Problem-Solving Companion To accompany Basic Engineering Circuit Analysis Ninth Edition J David Irwin Auburn University JOHN WILEY & SONS, INC Executive Editor Assistant Editor Marketing Manager Senior Production Editor Bill Zobrist Kelly Boyle Frank Lyman Jaime Perea Copyright © 2005, John Wiley & Sons, Inc All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 7504470 Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212)850-6008, e-mail: PERMREQ@WILEY.COM ISBN 0-471-74026-8 TABLE OF CONTENTS Preface……………………………………………………………………………………….2 Acknowledgement………………… ……………………………………………………….2 Chapter Problems………………… ……………………………….……………………………………………3 Solutions………………… …………………………………………………….………………………4 Chapter Problems………………… …………………………………………………….………………………7 Solutions………………… …………………………………………………….………………………9 Chapter Problems………………… …………………………………………………….………………………18 Solutions………………… …………………………………………………….………………………19 Chapter Problems………………… …………………………………………………….………………………25 Solutions………………… …………………………………………………….………………………26 Chapter Problems………………… …………………………………………………….………………………31 Solutions………………… …………………………………………………….………………………33 Chapter Problems………………… …………………………………………………….………………………42 Solutions………………… …………………………………………………….………………………44 Chapter Problems………………… …………………………………………………….………………………50 Solutions………………… …………………………………………………….………………………52 Chapter Problems………………… …………………………………………………….………………………66 Solutions………………… …………………………………………………….………………………67 Chapter Problems………………… …………………………………………………….………………………73 Solutions………………… …………………………………………………….………………………74 Chapter 10 Problems………………… …………………………………………………….………………………82 Solutions………………… …………………………………………………….………………………83 Chapter 11 Problems………………… …………………………………………………….………………………88 Solutions………………… …………………………………………………….………………………89 Chapter 12 Problems………………… …………………………………………………….………………………94 Solutions………………… …………………………………………………….………………………96 Chapter 13 Problems………………… …………………………………………………….………………………103 Solutions………………… …………………………………………………….………………………104 Chapter 14 Problems………………… …………………………………………………….………………………111 Solutions………………… …………………………………………………….………………………113 Chapter 15 Problems………………… …………………………………………………….………………………127 Solutions………………… …………………………………………………….………………………129 Chapter 16 Problems………………… …………………………………………………….………………………136 Solutions………………… …………………………………………………….………………………137 Appendix – Techniques for Solving Linear Independent Simultaneous Equations………………………… 146 STUDENT PROBLEM COMPANION To Accompany BASIC ENGINEERING CIRCUIT ANALYSIS, NINTH EDITION By J David Irwin and R Mark Nelms PREFACE This Student Problem Companion is designed to be used in conjunction with Basic Engineering Circuit Analysis, 8e, authored by J David Irwin and R Mark Nelms and published by John Wiley & Sons, Inc The material tracts directly the chapters in the book and is organized in the following manner For each chapter there is a set of problems that are representative of the end-of-chapter problems in the book Each of the problem sets could be thought of as a mini-quiz on the particular chapter The student is encouraged to try to work the problems first without any aid If they are unable to work the problems for any reason, the solutions to each of the problem sets are also included An analysis of the solution will hopefully clarify any issues that are not well understood Thus this companion document is prepared as a helpful adjunct to the book CHAPTER PROBLEMS 1.1 Determine whether the element in Fig 1.1 is absorbing or supplying power and how much -2A 12V + Fig 1.1 1.2 In Fig 1.2, element absorbs 24W of power Is element absorbing or supplying power and how much 12V + + 6V Fig 1.2 1.3 Given the network in Fig.1.3 find the value of the unknown voltage VX + 4V + 10V 2A 6A 2A 4A + 12V + - 8V - Fig 1.3 + - VX CHAPTER SOLUTIONS 1.1 One of the easiest ways to examine this problem is to compare