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The Analysis and Design of Linear Circuits Seventh Edition

1 Introduction

1.1 Exercise Solutions

Exercise 1–1. Given the pattern in the statement 1 kΩ = 1 kilohm = 1 × 103 ohms, fill in the blanks inthe following statements using the standard decimal prefixes.

(a) 10−3

is milli or m. 5.0 mW = 5 milliwatts = 5 × 10−3

watts.(b) d is deci or 10−1. 10.0 dB = 10.0 decibels = 1.0 bel.

(c) p is pico or 10−12. 3.6 ps = 3.6 picoseconds = 3.6× 10−12 seconds.

(d) micro is 10−6 or µ. Since we have less than one microfarad, we can also find expressions in terms of nanofarads with n being 10−9. 0.03 µF or 30 nF = 30 nanofarads = 30.0× 10−9 farads.

(e) 109 is giga or G. 6.6 GHz = 6.6 gigahertz = 6.6× 109 hertz.

Exercise 1–2. A device dissipates 100 W of power. How much energy is delivered to it in 10 seconds?Energy is the product of power and time. In this case, we have w = p t = 100 W× 10 s = 1000 J = 1 kJ.

Exercise 1–3. The graph is Figure 1–2(a) shows the charge q (t) flowing past a point in a wire as a functionof time.

(a) Find the current i(t) at t = 1, 2.5, 3.5, 4.5, and5.5 ms. Current is the time rate of change of charge,i = dq

dt. For each time, compute the slope of the plot and account for the units. At 1 s, we have

i = −20pC2ms

= −10 nA. Similarly for the other times, we have +40 nA, 0 nA, −20nA, and 0nA.

(b) Sketch the variation of i(t) versus time. The variations in i(t) are shown in Figure 1–2(b) in thetextbook and the plot is repeated below in Figure Ex1–3.

0 1 2 3 4 5 6−20

−10

0

10

20

30

40

Time, (ms)

C u r r e n t ( n A )

Figure Ex1–3

Exercise 1–4. The working variables of a set of two-terminal electrical devices are observed to be as follows:

Device 1 Device 2 Device 3 Device 4 Device 5

v +10 V ? −15 V +5 V ?i −3 A −3 A +10 mA ? −12 mA p ? +40 W ? +10 mW −120 mW

Device 1: p = vi = −30 W, (delivering power).Device 2: v = p/i = −13.3 V, (absorbing power).

Solution Manual Chapter 1 -1

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Device 3: p = vi = −150 mW, (delivering power).Device 4: i = p/v = +2 mA, (absorbing power).Device 5: v = p/i = +10 V, (delivering power).

Solution Manual Chapter 1 Page 1-2

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1.2 Problem Solutions

Problem 1–1. Express the following quantities to the nearest standard prefix using no more than threedigits.

(a). 1, 000, 000 Hz = 1 × 106 Hz = 1 MHz

(b). 102.5× 109

W = 103 GW

(c). 0.333× 10−7 s = 33.3× 10−9 s = 33.3 ns

(d). 10× 10−12 F = 10 pF

Problem 1–2. Express the following quantities to the nearest standard prefix using no more than threedigits.

(a). 0.000222 H = 222× 10−6 H = 222 µH

(b). 20.5× 105 J = 2.05× 106 J = 2.05 MJ

(c). 72.25× 103 C = 72.3 kC

(d). 3,264Ω = 3.264× 103 Ω = 3.26 kΩ

Problem 1–3. An ampere-hour (Ah) meter measures the time-integral of the current in a conductor. Duringan 8-hour period, a certain meter records 3300 Ah. Find the number of coulombs that flowed through themeter during the recording period.

By definition, 1 ampere = 1 coulomb/second and 1 hour = 3600 seconds. So 3300 Ah = 3300 ampere-hour= 3300 (coulomb/second)(hour)(3600 second/hour) = 11.88 MC.

Problem 1–4. Electric power companies measure energy consumption in kilowatt-hours, denoted kWh.One kilowatt-hour is the amount of energy transferred by 1 kW of power in a period of 1 hour. A power

company billing statement reports a user’s total energy usage to be 2500 kWh. Find the number of joulesused during the billing period.

