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APPENDIX A Exercises EA.1 Given ,68 and 32 21 jZjZ  we have: 310 21 jZZ  96 21 jZZ  123418122416 2 21 jjjjZZ  36.002.0 100 18241216 68 68 68 32 / 2 21 j jjj j j […]

APPENDIX C PC.1 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0. Then using Equation 3.5 in the text, we have tdx dxtitq t t 33 0)()( 0 […]

(b) If the diameter of the wire is doubled, the cross sectional area A is increased by a factor of four. Thus, the resistance is decreased by a factor of four to 2.5 .  P1.56 The resistance is proportional […]

CHAPTER 1 Exercises E1.1 Charge = Current  Time = (2 A)  (10 s) = 20 C E1.2 A )2cos(200 )200cos(2000.01 0t)0.01sin(20( )( )( tt dt d dt tdq ti  E1.3 Because i 2 has a positive value, […]

P10.65 P10.66 37 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, […]

We must be careful to choose the value of R small enough so z I remains positive for all values of source voltage and load current. (Keep in mind that the Zener diode cannot supply power.) From the circuit, we […]

CHAPTER 10 Exercises E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have 4 15 10 exp(0.600 / 0.026) 1 9.502 10 A     Then for 0.650 D v  V, we have 15 […]

P11.39 The equivalent circuit is: 7 36 3 8 10667.6 50 100 1050010 10500 10       i i i ooc voc I I V V A  50 iB iRR  200 oA oRR Thus, […]

CHAPTER 11 Exercises E11.1 (a) A noninverting amplifier has positive gain. Thus )2000sin(0.5)(50)()( ttvtvAtv ii vo π (b) An inverting amplifier has negative gain. Thus )2000sin(0.5)(50)()( ttvtvAtv ii vo π E11.2 7525 oc    L o v i […]

(d) The gain magnitude is constant and the phase is proportional to frequency so this amplifier produces no amplitude or phase distortion. P11.79 The sketch is The relationship between rise time and bandwidth is: B tr 35.0  Percentage tilt, […]

Substituting values and solving we find 582.1 GSQ V V and 582.3 GSQ V V. The correct root is 582.3 GSQ V V. (As a check, we see that the device does operate in saturation because we have 8 DSQ […]

CHAPTER 12 Exercises E12.1 (a) vGS  1 V and vDS  5 V: Because we have vGS < Vto , the FET is in cutoff. (b) vGS  3 V and vDS  0.5 V: Because vGS > Vto […]

(e) We are given )2000sin(520)()( ttiti s B  . From which we determine that A. 25 and A, 20 A, 15 maxmin μμμ  BBQBIII Then at the intersections of the load line with the respective characteristics, we determine […]

CHAPTER 13 Exercises E13.1 The emitter current is given by the Shockley equation:               1exp T BE ESEV v Ii For operation with 1exp have we , […]

P14.25 (a) 21 00 vRivo  R vv io 21   Since io is independent of the load, the output impedance is infinite. (b) The circuit diagram is: Writing KVL around loop #1, we have ininin RiRiv  0 […]

and the rate of change of the output is     tV dt tdv om o ωω cos The maximum rate of change of the output is   om oV dt tdv ω max Thus, we require […]

CHAPTER 14 Exercises E14.1 (a) A A AR v i  B B BR v i  B B A A B A FR v R v iii          B B A […]

dt di L dt di Me 2 2 1 2         dt td dt td e 1000exp2 1.0 1000exp 2.0 1           dt td […]

CHAPTER 15 Exercises E15.1 If one grasps the wire with the right hand and with the thumb pointing north, the fingers point west under the wire and curl around to point east above the wire. E15.2 If one places the […]

Nm 3980.0 60 2 1200 ref , rot    π ω m T At no load, we have 0 out  T and rot dev TT  . A 2857.0 393.1 3980.0 dev  φ K T IA […]

CHAPTER 16 Exercises E16.1 The input power to the dc motor is loss outsourcesource in PPIVP  Substituting values and solving for the source current we have 335074650220  source I A 8.184 source I Also we have %76.91 335074650 […]

CHAPTER 17 Exercises E17.1 From Equation 17.5, we have )240cos()()120cos()()cos()( gap    tKitKitKiBc b a Using the expressions given in the Exercise statement for the currents, we have )240cos()120cos( )120cos()240cos()cos()cos( gap       tKI […]

