ⎜ ⎟ = 3H ⎜ ⎟ =1+ + ⎜1 + 1⎟ ⎜1+1+1⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 2⎠ ⎝3 3 3⎠

(a)

Lequiv

(b)

For a network of this type having 3 stages,

Lequiv = 1 +

1 1 + 2+ 2 3+3

(2 )

2

(3)

2

+

1 3

=1+

(2)2 + (3)3 2(2 ) 3(3)2

Extending for the general case of N stages, 1 1 +K+ 1 1 1 1 1 1 1 + + + +K 2 2 3 3 3 N N 1 1 1 = 1+ + +K+ = N 2(1 / 2) 3(1 / 3) N(1/N)

Lequiv = 1 +

1

+

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Engineering Circuit Analysis, 7th Edition

40.

C equiv =

Chapter Seven Solutions

10 March 2006

(3 p )(0.25 p ) = 0.231 pF 3 p + 0.25 p

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Engineering Circuit Analysis, 7th Edition

41.

Lequiv =

Chapter Seven Solutions

10 March 2006

(2.3& n )(0.3& n ) = 0.2916& nH 2.6& n

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Engineering Circuit Analysis, 7th Edition

42.

Chapter Seven Solutions

(a)

Use 2 x 1μH in series with 4 x 1μH in parallel.

(b)

Use 2 x 1μH in parallel, in series with 4 x 1μH in parallel.

(c)

Use 5 x 1μH in parallel, in series with 4 x 1μH in parallel.

10 March 2006

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

43. (a)

R = 10Ω :10 10 10 = ∴ R eq =

(b)

(c)

10 10 55 + 10 + 10 10 = , 3 3 3

55 30 = 11.379Ω 3

L = 10H ∴ Leq = 11.379H 1 = 5.4545 1/ 30 + 1/10 + 1/ 20 10 ∴ Ceq = 5.4545 + = 8.788F 3 C = 10F :

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

44. (a)

oc :L eq = 6 1 + 3 = 3.857H sc : L eq = (3 2 + 1) 4 = 2.2

(b)

4 = 1.4194H

1 7 1 = , ceq = = 1.3125F 1/ 4 + 1/ 2 3 3 / 7 + 1/ 2 1 5 5 = , Ceq = 4 + = 4.833F sc : 1/ 5 + 1 6 6

oc :1 +

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

45. (a)

(b)

(c)

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

46.

is = 60e −200t mA, i1 (0) = 20mA

(a)

6 4 = 2.4H ∴ v = Leq is′ = 2.4 × 0.06(−200)e−200t

10 March 2006

or v = −28.8e −200t V

1 t 4.8 −200t (e −28.8e−200t dt + 0.02 = − 1) + 0.02 ∫ 6 o 200 = 24e−200t − 4mA(t > 0)

(b)

i1 =

(c)

i2 = is − i1 = 60e−200t − 24e−200t + 4 = 36e−200t + 4mA(t > 0)

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

47.

vs = 100e −80tV , v1 (0) = 20V

(a)

i = Ceq vs′ = 0.8 ×10−6 (−80)100e −80t = −6.4 × 10−3 e−80t A

(b)

v1 = 106 (−6.4 ×10−3 ) ∫ e −80t dt + 20 =

t

o

10 March 2006

6400 −80t (e − 1) + 20 80

∴ v1 = 80e−80t − 60V (c)

t 106 1600 −80t (−6.4 ×10−3 ) ∫ e −80t dt + 80 = (e − 1) + 80 v2 o 4 80 = 20e −80t + 60V

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

48. (a)

vc − vs v −v + 5 ×10−6 vc′ + c L = 0 20 10 t vL − vc 1 + vL dt + 2 = 0 10 8 × 10−3 ∫o

(b)

20i20 +

1 5 ×10−6

1 5 × 10−6

∫ (i

t

o

L

t

∫ (i o

20

− iL )dt + 12 = vs

− i20 )dt − 12 + 10iL + 8 × 10−3 iL′ = 0

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

49. vc (t ): 30mA: 0.03 × 20 = 0.6V, vc = 0.6V 9V: vc = 9V, 20mA: vc = −0.02 × 20 = 0.4V 0.04 cos103 t : vc = 0 ∴ vc (t ) = 9.2V vL (t ): 30mA, 20mA, 9V: vL = 0; 0.04 cos103 t : vL = −0.06 × 0.04 (−1000) sin103 t = 2.4sin103 tV

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Engineering Circuit Analysis, 7th Edition

50.