We have the following relationships: 1 kWh = 1000 watt-hours, 1 watt = 1 joule/second, and 1hour = 3600 seconds. So 2500 kWh = 2500 kilowatt-hours = 2500000 watt-hours = 2500000 (joules/sec-ond)(hours)(3600 second/hour) = 9× 109 J = 9 GJ.

Problem 1–5. Fill in the blanks in the following statements.

(a). To convert capacitance from picofarads to microfarads, multiply by 10−6. We have 1 pF = 1 × 10−12

F = (1× 10−6)× 10−6 F = 1× 10−6µF.

(b). To convert resistance from megohms to kilohms, multiply by 103. We have 1 MΩ = 1 × 106 Ω =(1× 103)× 103 Ω = 1× 103 kΩ.

(c). To convert voltage from millivolts to volts, multiply by 10−3. We have 1 mV = 1× 10−3 V.

(d). To convert energy from megajoules to joules, multiply by 106. We have 1 MJ = 1× 106 J.

Problem 1–6. A wire carries a constant current of 30 mA. How many coulombs flow past a given point inthe wire in 5 s?

We know that 1 ampere is equivalent to 1 coulomb/second. Since the current is constant, if we multiplythe current by the time, we get the charge flowing past a point over that period of time. We can calculateq = i × t = 30 mA × 5 s = (30 millicoulombs/second)(5 seconds) = 150 mC.


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Problem 1–7. The net positive charge flowing through a device is q (t) = 20 + 4t mC. Find the currentthrough the device.

The current through the device is the derivative of the charge, i = dq/dt.

i = dq

dt =

d

dt(20 + 4t) mC = 4 mA

The following MATLAB code calculates the same answer.

syms t real

q t = 2 0 + 4*t;

it = diff(qt,t)

Problem 1–8. Figure P1–8 shows a plot of the net positive charge flowing in a wire versus time. Sketchthe corresponding current during the same period of time.

Take the derivative of the charge waveform to find the current. The charge waveform is piecewise linear,so calculate the slope of each segment to find the current values. The following table presents the results.

Start Time (s) Stop Time (s) Start Charge (C) End Charge (C) Current (A)0 2 10 30 +102 3 30 -10 -403 5 -10 -20 -55 6 -20 30 +50

The following MATLAB code plots the original charge waveform and the corresponding current.

syms t

qt = (10+ 10*t)*(heaviside(t)-heaviside(t-2)) ...

+ (110-40*t)*(heaviside(t-2)-heaviside(t-3)) ...

+ (5-5*t)*(heaviside(t-3)-heaviside(t-5)) ...

+ (-270+50*t)*(heaviside(t-5)-heaviside(t-6));

tt = 0:0.01:6;

qtt = subs(qt,t,tt);

figure; plot(tt,qtt,'b','LineWidth',3)

xlabel('Time (s)'); ylabel('Charge (C)')

grid on

it = diff(qt,t);

itt = subs(it,t,tt);

figure; plot(tt,itt,'g','LineWidth',3)

xlabel('Time (s)'); ylabel('Current (A)')

grid on; axis([0 6 -60 60])

Figure P1–8 displays the resulting plots.

0 1 2 3 4 5 6−20

−15

−10

−5

0

5

10

15

20

25

30

Time (s)

C h a r g e ( C )

0 1 2 3 4 5 6−60

−40

−20

0

20

40

60

Time (s)

C u r r e n t ( A )

Figure P1–8


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Problem 1–9. The net negative charge flowing through a device varies as q (t) = 3t2 C. Find the currentthrough the device at t = 0 s, t = 0.5 s, and t = 1 s.

The current is the derivative of the charge with respect to time, i = dq/dt.

i(t) = dq

dt =

d

dt(3t2) C = 6t A

Evaluate i(t) at 0, 0.5, and 1 s to find the corresponding currents. We have i(0) = 0 A, i(0.5) = 3 A, andi(1) = 6 A. Note that since the negative charge is specified in the problem statement, the current flows inthe opposite direction as the charge flow.


syms t

q t = 3*tˆ2;

it = diff(qt,t)

tt = [0, 0.5, 1];

itt = subs(it,t,tt)

Problem 1–10. A cell-phone charger outputs 9.6 V and is protected by a 50 mA fuse. A 1.5 W cell phoneis connected to it to be charged. Will the fuse blow?