  15.044.106.05.7 15.044.106.05.7 20.008.0 jj jj jZ s    6193.0468.1 j   87.22594.1    lagging %14.9287.22cosfactor power      87.220.276 87.22594.1 0440     s s sZ V I […]

P2.18 (a) For a series combination 321 /1/1/1 GGG  (b) For a parallel combination of conductances 321 GGGGeq  1 Geq  P2.19 To supply the loads in such a way that turning one load on or off does […]

%50%100  s L P P η On the other hand, for t LRR 9 , we have       %90%100 9 9.0 10 2 22      s L t t L […]

CHAPTER 2 Exercises E2.1 (a) R 2 , R 3 , and R 4 are in parallel. Furthermore R 1 is in series with the combination of the other resistors. Thus we have:    3 /1/1/1 1 432 […]

102040 21  ii 04020 21  ii Solving, we find 3333.0 1 i and 1667.0 2 i . Thus,   V 333.320 21  iiv . P2.72 The mesh currents and corresponding equations are: mA 30 A i […]

CHAPTER 3 Exercises E3.1 V )10sin(5.0)102/()10sin(10/)()( 5656 ttCtqtv   A )10cos(1.0)10cos()105.0)(102()( 5556 tt dt dv Cti   E3.2 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = […]

P3.46        1 0 10 1 t LLL idttv L ti A 20010)6( 6 0 3 1  dtiL       W 200666  LL ivp J 600)6()6( 2 2 […]

CHAPTER 4 Exercises E4.1 The voltage across the circuit is given by Equation 4.8: )/exp()( RCtVtv i C  in which Vi is the initial voltage. At the time t 1% for which the voltage reaches 1% of the initial […]

Thus, 1 2 K and the current (in amperes) is given by     0 for 20exp1 0 for 0   tt tti P4.34 The general form of the solution is     LtRKKtiL  […]

Since we have 0 ωα  , this is the under-damped case. The natural frequency is given by Equation 4.76 in the text: 322 010660.8  αωω n The complementary solution is given in Equation 4.77 in the text:  […]

P5.45* The peak value of   tiL is five times larger than the source current! This is possible because current in the capacitance balances the current in the inductance (i.e., 0 C L II ).   mA 9050 […]

1 CHAPTER 5 Exercises E5.1 (a) We are given )30200cos(150)(   ttv  . The angular frequency is the coefficient of t so we have radian/s 200   . Then Hz 1002/  f ms 10/1  fT […]

     120cos200  ttvcn                 90cos3200 30cos3200 150cos3200    ttv ttv ttv bc ab    ca P5.90 […]

P6.91 (a) Applying the voltage-division principle, we have fCjR R R fH    2/1 1 )( 1 2 2 (b) A MATLAB program to produce the desired plot is R1 = 9000; R2 = 1000; C = 1e-8; […]

The open-circuit voltage is given by       L s L soctRRR R tvtvtv   In terms of phasors, this becomes: L s L st RRR R  VV (1) Zeroing the source, we find […]

CHAPTER 6 Exercises E6.1 (a) The frequency of )20002cos(2)( in ttv   is 2000 Hz. For this frequency .602)(   fH Thus,  60402602)( in out  VV fH and we have ).6020002cos(4)( out   ttv  […]

(c) At very low frequencies, with the capacitance considered to be an open circuit, no current flows and )( fH becomes very small in magnitude as shown in the plot. (d) At very high frequencies with the inductance considered as […]

BABAY  CBCBZ  P7.78 By inspection, we see that XA  The Karnaugh maps for B and C are: YXYXB  XYZZYXZYXZYXC  P7.79* (a) DBBCAF  38 © 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights […]

CHAPTER 7 Exercises E7.1 (a) For the whole part, we have: Quotient Remainders 23/2 11 1 11/2 5 1 5/2 2 1 2/2 1 0 1/2 0 1 Reading the remainders in reverse order, we obtain: 2310 = 101112 For […]

P7.52 Applying De Morgan’s Laws to the output of the circuit, we have )()())(( DCBADCBA  Thus, the circuit implemented with NOR gates is P7.53* The truth table is: A B A  B 0 0 0 0 1 1 […]

CHAPTER 8 Exercises E8.1 The number of bits in the memory addresses is the same as the address bus width, which is 20. Thus, the number of unique addresses is 220 = 1,048,576 = 1024  1024 = 1024K. E8.2 […]

CHAPTER 9 Exercises E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown i Figure 9.2 in the book. Thus the input voltage is in sensor in sensor in RR R vv   […]