(Solution Manual) Introduction to Electric Circuits

Dorf,Svoboda, 6

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Solution Manual

6th Edition

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Solution Manual for Introduction to Electric Circuits, The downloadable file consists of 670 pages, and the total size of the file is 7.06 MB. Purchase now only for highly discounted price.

Author(s): Dorf, Svoboda

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1 Solution Manual to accompany Introduction to Electric Circuits, 6e By R. C. Dorf and J. A. Svoboda 2 Table of Contents Chapter 1 Electric Circuit Variables Chapter 2 Circuit Elements Chapter 3 Resistive Circuits Chapter 4 Methods of Analysis of Resistive Circuits Chapter 5 Circuit Theorems Chapter 6 The Operational Amplifier Chapter 7 Energy Storage Elements Chapter 8 The Complete Response of RL and RC Circuits Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements Chapter 10 Sinusoidal Steady-State Analysis Chapter 11 AC Steady-State Power Chapter 12 Three -Phase Circuits Chapter 13 Frequency Response Chapter 14 The Laplace Transform Chapter 15 Fourier Series and Fourier Transform Chapter 16 Filter Circuits Chapter 17 Two-Port and Three-Port Networks Errata for Introduction to Electric Circuits, 6th Edition Errata for Introduction to Electric Circuits, 6th Edition Page 18, voltage reference direction should be + on the right in part B: Page 28, caption for Figure 2.3 -1: "current" instead of "cuurent" Page 41, line 2: "voltage or current" instead of "voltage or circuit" Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit. Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..." Page 164, Table 5.5 -1: method 2, part c, one should insert the phrase "Zero all independent sources, then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read: "Zero all independent sources, then connect a 1 -A source from terminal b to terminal a. Determine Vab. Then Rt = Vab/1." Page 340, Problem P8.3 -5: The answer should be . Page 340, Problem P8.3 -6: The answer should be . Page 341, Problem P.8.4 -1: The answer should be Page 546, line 4: The angle is instead of . Page 554, Problem 12.4.1 Missing parenthesis: Page 687, Equation 15.5 -2: Partial t in exponent: Errata for Introduction to Electric Circuits, 6th Edition Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2 (s) and H c(s) = V1(s) / Vs(s). 1-1 t Exercises Ex. 1.3-1 i t Chapter 1 – Electric Circuit Variables  8t 2  4 t A t 8 t q(t)   i d  q(0)   (8 2  4 ) d  0   3 2 2  8 t3 2 t 2 C 0 0 Ex. 1.3-3 3 0 3 q t    t i   d  q 0   t 4 sin 3 d  0  4 cos 3 t  4 cos 3 t  4 C 0 0 3 0 3 3 Ex. 1.3-4 0 t0 i t   dqt  i(t)  2 0t2 dt 2e2t 2 t 2  Ex. 1.4-1 i1 = 45 A = 45  10-6 A < i2 = 0.03 mA = .03  10-3 A = 3  10-5 A < i3 = 25  10-4 A Ex. 1.4-2  q  i t = 4000 A 0.001 s   4 C Ex. 1.4-3 i   q  t  45109 5103  9 106 = 9 A Ex. 1.4-4 i = 10 billion electron  1.602 1019 C  = 10109 electron  1.602 1019 C     s  electron   s  electron  = 1010 1.602 1019 electron C  1.602 109 C  s s 1.602 nA electron Ex. 1.6-1 1-2 0 0 0 0 t t 2 2  (a) The element voltage and current do not adhere to the passive convention in Figures 1.6 -1B and 1.6-1C so the product of the element voltage and current is the power supplied by these elements. (b) The element voltage and current adhere to the passive convention in Figures 1.6-1A and 1.6-1D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) The element voltage and current do not adher e to the passive convention in Figure 1.6 -1B, so the product of the element voltage and current is the power delivered by this element: (2 V)(6 A) = 12 W. The power received by the element is the negative of the power delivered by the element, -12 W. (d) The element voltage and current do not adhere to the passive convention in Figure 1.6 -1B, so the product of the element voltage and current is the power supplied by this element: (2 V)(6 A) = 12 W. (e) The element voltage and current adhere to the passive conv ention in Figure 1.6-1D, so the product of the element voltage and current is the power delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the element is the negative of the power delivered to the element, -12 W. Problems Section 1-3 Electric Circuits and Current Flow P1.3-1 i t  d 4 1 e5t  20 e5t A dt P1.3-2 q t   t i   d  q 0  t 4 1 e5  d  0  t 4 d  t 4 e5d  4 t  4 e5t  4 C 5 5 P1.3-3 t t q t    i   d   0 d  0 C for t  2 so q(2) = 0. q t   t i   d  q 2  t 2 d  2 t  2 t  4 C for 2  t  4. In particular, q(4) = 4 C. q t   t i   d  q 4  t 1 d  4   t  4  8  t C for 4  t  8. In particular, q(8) = 0 C. 4 4 q t   8 i   d  q 8  8 0 d  0  0 C 4 for 8  t . 2 P1.3-4 1-3 i = 600 A = 600 C s Silver deposited = 600 C20 min60 s 1.118 mg = 8.05105 mg=805 g s min C Section 1-6 Power and Energy P1.6-1 a.) q   i dt = it = 10 A2 hrs3600 s/hr = 7.2104 C b.) P  v i  110 V10 A  1100 W c.) Cost = 0.06 $ 1.1kW  2 hrs = 0.132 $ kWhr P1.6-2 P = 6 V10 mA  0.06 W  t w  P 200 Ws 0.06 W  3.33103 s P1.6-3 for 0  t  10 s: v = 30 V and i = 30 t 15  2t A  P  30(2t)  60t W for 10  t  15 s: v t    25 t  b 5  v 10  30 V  b  80 V v(t)  5t  80 and i(t)  2t A P  2t 5t 80  10t 2 160t W for 15 t  25 s: v  5 V and i(t)   30 t  b A 10 i(25)  0  b = 75 i(t)  3t 75 A  P  53t 75  15t 375 W 1-4 0 10 15 t Energy   P dt  10 60t dt  15 160t 10t 2 dt   25 37515t  dt  30t 2 10  80t 2  10t3 15  375t  15t 2 25  5833.3 J 0 3 10 2 15 P1.6-4 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). w   Pdt   vi d536002 110.5   d  22 t 0.5 53600  2 0 0 3600  3600 0 = 441103 J  441 kJ b.) Cost = 441kJ 1 hr  10¢  1.23¢ 3600s kWhr P1.6-5 p t   1 cos 3t sin 3t   1 sin 6 t 3 6 p 0.5  1 sin 3  0.0235 W 6 p 1  1 sin 6  0.0466 W 6 1-5 Here is a MATLAB program to plot p(t): clear t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time v=4*cos(3*t); % device voltage i=(1/12)*sin(3*t); % device current for k=1:length(t) p(k)=v(k)*i(k); % power end plot(t,p) xlabel('time, s'); ylabel('power, W') P1.6-6 p t   16 sin 3t sin 3t   8cos 0  cos 6 t   8  8 cos 6 t W Here is a MATLAB program to plot p(t): clear t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time v=8*sin(3*t); % device voltage i=2*sin(3*t); % device current for k=1:length(t) p(k)=v(k)*i(k); % power end plot(t,p) xlabel('time, s'); ylabel('power, W') 1-6 P1.6-7 p t   4 1 e2t  2 e2t  81 e2t e2t Here is a MATLAB program to plot p(t): clear t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time v=4*(1 -exp(-2*t)); % device voltage i=2*exp( -2*t); % device current for k=1:length(t) p(k)=v(k)*i(k); % power end plot(t,p) xlabel('time, s'); ylabel('power, W') P1.6-8 P  V I =3 0.2=0.6 W w  P  t  0.6  5 60=180 J 1-7 Verification Problems VP 1-1 Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W = 0 W The element voltages and currents satisfy conservation of energy and may be correct. VP 1-2 Notice that the element voltage and current of some branches do not adhere to the passive convention. The sum of the powers absorbed by each branch are: -(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A) = -9 W + 6 W + 6 W + 12 W + 9 W -12 W  0 W The element voltages and currents do not satisfy conservation of energy and cannot be correct. Design Problems DP 1-1 The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25) = 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust the estimates of the maximum voltage and current and a Grade A device otherwise. 