If the current to the cell phone exceeds 50 mA, then the fuse will blow. The current is the power dividedby voltage, i = p/v = (1.5 W)/(9.6 V) = 156.25 mA. The current is greater than 50 mA, so the fuse willblow.


p = 1.5;

v = 9.6;

fuse = 50e-3;

i = p/v

FuseBlows = i>fuse

Problem 1–11. For 0 ≤ t ≤ 5 s, the current through a device is i(t) = 4t A. For 5 < t ≤ 10 s, the currentis i(t) = 40 − 4t A, and i(t) = 0 A for t > 10 s. Sketch i(t) versus time and find the total charge flowingthrough the device between t = 0 s and t = 10 s.

The total charge flowing through the device is the integral of the current over time.

q =

100

i(t) dt =

50

4t dt +

105

(40− 4t) dt

= 2t250

+ (40t− 2t2)105

= (50− 0) + [(400 − 200)− (200− 50)]

= 100 C

The following MATLAB code plots the current versus time and calculates the total charge flowing through

the device.

syms t

i t = 4*t*(heaviside(t)-heaviside(t-5)) ...

+ (40-4*t)*(heaviside(t-5)-heaviside(t-10));

qTotal = int(it,t,0,10)

tt = 0:0.01:10;

itt = subs(it,t,tt);

plot(tt,itt,'b','LineWidth',3)

grid on

xlabel('Time (s)')

ylabel('Current (A)')


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Figure P1–11 displays the resulting plot.

0 1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

14

16

18

20

Time (s)

C u r r e n t ( A )

Figure P1–11

Problem 1–12. The charge flowing through a device is q (t) = 1 − e−1000t µC. How long will it take the

current to reach 200 µA?The current is the derivative of the charge with respect to time, i(t) = d

dtq (t). Compute the current and

then solve for the time in terms of the current. Substitute in a current of 200 µA to find the correspondingtime.

i(t) = d

dtq (t) =

d

dt(1− e−1000t) =

e−1000t

1000

Solve for t in terms of i(t):

e−1000t

1000 = i(t)

e−1000t = 1000i(t)

−1000t = ln [1000i(t)]

t = −1

1000 ln [1000i(t)]

Substitute i(t) = 200 µA to get t = 1.6094 ms. The following MATLAB code calculates the same answer.

syms t

qt = 1e-6*(1-exp(-1000*t));

it = diff(qt,t)

Time200 = solve(it-200e-6,t);

Time200 = vpa(Time200,5)

Problem 1–13. The 12-V automobile battery in Figure P1–13 has an output capacity of 100 ampere-hours(Ah) when connected to a head lamp that absorbs 200 watts of power. The car engine is not running andtherefore not charging the battery. Assume the battery voltage remains constant.

(a). Find the current supplied by the battery and determine how long can the battery power the headlight.

The current is the power divided by the voltage, i = p/v = 200W/12V = 16.667 A. Divide thecapacity of the battery by the current to determine how long the battery will power the headlight,t = 100Ah/16.667A = 6 hours.


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(b). A 100 W device is connected through the utility port. How long can the battery power both theheadlight and the device?

The current is the power divided by the voltage, i = p/v = (200 + 100)W/12 V = 25 A. Note that thepower requirement increased by 50%, so the current increased by 50% as well. Divide the capacity of thebattery by the current to determine how long the battery will power the headlight, t = 100Ah/25A = 4hours.

The following MATLAB code calculates the same answers.

v = 12;

p = 200;

i = p/v

cap = 100;

t = cap/i

p2 = p+100;

i2 = p2/v

t2 = cap/i2

Problem 1–14. An incandescent lamp absorbs 100 W when connected to a 120-V source. A energy-efficient

compact fluorescent lamp (CFL) producing the same amount of light absorbs 16 W when connected to thesame source. How much cheaper is it to operate the CFL versus the incandescent bulb over 1000 hours whenelectricity costs 7.8 cents/kWh?