Chapter Seven Solutions

10 March 2006

We begin by selecting the bottom node as the reference and assigning four nodal voltages: V4

V2

V1

V3

Ref.

(

)

[1]

0.8V1 + 0.2 V2 – V4 = 0

[2]

t V1 - V2 + 0.02 × 103 ∫ V4 − 40e − 20t dt ′ 0 50

1, 4 Supernode:

20×10-3 e-20t =

and:

V1 – V4 = 0.2 Vx or

Node 2:

0 =

V2 - V1 V - 40e −20t dV2 + 2 + 10- 6 50 100 dt

[3]

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Engineering Circuit Analysis, 7th Edition

51.

(a)

Chapter Seven Solutions

10 March 2006

R i = ∞, R o = 0, A = ∞∴ vi = 0 ∴ i = Cvs′ also 0 + Ri + vo = 0 ∴ vo = − RCvs′ 1 idt + vi c∫ −1 1+ A vo = −Avi ∴ vi = vo ∴ i = vi A R 1 1 1 1+ A v ∴ vs = ∫ idt − vo = − vo + − o dt ∫ c A A RC A 1+ A 1+ A ∴ Avs′ = −vo′ − vo or vo′ + vo + Avs′ = 0 RC RC −vi + Ri − Avi = 0, vs =

(b)

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Engineering Circuit Analysis, 7th Edition

52.

Chapter Seven Solutions

10 March 2006

Place a current source in parallel with a 1-MΩ resistor on the positive input of a buffer with output voltage, v. This feeds into an integrator stage with input resistor, R2, of 1-MΩ and feedback capacitor, Cf, of 1 μF. dv c f ions i=Cf = 1.602 × 10 −19 × dt sec

0=

0=

Va − V 1 × 10

6

+Cf

dv c f dt

=

Va − V 1 × 10

6

+ 1.602 × 10 −19

ions sec

dv c f −V −V −19 ions +Cf = 1 . 602 10 + × R2 dt sec 1 × 10 6

Integrating current with respect to t,

1 R2

∫0 vdt ' = C f (Vc t

f

)

− Vc f (0)

1.602 × 10 −19 × ions = C f Vc f R2 − R1 −1 × 1.602 × 10 −19 × ions ⇒ Vout = × 1.602 × 10 −19 × ions Vc f = Va − Vout ⇒ Vout = R2 C f Cf R1 = 1 MΩ, Cf = 1μF

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Engineering Circuit Analysis, 7th Edition

53. (a)

Chapter Seven Solutions

10 March 2006

R = 0.5MΩ, C = 2 μ F, R i = ∞, R o = 0, vo = cos10t − 1V vo ⎞ 1⎞ 1 t⎛ ⎛ Eq. (16) is: ⎜ 1 + ⎟ vo = − ⎜ vs + ⎟ dt − vc (0) ∫ o RC ⎝ A⎠ ⎝ A⎠ vo ⎞ ⎛ 1⎞ 1⎞ 1 ⎛ 1⎞ 1 ⎛ ⎛ ∴ ⎜1 + ⎟ vo′ = − ⎜ vs + ⎟ ∴ ⎜ 1 + ⎟ (−10sin10t ) = −1⎜ vs + cos10t − ⎟ A⎠ RC ⎝ A⎠ ⎝ A⎠ A ⎝ A⎠ ⎝ 1⎞ 1 1 ⎛ ∴ vs = ⎜1 + ⎟10sin10t + − cos10t Let A = 2000 A A ⎝ A⎠ ∴ vs = 10.005sin10t + 0.0005 − 0.0005cos10t

(b)

Let A = ∞∴ vs = 10sin10tV

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Engineering Circuit Analysis, 7th Edition

54.