1-8 DP1-2 p t   20 1 e8t  0.03e8t  0.6 1 e8t e8t Here is a MATLAB program to plot p(t): clear t0=0; % initial time tf=1; % final time dt=0.02; % time increment t=t0:dt:tf; % time v=20*(1 -exp(-8*t)); % device voltage i=.030*exp( -8*t); % device current for k=1:length(t) p(k)=v(k)*i(k); % power end plot(t,p) xlabel('time, s'); ylabel('power, W') Here is the plot: The circuit element must be able to absorb 0.15 W. 2-1 Exercises Ex. 2.3-1 Chapter 2 - Circuit Elements m i1  i2   mi1  mi2 m a i1   a mi1   superposition is satisfied homogeneity is satisfied Therefore the element is linear. Ex. 2.3-2 m i1  i2   b  mi1  mi2  b  mi1  b  mi2  b  superposition is not satisfied Therefore the element is not linear. Ex. 2.5-1 v 2 10 2 P    1 W R 100 Ex. 2.5-2 v2 (10 cos t)2 2 P    10 cos t W R 10 Ex. 2.8-1 ic  1.2 A, vd  24 V id  4 (1.2)   4.8 A id and vd adhere to the passive convention so P  vd id  (24) (4.8)  115.2 W is the power received by the dependent source Ex. 2.8-2 2-2 vc  2 V, id  4 vc  8 A and vd  2.2 V id and vd adhere to the passive convention so P  vd id  (2.2) (8)  17.6 W is the power received by the dependent source. The power supplied by the dependent source is 17.6 W. Ex. 2.8-3 ic  1.25 A, vd  2 ic  2.5 V and id  1.75 A id and vd adhere to the passive convention so P  vd id  (2.5) (1.75)  4.375 W is the power received by the dependent source. Ex. 2.9-1 2-3   45 , I  2 mA, R  20 k p a    a R  45 (20 k)  2.5 k 360 p 360 3 3 vm  (210 )(2.510 )  5 V Ex. 2.9-2 v 10 V, i  280 A, k 1 A for AD590 K i  K    i  kT  T  (280A)1 280 K k  A     Ex. 2.10-1 At t = 4 s both switches are open, so i = 0 A. Ex. 2.10.2 At t = 4 s the switch is in the up position, so v  i R= (2 mA)(3 k ) = 6V . At t = 6 s the switch is in the down position, so v = 0 V. Problems Section 2-3 Engineering and Linear Models P2.3-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.3-2 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is v  0.12 i . The element is indeed linear. (b) When i = 40 mA, v = (0.12 V/A)(40 mA) = (0.12 V/A)(0.04 A) = 4.8 mV (c) When v = 4 V, i 4 0.12  33 A = 33 A. 2-4 P2.3-3 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is v  256.5i . The element is indeed linear. (b) When i = 4 mA, v = (256.5 V/A)(4 mA) = (256.5 V/A)(0.004 A) = 1.026 V (c) When v = 12 V, i 12 256.5  0.04678 A = 46.78 mA. P2.3-4 Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v  3(2i) + 5 = 11.. Hence, the property of homogeneity is not satisfied. The element is not linear. Section 2-5 Resistors P2.5-1 P2.5-2 i  is  3 mA and v  24 V R  v  24  8000  8 k i .003 P = (3103 )24 = 72103  72 mW P2.5-3 v  vs =10 V and R  5  i  v  10 2 A R 5 v and i adhere to the passive convention  p  v i  210  20 W is the power absorbed by the resistor i  is  3 A and v  Ri = 7  3 = 21 V v and i adhere to the passive convention  P  v i  21  3 = 63 W is the power absorbed by the resistor. P2.5-4 2-5 R 2 v  vs  24 V and i  2 A R  v  24  12  i 2 p  vi  242 = 48 W P2.5-5 v1  v2  vs  150 V; R1  50 ; R2  25  v 1 and i1 adhere to the passive convention so i  v 1  150  3 A 1 R 50 1 v and i do not adhere to the passive convention so i  v2   150  6 A 2 2 2 25 The power absorbed by R1 is P1  v1 i1  150 3  450 W The power absorbed by R2 is P2   v 2i 2  150(6)   P2.5-6 900 W i 1  i 2  i s  2 A ; R1=4  and R2  8  v 1 and i 1 do not adhere to the passive convention so v 1 R 1 i 1 428 V. The power absorbed by R 1 is P1 v 1i 1 (8)(2)  16 W. v2 and i 2 do adhere to the passive convention so v2 R 2 i 2 8 2  16 V . The power absorbed by R 2 is P2  v 2i 2  16  2  32 W. P2.5-7 Model the heater as a resistor, then v2 v2 (250)2 with a 250 V source: P   R   62.5  R P 1000 v2 (210)2 with a 210 V source: P   705.6 W R 62.5 P2.5-8 2-6 The current required by the mine lights is: i  P  5000  125 A v 120 3 Power loss in the wire is : i2 R Thus the maximum resistance of the copper wire allowed is R  0.05P i2  0.055000 (125/3)2  0.144  now since the length of the wire is L  2100  200 m 20,000 cm thus R  L / A with  = 1.7106cm from Table 2.51 L 1.710620,000 2 A    0.236 cm R 0.144 Section 2-6 Independent Sources P2.6-1 (a) i  vs R  15  3 A and P 5  R i 2 5 3 2  45 W (b) i and P do not depend on is . The values of i and P are 3 A and 45 W, both when i s  3 A and when i s  5 A. P2.6-2 (a) v  R i s  5 2  10 V and P  v2 R  102 5  20 W (b) v and P do not depend on v s . The values of v and P are 10V and 20 W both when v s  10 V and when v s  5 V P2.6-3 2-7 Consider the current source: i s and v s do not adhere to the passive convention, so Pcs i sv s 312  36 W is the power supplied by the current source. Consider the voltage source: i s and v s do adhere to the passive convention, so Pvs  i s vs 312  36 W is the power absorbed by the voltage source.  