Find the energy in kWh used by each type of bulb and then calculate the corresponding costs. Forthe incandescent lamp, we have w = (100 W)(1000 h) = 100 kWh. The corresponsding cost is (100 kWh)(0.078 dollars/kWh) = $7.80. For the CFL, we have w = (16 W)(1000 h) = 16 kWh. The corresponsdingcost is (16 kWh) (0.078 dollars/kWh) = $1.25. While operating for 1000 hours, the CFL saves $6.55. Thefollowing MATLAB code calculates the same answer.

V = 120;

T = 1000;

P incand = 100;

rate = 0.078;

kWh incand = P incand*T/1000;

cost incand = rate*kWh incand

P fluor = 16;

kWh fluor = P fluor*T/1000;

cost fluor = rate*kWh fluor

cost difference = cost incand - cost fluor

Problem 1–15. The current through a device is zero for t < 0 and is i(t) = 3e−2t A for t ≥ 0. Find thecharge q (t) flowing through the device for t ≥ 0.

The charge flowing through the device is the integral of the current over time.

q (t) =

t0

i(τ ) dτ =

t0

3e−2τ dτ = −3

2e−2τ

t0

= −3

2

e−2t − 1

=

3

2

1− e−2t

C, t ≥ 0


syms t tau

i t = 3*exp(-2*t);

itau = subs(it,t,tau)

qt = simple(int(itau,tau,0,t))


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Problem 1–16. A string of holiday lights is protected by a 5-A fuse and has 25 bulbs, each of which israted at 7 W. How many strings can be connected end-to-end across a 120 V circuit without blowing a fuse?

The current is the power divided by the voltage, i = p/v. Each string of lights increases the power by25× 7 W = 175 W. Correspondingly, the current increases by (175 W)/(120 V) = 1.4583 A. Since each fusecan handle up to 5 A, divide the fuse rating by the current required for each string and round down to getthe maximum number of strings. We have (5 A)/(1.4583 A) = 3.4286. The maximum number of strings isthree.

Problem 1–17. The i-v relationship for a photocell when illuminated is i = ev − 10 A. For v = −2, 2 and3 V, find the device power and state whether it is absorbing or delivering power.

For each voltage, substitute into the expression for current and solve for the current. Multiply the currentand voltage to find power. If the power is positive, the photocell is absorbing power. If the power is negative,the photocell is delivering power. The following table summarizes the results of the calculations.

v (V) i (A) p (W) Absorbing/Delivering-2 -9.8647 19.7293 Absorbing2 -2.6109 -5.2219 Delivering3 10.0855 30.2566 Absorbing


v = [ - 2 2 3 ]

iv = exp(v)-10

p = v .*iv

pAbsorbs = p>0

Results = [v' iv' p' pAbsorbs']

Problem 1–18. A new 6 V Alkaline lantern battery delivers 237.5 kJ of energy during its lifetime. Howlong will the battery last in an application that draws 15 mA continuously. Assume the battery voltage isconstant.

The power delivered is the product of the voltage and current, p = vi = (6v)(15 mA) = 90mW. A wattis a joule per second, so the application draws 90 mJ/s. Divide the capacity of the battery by the rate toget the total time, (237.5 kJ)/(90 mJ/s) = 2.6389 Ms = 733.02 h = 30.54 days.

Problem 1–19. The maximum power the device can dissipate is 0.25 W. Determine the maximum currentallowed by the device power rating when the voltage is 9 V.

The maximum current will be the maximum power divided by the voltage, iMax = pMax/v = (0.25 W)/(9V) = 27.778 mA.

Problem 1–20. Traffic lights are being converted from incandescent bulbs to LED arrays to save operatingand maintenance costs. Typically each incandescent light uses three 100-W bulbs, one for each color R, Y,G. A competing LED array consists of 61 LEDs with each LED requiring 9 V and drawing 20 mA of current.There are three arrays per light - R, Y, G. A small city has 1560 traffic signals. Since one light is always on24/7, how much can a city save in one year if the city buys their electricity at 7.2 cents per kWh?