it with the diagram that illustrates the sign convention for power as shown below in Fig S1.1(b) -2A i(t) + 12V v(t) + - Fig S1.1(a) Fig S1.1(b) We know that if we simply arrange our variables in the problem to match those in the diagram on the right, then p(t) = i(t) v(t) and the resultant sign will indicate if the element is absorbing (+ sign) or supplying (- sign) power If we reverse the direction of the current, we must change the sign and if we reverse the direction of the voltage we must change the sign also Therefore, if we make the diagram in Fig S1.1(a) to look like that in Fig S1.1(b), the resulting diagram is shown in Fig S1.1(c) 2A + (-12V) Fig S1.1(c) Now the power is calculated as P = (2) (-12) = -24W And the negative sign indicates that the element is supplying power 1.2 Recall that the diagram for the passive sign convention for power is shown in Fig S1.2(a) and if p = vi is positive the element is absorbing power and if p is negative, power is being supplied by the element i + v - Fig S1.2(a) If we now isolate the element and examine it, since it is absorbing power, the current must enter the positive terminal of this element Then P = VI 24 = 6(I) I = 4A The current entering the positive terminal of element is the same as that leaving the positive terminal of element If we now isolate our discussion on element 1, we find that the voltage across the element is 6V and the current of 4A emanates from the positive terminal If we reverse the current, and change its sign, so that the isolated element looks like the one in Fig S1.2(a), then P = (6) (-4) = -24W And element is supplying 24W of power 1.3 By employing the sign convention for power, we can determine whether each element in the diagram is absorbing or supplying power Then we can apply the principle of the conservation of energy which means that the power supplied must be equal to the power absorbed If we now isolate each element and compare it to that shown in Fig S1.3(a) for the sign convention for power, we can determine if the elements are absorbing or supplying power i + P = Vi V - Fig S1.3(a) For the 12V source and the current through it to be arranged as shown in Fig S1.3(a), the current must be reversed and its sign changed Therefore P12V = (12) (-6) = -72W Treating the remaining elements in a similar manner yields P1 = (4) (6) = 24W P2 = (2) (10) = 20W P3 = (8) (4) = 32W PVX = (VX) (2) = 2VX Applying the principle of the conservation of energy, we obtain And -72 + 24 + 20 + 32 + 2VX = VX = -2V CHAPTER PROBLEMS 2.1 Determine the voltages V1 and V2 in the network in Fig 2.1 using voltage division 2kΩ 12v +- + 4kΩ 3kΩ + V2 - 2kΩ V1 - Fig 2.1 2.2 Find the currents I1 and I0 in the circuit in Fig 2.2 using current division 2kΩ I1 6kΩ 3kΩ 12kΩ 9mA I0 Fig 2.2 2.3 Find the resistance of the network in Fig 2.3 at the terminals A-B 8kΩ 10kΩ 2kΩ A 12kΩ 3kΩ 4kΩ 12kΩ 6kΩ 18kΩ B 6kΩ 3kΩ Fig 2.3 2.4 Find the resistance of the network shown in Fig 2.4 at the terminals A-B A B 4kΩ 6kΩ 2kΩ 12kΩ 12kΩ Fig 2.4 18kΩ 12kΩ 2.5 Find all the currents and voltages in the network in Fig 2.5 2kΩ A 10kΩ B I1 48V + - 4kΩ + V1 I2 6kΩ I3 I4 + V2 - 2kΩ 3kΩ I5 I6 + V3 - 4kΩ Fig 2.5 2.6 In the network in Fig 2.6, the current in the 4kΩ resistor is 3mA Find the input voltage VS 2kΩ VS +- 2kΩ 1kΩ 4kΩ 3mA Fig 2.6 9kΩ 6kΩ 3kΩ 154 Then A-1 is called the inverse of A It can be shown that the inverse of the matrix A is equal to the adjoint divided by the determinant (written here as |A|); that is A −1 = adj A A (A.21) Example A.10 Given ⎡2 ⎤ B = ⎢⎢1 3⎥⎥ ⎢⎣3 2⎥⎦ Find B-1 Solution 3 ⎡3 1⎤ −1 +3⎢ ⎥ 2 ⎣2 3⎦ = − + 21 = 18 B =2 and ⎡ −5 ⎤ adj B = ⎢⎢ − 5⎥⎥ ⎢⎣− ⎥⎦ Therefore, ⎡ −5 ⎤ ⎢ − 5⎥⎥ B = ⎢ 18 ⎢⎣− ⎥⎦ −1 We now have the tools necessary to solve Eqs (A.1) using matrices The following example illustrates the approach Example A.11 The node equations for