Chapter Seven Solutions

10 March 2006

Create a op-amp based differentiator using an ideal op amp with input capacitor C1 and feedback resistor Rf followed by inverter stage with unity gain. 1mV dvs R = 60 × / min R f C1 dt rpm R RfC1=60 so choose Rf = 6 MΩ and C1 = 10 μF. Vout = +

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Engineering Circuit Analysis, 7th Edition

55.

(a)

0=

Chapter Seven Solutions

10 March 2006

Va − Vout 1 + vdt Rf L∫

Va = V = 0,∴

− Rf Vout 1 = ⇒ = V v dt L out L Rf L∫

t

∫0 v s dt '

(b) In practice, capacitors are usually used as capacitor values are more readily available than inductor values.

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Engineering Circuit Analysis, 7th Edition

56.

Chapter Seven Solutions

10 March 2006

One possible solution:

vout = −

1 R1C f

∫v

in

dt

we want vout = 1 V for vin = 1 mV over 1 s.

vin

In other words, 1 = −

1 R1C f

∫

1

0

10−3 dt = −

10−3 R1C f

Neglecting the sign (we can reverse terminals of output connection if needed), we therefore need R1Cf = 10–3. Arbitrarily selecting Cf = 1 μF, we find R1 = 1 kΩ.

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Engineering Circuit Analysis, 7th Edition

57.

Chapter Seven Solutions

10 March 2006

One possible solution of many:

dvin dt ⎛ dv ⎞ 100 mV maximum ⎜ in ⎟ = . 60s ⎝ dt ⎠

vout = − RC

vin

⎛ 100 mV ⎞ In other words, vout = 1 V = RC ⎜ ⎟ ⎝ 60s ⎠ or RC = 600

Arbitrarily selecting C = 1000 μF, we find that R = 600 kΩ.

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Engineering Circuit Analysis, 7th Edition

58.

Chapter Seven Solutions

10 March 2006

One possible solution of many:

dvin dt ⎛ dv ⎞ 100 mV At 1 litre/s, ⎜ in ⎟ = . s ⎝ dt ⎠

vout = − RC

vin

⎛ 100 mV ⎞ In other words, vout = 1 V = RC ⎜ ⎟ ⎝ 1s ⎠ or RC = 10

Arbitrarily selecting C = 10 μF, we find that R = 1 MΩ.

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Engineering Circuit Analysis, 7th Edition

59.

Chapter Seven Solutions

10 March 2006

One possible solution:

vin

The power into a 1 Ω load is I2, therefore energy = W = I2Δt. vout =

1 R1C f

∫I

2

dt

we want vout = 1 mV for vin = 1 mV (corresponding to 1 A 2 ).

(

)

Thus, 10−3 = RC 10−3 , so RC = 1 Arbitrarily selecting C = 1 μF, we find that we need R = 1 MΩ.

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Engineering Circuit Analysis, 7th Edition

60.

Chapter Seven Solutions

10 March 2006

One possible solution of many:

vout = − RC

dvin dt

vin

Input: 1 mV = 1 mph, 1 mile = 1609 metres. Thus, on the input side, we see 1 mV corresponding to 1609/3600 m/s. Output: 1 mV per m/s2. Therefore, vout = 2.237 RC = 1 so RC = 0.447 Arbitrarily selecting C = 1 μF, we find that R = 447 kΩ.

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

61. (a)

(b)

(c)

20v20 +

1 5 × 10−6

1 5 × 10−6

∫ (v

t

o

c

t

∫ (v o

20

− vc )dt + 12 = is

− v20 )dt − 12 + 10vc + 8 × 10−3 vc′ = 0

iL − is i −i + 5 × 10−6 iL′ + L c = 0 20 10 t ic − iL 1 + i dt + 2 = 0 −3 ∫o c 10 8 ×10

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

62.

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

63.

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

64. (a)

(b)

“Let is = 100e-80t A and i1 (0) = 20 A in the circuit of (new) Fig. 7.62. (a) Determine v(t) for all t. (b) Find i1(t) for t ≥ 0. (c) Find v2(t) for t ≥ 0.”