The voltage source supplies 36 W. P2.6-4 Consider the current source: i s and vs adhere to the passive convention so Pcs  i s vs 312  36 W is the power absorbed by the current source. Current source supplies  36 W. Consider the voltage source: i s and vs do not adhere to the passive convention so Pvs  i s vs  31236 W is the power supplied by the voltage source. P2.6-5 (a) P  v i  (2 cos t) (10 cos t)  20 cos2 t mW 1 (b) w P dt  1 20 cos2 t dt= 20 1 t  1 sin 2t     10  5 sin 2 mJ 0 0  2 4     0 1 Section 2-7 Voltmeters and Ammeters P2.7-1 2-8 (a) R  v  5  10  i 0.5 (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W. P2.7-2 The voltmeter current is zero so the ammeter current is equal to the current source current except for the reference direction: i = -2 A The voltage v is the voltage of the current source. The power supplied by the current source is 40 W so 40  2 v  v  20 V Section 2-8 Dependent Sources P2.8-1 2-9 r  vb  8  4  i a 2 P2.8-2 v  8 V ; g v  i  2 A ; g  i a  2  0.25 A b b a vb 8 V P2.8-3 i  8 A ; d i  i  32A ; d  i a  32  4 A b b a i b 8 A P2.8-4 v  2 V ; b v  v  8 V ; b  vb  8  4 V a a b va 2 V Section 2 -9 Transducers P2.9-1 P2.9-2 a =  ,  = 360 vm 360 Rp I  = (360)(23V) = 75.27 (100 k)(1.1 mA) AD590 : k =1 A ,  K v=20 V (voltage condition satisfied) 4  A < i < 13  A  i   4K< T <13K T   k  Section 2-10 Switches P2.10-1 2-10 At t = 1 s the left switch is open and the right switch is closed so the voltage across the resistor is 10 V. i = v = 10  2 mA R 5103 At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V. i = v = 15  3 mA R 5103 P2.10-2 At t = 1 s the current in the resistor is 3 mA so v = 15 V. At t = 4 s the current in the resistor is 0 A so v = 0 V. Verification Problems VP2-1 vo =40 V and i s =  (2)  2 A. (Notice that the ammeter measures  i s rather than i s .) So vo  40  20 V i s 2 A Your lab partner is wrong. VP2-2 We expect the resistor current to be i  vs R  12 25 = 0.48 A. The power absorbed by this resistor will be P  i vs = (0.48) (12) = 5.76 W. A half watt resistor can't absorb this much power. You should not try another resistor. 2-11 1 2 3 Design Problems DP2-1 1.) 10  0.04  R R  10 0.04  250    2.) 102 R  1  2 R  200   Therefore 200 < R < 250 . For example, R = 225 . DP2-2 1.) 2R  40  2.) 22 R  15  R  20  R  15  3.75  4 Therefore 20 < R < 3.75 . These conditions cannot satisfied simultaneously. DP2-3 P  30 mA2  1000   .032 1000  0.9 W  1 W P  30 mA2  2000   .032 2000  1.8 W  2 W P  30 mA2  4000   .032 4000  3.6 W  4 W 3-1 Chapter 3 – Resistive Circuits Exercises Ex 3.3-1 Apply KCL at node a to get 2 + 1 + i3 = 0  i3 = -3 A Apply KCL at node c to get 2 + 1 = i4  i4 = 3 A Apply KCL at node b to get i3 + i6 = 1  -3 + i6 = 1  i6 = 4 A Apply KVL to the loop consisting of elements A and B to get -v2 – 3 = 0  v2 = -3 V Apply KVL to the loop consisting of elements C, E, D, and A to get 3 + 6 + v4 – 3 = 0  v4 = -6 V Apply KVL to the loop consisting of elements E and F to get v6 – 6 = 0  v6 = 6 V Check: The sum of the power supplied by all branches is -(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0 Ex 3.3-2 3-2 Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0  i = 4 A Ex 3.3-3 18  0 12  v  0  v  30 V and i  2 v  3   i  9 A a a m 5 a m Ex 3.3-4 v 10  4v  8  0  v  18  6 V and v  4 v  24 V a a a 3 m a Ex 3.4-1 From voltage division v  12  3  = 3V 3  39   then i = v3 = 1A 3 The power absorbed by the resistors is: 12 6  12 3  12 3  12 W The power supplied by the source is (12)(1) = 12 W.