Solution Manual Engineering Circuit Analysis 7th Edition By William H. Hayt

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1. (a)

12 μs (b) 750 mJ (c) 1.13 kΩ

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

Chapter Two Solutions

10 March 2006

(g) 39 pA (h) 49 kΩ (i) 11.73 pA

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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

Chapter Two Solutions

10 March 2006

(i) 32 mm

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3. (a)

Chapter Two Solutions

⎛ 745.7 W ⎞ ⎟ = 298.3 ⎝ 1 hp ⎠

( 400 Hp ) ⎜

10 March 2006

kW

⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ (b) 12 ft = (12 ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.658 m ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ (c) 2.54 cm =

25.4 mm

⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = 70.69 ⎝ 1 Btu ⎠ (e) 285.4´10-15 s =

kJ

285.4 fs

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1.

Chapter Five Solutions

10 March 2006

Define percent error as 100 [ex – (1 + x)]/ ex x 0.001 0.005 0.01 0.05 0.10 0.50 1.00 5.00

1+x 1.001 1.005 1.01 1.05 1.10 1.50 2.00 6.00

ex 1.001 1.005 1.010 1.051 1.105 1.649 2.718 148.4

% error 5×10-5 1×10-3 5×10-3 0.1 0.5 9 26 96

Of course, “reasonable” is a very subjective term. However, if we choose x < 0.1, we ensure that the error is less than 1%.

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14.

Chapter Six Solutions

10 March 2006

In order to deliver 150 mW to the 10-kΩ resistor, we need vout = (0.15)(10 × 103 ) = 38.73 V. Writing a nodal equation at the inverting input, we find 5 5 − vout 0 = + R 1000 Using vout = 38.73, we find that R = 148.2 Ω.

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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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64.

Chapter Eight Solutions

10 March 2006

(a) 0 W (b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is 12 || 11 = 5.739 kΩ. Thus, the circuit time constant is L/R = 1.307 μs. The final value of the total current flowing into the parallel inductor combination is 50/12 mA = 4.167 mA. This will be divided between the two inductors, so that i(∞) = (4.167)(30)/ (30 + 10) = 3.125 mA. We may therefore write i(t) = 3.125[1 – e-10 we find 2.810 A.

6t/ 1.307

] A. Solving at t = 3 μs,

(c) PSpice verification

We see from the Probe output that our hand calculations are correct by verifying using the cursor tool at t = 3 μs.

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60. (a)

α=

Chapter Nine Solutions

10 March 2006

1 8 × 106 8 ×106 × 13 2 = = 1000, ω = = 26 × 106 o 3 2RC 2 × 4 ×10 4

∴ωd = 26 − 1 × 103 = 5000, vc (0) = 8V iL (0) = 8mA, vc , f = 0 ∴ vc = e−1000t (A1 cos1000t + A 2 sin 5000t ) 1 8 − 0.008) = 0 ic (0+ ) = 8 × 106 (0.01 − C 4000 ∴ 5000A 2 − 1000 × 8 = 0, A 2 = 1.6 ∴ A1 = 8; vc′ (0+ ) =

So vc(t) = e-1000t (8 cos 1000t + 1.6 sin 1000t) V, t > 0 (b)

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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):

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Chapter Eleven Solutions

41.

I = 4∠35° A rms

(a)

V = 20I + 80∠35° Vrms, Ps , gen = 80 × 10 cos 35° = 655.3 W

(b)

PR = I R = 16 × 20 = 320 W

(c)

PLoad = 655.3 − 320 = 335.3 W

(d)

APs , gen = 80 × 10 = 800 VA

(e)

APR = PR = 320 VA

(f)

I L = 10∠0° − 4∠35° = 7.104∠ − 18.84° A rms

10 March 2006

2

∴ APL = 80 × 7.104 = 568.3 V A (g)

PFL = cos θ L =

PL 335.3 = = 0.599 APL 568.3

since I L lags V,

PFL is lagging

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