To solve this problem, compare two lights and then scale the problem to the number of lights in the city.The incandescent light always has on one 100-W bulb operating 24 hours per day for 365 days, which yields

a total of 876 kWh at a cost of $63.072. The LED light always has on one array, using a total power of (61)(9 V)(20 mA) = 10.98 W. Over one year, the LED lights uses 96.185 kWh of energy at a cost of $6.9253.The savings per light per year is $56.147, which, for a total of 1560 lights, translates into a city-wide savingof $87,589 per year. The following MATLAB code calculates the same answer.

p incand = 100/1000; % Incandescent power in kW

rate = 0.072; % Dollars per kWh

hours = 24*365; % Hours per year

cost incand = p incand*hours*rate

v le d = 9; % LED voltage


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i led = 20e-3; % LED current

n l ed = 61 ; % Number of LED lights per array

p le d = n le d*v led*i led/1000 % Power in kW

cost led = p led*hours*rate

savings light = cost incand - cost led

lights = 1560;

savings year = lights*savings light

Problem 1–21. Two electrical devices are connected as shown in Figure P1-21. Using the reference marksshown in the figure, find the power transferred and state whether the power is transferred from A to B or Bto A when

(a). v = +11 V and i = −1.1 A

(b). v = +80 V and i = +20 mA

(c). v = −120 V and i = −12 mA

(d). v = −1.5 V and i = −600 µA

The passive sign convention applies to Element B, so if the power is positive the transfer is from A toB, and if the power is negative, the transfer is from B to A. For each case, calculate the power p = iv anddetermine the direction of the power flow.

The following table summarizes the results of the calculations.

Case v i p Power Transfer(a) +11 V -1.1 A -12.1 W B to A(b) +80 V +20 mA +1.6 W A to B(c) -120 V -12 mA + 1.44 W A to B(d) -1.5 V -600 µA +900 µW A to B


v = [11 80 -120 -1.5];

i = [-1.1 20e-3 -12e-3 -600e-6];

p = v .*i

AtoB = p>0

Results = [v' i' p' AtoB']

Problem 1–22. Figure P1–22 shows an electric circuit with a voltage and a current variable assigned toeach of the six devices. The device voltages and currents are observed to be

v (V) i (A)Device 1 15 −1

Device 2 5 1Device 3 10 2Device 4 −10 −1Device 5 20 −3Device 6 20 2

Find the power associated with each device and state whether the device is absorbing or delivering power.Use the power balance to check your work.

The power associated with each device is the product of the voltage and currnet, p = vi. If the power ispositive, the device is absorbing power. If the power is negative, the device is delivering power. The originaltable is expanded below to include power and the direction of power flow.


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v (V) i (A) p (W) Absorbing/DeliveringDevice 1 15 −1 −15 DeliveringDevice 2 5 1 5 AbsorbingDevice 3 10 2 20 AbsorbingDevice 4 −10 −1 10 AbsorbingDevice 5 20 −3 −60 DeliveringDevice 6 20 2 40 Absorbing

For the power to balance in this system, the sum of the individual device powers should be zero. Wehave −15 + 5 + 20 + 10 − 60 + 40 = 0, so the power balances.

The following MATLAB code calculates the same results.

v = [1 5 5 10 -10 2 0 2 0];

i = [ - 1 1 2 - 1 - 3 2 ] ;

p = v .*i

Absorbing = p>0

Balance = sum(p)

Results = [v' i' p' Absorbing']

Problem 1–23. Figure P1–22 shows an electric circuit with a voltage and a current variable assigned toeach of the six devices. Use power balance to find v4 when v1 = 20 V, i1 = −2 A, p2 = 20 W, p3 = 10 W,i4 = 1 A, and p5 = p6 = 2.5 W. Is device 4 absorbing or delivering power?

First, calculate the power associated with device 1, p1 = v1i1 = −40 W. The power must balance in thecircuit, so the sum of all of the device powers is zero. Therefore, we can solve for p4 and v4 as follows:

p4 = − p1 − p2 − p3 − p5 − p6 = 40− 20− 10− 2.5− 2.5 = 5 W

v4 = p4/i4 = 5/1 = 5 V

The power is positive, so the device is absorbing power.

Problem 1–24. Suppose in Figure P1–22 a ground is connected to the minus (−) side of device 6 andanother to the junction of devices 2, 3 and 4. Further, assume that the voltage v4 is 5 V and v1 is 10 V.What are the voltages v2, v3, v5 and v6?