(c)

(a)

L eq = 1 4 = 0.8μ H∴ v(t ) = Leq is′ = 0.8 × 10−6 × 100(−80)r −80t V ∴ v(t) = −6.43-80t mV t

(b)

i1 (t ) = 106 ∫ −6.4 × 10−3 e −80t dt + 20 ∴ i1 (t ) =

(c)

i2 (t ) = is − i1 (t ) ∴ i2 (t ) = 20e−80t + 60A

o

6400 −80t (e − 1) = 80e−80t − 60A 80

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Engineering Circuit Analysis, 7th Edition

Chapter Seven Solutions

10 March 2006

65.

In creating the dual of the original circuit, we have lost both vs and vout. However, we may write the dual of the original transfer function: iout/ is. Performing nodal analysis, iS =

1 t V1dt ′ + G in (V1 - V2 ) L1 ∫0

iout = Aid = GfV2 + Gin (V2 – V1)

[1] [2]

Dividing, we find that

iout = iS

G in (V2 - V1 ) + G f V2 1 t V1dt ′ + G in (V1 - V2 ) L1 ∫0

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Engineering Circuit Analysis, 7th Edition

66.

IL = 4/10 = 400 mA. W =

Chapter Seven Solutions

10 March 2006

1 2 LI L = 160 mJ 2

PSpice verification:

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Engineering Circuit Analysis, 7th Edition

67.

IL = 4/(4/3) = 3 A. W =

Chapter Seven Solutions

10 March 2006

1 2 LI L = 31.5 J 2

PSpice verification:

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Engineering Circuit Analysis, 7th Edition

68.

Chapter Seven Solutions

10 March 2006

We choose the bottom node as the reference node, and label the nodal voltage at the top of the dependent source VA. Then, by KCL, VA − 4 VA VA V + + = 0.8 A 100 20 25 25 Solving, we find that VA = 588 mV. Therefore, VC, the voltage on the capacitor, is 588 mV (no DC current can flow through the 75 Ω resistor due to the presence of the capacitor.) Hence, the energy stored in the capacitor is

(

)

1 1 2 CV 2 = 10−3 ( 0.588 ) = 173 μJ 2 2

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Engineering Circuit Analysis, 7th Edition

69.

Chapter Seven Solutions

10 March 2006

By inspection, noting that the capacitor is acting as an open circuit, the current through the 4 kΩ resistor is 8 mA. Thus, Vc = (8)(4) = 32 V. Hence, the energy stored in the capacitor =

(

)

1 1 2 CV 2 = 5 × 10−6 ( 32 ) = 2.56 mJ 2 2

PSpice verification:

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Engineering Circuit Analysis, 7th Edition

70.

Chapter Seven Solutions

10 March 2006

C1 = 5 nF, Rf = 100 MΩ. vout = − R f C1

(

)( )

dvs = − 5 × 10−9 108 ( 30 cos100t ) = −15cos10t V dt

Verifying with PSpice, choosing the LF411 and ±18 V supplies:

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Engineering Circuit Analysis, 7th Edition

71.

Chapter Seven Solutions

10 March 2006

PSpice verification w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2)]2 = 220.8 μJ. This is in agreement with the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7th Edition

72.

Chapter Seven Solutions

10 March 2006

PSpice verification w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2)]2 = 669.2 pJ. This is in agreement with the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7th Edition

73.

Chapter Seven Solutions

10 March 2006

Va − V s 1 + ∫ v L f dt R1 L − Vs 1 Va = Vb = 0, 0= + ∫ v L f dt R1 L 0=

V L f = Va − Vout = 0 − Vout =

Vout = −

(

L dVs R1 dt

)

L f dVs Lf d =− A cos 2π 10 3 t ⇒ L f = 2 R1 ; Let _R = 1 Ω and L = 1 H. R1 dt R1 dt

PSpice Verification: clearly, something rather odd is occuring in the simulation of this particular circuit, since the output is not a pure sinusoid, but a combination of several sinusoids.

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Engineering Circuit Analysis, 7th Edition

74.

Chapter Seven Solutions

10 March 2006

PSpice verification w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2) - 7]2 = 184.2 μJ. This is in reasonable agreement with the PSpice simulation results shown below.

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Engineering Circuit Analysis, 7th Edition

75.

Chapter Seven Solutions

10 March 2006

PSpice verification w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2) - 7]2 = 558.3 pJ. This is in agreement with the PSpice simulation results shown below.

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