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CHAPTER TWO SOLUTIONS
1.

(a) 12 µs
(b) 750 mJ
(c) 1.13 kΩ

(d) 3.5 Gbits
(e) 6.5 nm
(f) 13.56 MHz

Engineering Circuit Analysis, 6th Edition

(g) 39 pA
(h) 49 kΩ
(i) 11.73 pA

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

2.

(a) 1 MW
(b) 12.35 mm
(c) 47. kW
(d) 5.46 mA

(e) 33 µJ
(f) 5.33 nW
(g) 1 ns
(h) 5.555 MW

EngineeringCircuit Analysis, 6th Edition

(i) 32 mm

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

3.

Motor power = 175 Hp
(a) With 100% efficient mechanical to electrical power conversion,
(175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW
(b) Running for 3 hours,
Energy = (130.5×103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ
(c) A single battery has 430 kW-hr capacity. Werequire
(130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

4.

The 400-mJ pulse lasts 20 ns.
(a) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)

400

t (ns)
20

Then P = 400×10-3/20×10-9 = 20 MW.
(b) At 20pulses per second, the average power is
Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

5.

The 1-mJ pulse lasts 75 fs.
(c) To compute the peak power, we assume the pulse shape is square:
Energy (mJ)

1

t (fs)
75

Then P = 1×10-3/75×10-15 = 13.33 GW.
(d) At100 pulses per second, the average power is
Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

6.

The power drawn from the battery is (not quite drawn to scale):
P (W)

10

6

t (min)
5

7

17

24

(a) Total energy (in J) expended is
[6(5) + 0(2) +0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ.
(b) The average power in Btu/hr is
(6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

7.

Total charge q = 18t2 – 2t4 C.
(a) q(2 s) = 40 C.
(b) To find the maximum charge within 0 ≤ t ≤ 3 s, we need to take the firstand
second derivitives:
dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s
d2q/dt2 = 36 – 24t2
substituting t = 2.121 s into the expression for d2q/dt2, we obtain a value of
–14.9, so that this root represents a maximum.
Thus, we find a maximum charge q = 40.5 C at t = 2.121 s.
(c) The rate of charge accumulation at t = 8 s is
dq/dt|t = 0.8 = 36(0.8) – 8(0.8)3 = 24.7 C/s.
(d) See Fig.(a) and (b).

(b)

(a)

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS
8.

Referring to Fig. 2.6c,

- 2 + 3e −5t A,
i1 (t ) = 
3t
 - 2 + 3e A,

t0

Thus,
(a) i1(-0.2) = 6.155 A
(b) i1 (0.2) = 3.466 A
(c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately:
fort < 0, - 2 + 3e-5t = 0 leads to t = -0.2 ln (2/3) = +2.027 s (impossible)
for t > 0, -2 + 3e3t = 0 leads to t = (1/3) ln (2/3) = -0.135 s (impossible)
Therefore, the current is never negative.
(d) The total charge passed left to right in the interval 0.08 < t < 0.1 s is
0.1

=



=

q(t)

∫ [− 2 + 3e ]dt

i (t )dt

− 0.08 1
0

−5t

− 0.08

= − 2 + 3e −5t

0
-0.08

+∫ [− 2 + 3e ]dt
0.1

3t

0

+ − 2 + 3e3t

0.1
0

= 0.1351 + 0.1499
= 285 mC

Engineering Circuit Analysis, 6th Edition

Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved.

CHAPTER TWO SOLUTIONS

9.

Referring to Fig. 2.28,

(a) The average current over one period (10 s) is
iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10

= 800 mA

(b) The total charge transferred...
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