Start at the junction of devices 2, 3 and 4, where the voltages is zero because it is connected to ground.Since v4 = 5 V, there is a 5-V drop across device 4 from left to right. Therefore, the voltage at the junctionof devices 4, 5 and 6 is −5 V. Devices 5 and 6 are grounded at their negative sides, so the voltage across eachis −5 V. Device 3 is grounded on both sides, so its voltage is zero. The negative side of device 1 is groundedand v1 = 10 V, so the voltage at the junction of devices 1 and 2 is 10 V. The negative side of device 2 isgrounded, so its voltage is also 10 V. In summary, we v2 = 10 V, v3 = 0 V, v5 = −5 V, and v6 = −5 V.


v4 = 5;

v1 = 10;

v5 = -v4

v6 = v5

v 3 = 0

v2 = v1

Problem 1–25. For t ≥ 0 the voltage across and power absorbed by a two-terminal device are v(t) = 2e−t

V and p(t) = 40e−2t mW. Find the total charge delivered to the device for t ≥ 0.

First find the current i(t) = p(t)/v(t) and then integrate the current to find the charge.


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i(t) = p(t)

v(t) =

40e−2t mW

2e−t V = 20e−t mA

q (t) =

t0

i(τ ) dτ =

t0

20e−τ dτ = −20e−τ t

0= −20

e−t

− 1

= 20

1− e−t

mC

To find the total charge delivered to the device for t ≥ 0, evaluate q (t) in the limit as t →∞.

q Total = limt→∞

q (t) = limt→∞

20

1− e−t

= 20(1− 0) = 20 mC


syms t

v t = 2*exp(-t);

pt = 40e-3*exp(-2*t);

it = simple(pt/vt)

q = simple(int(it,t,0,inf))

Problem 1–26. Repeat Problem 1–22 using MATLAB to perform the calculations. Create a vector for thevoltage values, v = [15 5 10 -10 20 20], and a vector for the current values, i = [ - 1 1 2 - 1 - 3 2 ].Compute the corresponding vector for the power values, p, using element-by-element multiplication (.*) andthen use the sum command to verify the power balance.

The following MATLAB code provides the solution.

de vice = [ 1 2 3 4 5 6] ;

v = [1 5 5 10 -10 2 0 2 0];

i = [ - 1 1 2 - 1 - 3 2 ] ;

p = v .*i

Absorbing = p>0

Balance = sum(p)

Results = [device' v' i' p' Absorbing']

The corresponding MATLAB output is shown below.

Balance =

0.0000e+000

Results =

1.0000e+000 15.0000e+000 -1.0000e+000 -15.0000e+000 0.0000e+000

2.0000e+000 5.0000e+000 1.0000e+000 5.0000e+000 1.0000e+000

3.000 0e+0 00 10 .000 0e+0 00 2 .000 0e+0 00 20 .000 0e+0 00 1 .0000 e+0 00

4.0000e+000 -10.0000e+000 -1.0000e+000 10.0000e+000 1.0000e+000

5.0000e+000 20.0000e+000 -3.0000e+000 -60.0000e+000 0.0000e+000

6.000 0e+0 00 20 .000 0e+0 00 2 .000 0e+0 00 40 .000 0e+0 00 1 .0000 e+0 00

The power balance is zero, as expected, and the other results match those in Problem 1–22.

Problem 1–27. Using the passive sign convention, the voltage across a device is v(t) = 170 cos(377t) Vand the current through the device is i(t) = 2 sin(377t) A. Using MATLAB, create a short script (m-file) toassign a value to the time variable, t, and then calculate the voltage, current, and power at that time. Runthe script for t = 5 ms and t = 10 ms and for each result state whether the device is absorbing or deliveringpower.

The following MATLAB code provides the solution.

t = 5e-3

vt = 170*cos(377*t)


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i t = 2*sin(377*t)

pt = vt*it

Absorbing = pt>0

t = 10e-3

vt = 170*cos(377*t)

i t = 2*sin(377*t)

pt = vt*it

Absorbing = pt>0


t = 5.0000e-003

vt = -52.5401e+000

it = 1.9021e+000

pt = -99.9357e+000

Absorbing = 0

t = 10.0000e-003

vt = -137.5240e+000

it = -1.1757e+000

pt = 161.6889e+000

Absorbing = 1

The following table summarizes the results.

t (ms) v(t) (V) i(t) (A) p(t) (W) Absorbing/Delivering5 -52.54 1.90 -99.94 Delivering

10 -137.52 -1.18 161.69 Absorbing

Problem 1–28. Power Ratio in dB (A). A stereo amplifier takes the output of a CD player, for example,and increases the power to an audible level. Suppose the output of the CD player is 50 mW and the desiredaudible output is 100 W per stereo channel, find the power ratio of the amplifier per channel in decibels(dB), where the power ratio in dB is

PRdB = 10 log10

p2 p1

The power values are given, so substitute into the equation for the power ratio and calculate.

PRdB = 10log10

p2 p1

= 10log10

100

0.05

= 10 log10 (2000) = (10)(3.301) = 33.01dB

Problem 1–29. AC to DC Converter (A). A manufacturer’s data sheet for the converter in FigureP1-29 states that the output voltage is vdc = 5 V when the input voltage vac = 120 V. When the load drawsa current idc = 40 A the input power is iac = 300 W. Find the efficiency of the converter.

The efficiency of the converter is the percentage of input power that is delivered to the load. The powerdelivered to the load is the product of the voltage and current, pLoad = vdcidc = (5 V)(40 A) = 200 W. Thepower input to the converter is 300 W, so the efficiency is (200 W)/(300 W) = 66.67%.

Problem 1–30. Charge-Storage Device (A). A capacitor is a two-terminal device that can store electric

charge. In a linear capacitor the amount of charge stored is proportional to the voltage across the device.For a particular device the proportionality is q (t) = 10−7v(t). If v(t) = 0 for t < 0 and v(t) = 10(1−e−5000t)for t ≥ 0, find the energy stored in the device at t = 200 µs.

Take the derivative of the expression for charge to find the expression for current.

i(t) = d

dt

10−7v(t)

=

d

dt

10−6(1− e−5000t)

= 10−6

5000 e−5000t

=

1

200 e−5000t

Multiply the expressions for voltage and current to find the expression for power.

p(t) = v(t) i(t) =

10(1− e−5000t) 1

200 e−5000t

=

1

20

e−5000t − e−10000t


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Integrate the power from t = 0 s to t = 200 µs to determine the energy stored in the circuit.

w =

200µs0

p(t) dt =

200µs0

1

20

e−5000t − e−10000t

dt

= 1

20

−

e−5000t

5000 +

e−10000t

10000

200µs

0

= 1

20

−60.042× 10−6 + 100× 10−6

= 1.9979 µJ


syms t

tt = 200e-6;

C = 1e-7;

vt = 10*(1-exp(-5000*t))

q t = C*vt

it = diff(qt,t)

pt = it*vt

wt = double(int(pt,t,0,tt))


vt =

10 - 10/exp(5000*t)

qt =

1/1000000 - 1/(1000000*exp(5000*t))

it =

1/(200*exp(5000*t))

pt =

-(10/exp(5000*t) - 10)/(200*exp(5000*t))

wt =

1.9979e-006

Problem 1–31. Computer Data Sheet (A). A manufacturer’s data sheet for a notebook computer liststhe power supply requirements as 7.5 A @ 5 V, 2 A @ 15 V, 2.5 A @ -15 V, 2.25 A @ -5 V and 0.5 A @12 V. The data sheet also states that the overall power consumption is 115 W. Are these data consistent?Explain.

Compute the power associated with each of the five requirements by multiplying the voltage and currenttogether. The resulting values are 37.5 W, 30 W, -37.5 W, -11.25 W, and 6 W. In this case, it is notreasonable for the computer to supply power back to the power supply, so the negative powers are not acorrect interpretation of the passive sign convention. We should take the absolute value of the individualpowers to determine the total power requirement for the computer. Summing the magnitudes of the powersyields a total of 122.25 W, which is greater than the stated power consumption of 115 W. The stated powerconsumption is probably a reasonable value, based on the performance of the computer and the fact that,typically, the power supply with not have to deliver the maximum values constantly.


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