a network are 2V1 + 3V2 + V3 = 155 V1 + 2V2 + 3V3 = 3V1 + V2 +2V3 = Let us solve this set of equations using matrix analysis Solution Note that this set of simultaneous equations can be written as a single matrix equation in the form ⎡2 ⎤ ⎡ V1 ⎤ ⎡9⎤ ⎢1 ⎥ ⎢ V ⎥ = ⎢6 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣3 2⎥⎦ ⎢⎣ V3 ⎥⎦ ⎢⎣8⎥⎦ or AV = I Multiplying both sides of the preceding equation through A-1 yields A-1AV = A-1I or V = A-1I A-1 was calculated in Example A.10 Employing that inverse here, we obtain ⎡ − ⎤ ⎡9 ⎤ ⎢ − 5⎥⎥ ⎢⎢6⎥⎥ V= ⎢ 18 ⎢⎣− ⎥⎦ ⎢⎣8⎥⎦ or ⎡ V1 ⎤ ⎢V ⎥ = ⎢ ⎥ 18 ⎢⎣ V3 ⎥⎦ ⎡35⎤ ⎢29⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ and hence, V1 = 35 29 , V2 = , and V3 = 18 18 18 E–1 EQUATIONS CHAPTER Electric current-charge relationship i(t) = dq(t) dt q(t) = or Power t 3-q dw = p = vi dt i(x) dx Voltage-energy relationship v = Energy dw dq ¢w = t2 3t1 p dt = t2 3t1 vi dt CHAPTER Ohm’s law Two series resistors & voltage divider v(t)=R*i(t) or i(t)=Gv(t) where G = R i(t) = Power p(t)=v(t)i(t) = Ri2(t) = v2(t) R v(t) R1 + R2 vR1 = R1 v(t) R1 + R2 vR2 = R2 v(t) R1 + R2 Kirchhoff’s Current Law (KCL) i(t) N a ij(t) = + j=1 R1 vR – v(t) Kirchhoff’s Voltage Law (KVL) + N a vj(t) = R2 j=1 vR – Multiple series resistors & voltage divider RS = R1 + R2 + p + RN i(t) = vRi = v(t) RS Ri v(t) RS + vR1 – R1 + vR2 R2 i(t) – + vR3 – R3 + R4 vR4 i(t) – v(t) + R5 RN –v RN vR5 – + v(t) RS = R1 + R2 + R3 + … + RN E–2 Two parallel resistors & current divider Rp = R1 R2 R1 + R2 + R1 R2 v(t) = Rp i(t) = i(t) R1 + R2 i(t) R2 i1(t) = i(t) R1 + R2 R2 v(t) R1 i1(t) i2(t) – R1 i2(t) = i(t) R1 + R2 Multiple parallel resistors & current divider N 1 = a Rp R i=1 i ij(t) = Rp Rj + v(t) io(t) io(t) i1(t) i2(t) iN(t) R1 R2 RN + v(t) io(t) – Rp – Delta-to-wye resistance conversion Ra = R1 R2 R1 + R2 + R3 Rb = R2 R3 R1 + R2 + R3 Rc = R1 R3 R1 + R2 + R3 Delta-to-wye resistance conversion (Special case: Identical resistors) R Y = 13 R ¢ Wye-to-delta resistance conversion R1 = Ra Rb + Rb Rc + Ra Rc Rb R2 = Ra Rb + Rb Rc + Ra Rc Rc R3 = Ra Rb + Rb Rc + Ra Rc Ra a a Ra R1 R2 Rc c b c R3 Wye-to-delta resistance conversion (Special case: Identical resistors) R ¢ = 3R Y Rb b E–3 CHAPTER CHAPTER Ohm’s law expressed in node voltages i = Ideal op amp vm - v N R i+ = i- = v+ = v- R Node m i + vm Node N + +(vm–vN)- v+ i+ v– i– vN – – Ohm’s law expressed in loop currents v3 = Ai1 - i2 B R3 A + v1 vS2 B – R1 i1 vS1 + v3 R3 – R2 F C – v + E + i2 v4 R4 R5 – – v + D CHAPTER Equivalent circuit forms R1 R1 + R2 R2 R1 R1 R2 ———— R1 + R2 R2 V1 V1 – V2 V2 ∞ E–4 Equivalent circuit forms (continued) I1 I2 I1 – I2 + VS R Io = IS R Vo = VS IS – Thévenin & Norton equivalent circuits A i RTh voc i + A + Circuit B vo voc = RTh isc isc RTh vo – – B B i = isc - vo = voc - RTh i Circuit B vo RTh Maximum power transfer theorem (Thévenin v and R fixed, load RL variable) RL = R Pload = R i v2 4R v RL v & R fixed, RL variable CHAPTER Parallel-plate capacitor-Capacitance C = Charge stored on a capacitor eo A d q = Cv Current-voltage relationship of a capacitor dq i = —— dt d A + + q(t) – v(t) Dielectric (a) – i = C C v(t) = (b) dv dt t i(x) dx C 3-q Energy stored in a capacitor N 1 = a CS C i=1 i q2(t) = 21 Cv2(t) J C wC(t) = Current-voltage relationship of an inductor Capacitors connected in parallel N di(t) v(t) = L dt i(t) = E–5 Capacitors connected in series Cp = a Ci i=1 t v(x) dx L 3-q Inductors connected in series N LS = a Li i(t) i=1 + v(t) L – Inductors connected in parallel N 1 = a Lp L i=1 i Energy stored in an inductor wL(t) = 21 Li2(t) J CHAPTER First-order circuits Second-order circuits The unit step function Characteristic equation of a second-order circuit u(t) = b t t s2 + 20 s + 20 = Roots of the characteristic equation General form of the step response of a first-order circuit excited at t = t0 x(t) = x(q) + Cx At0 B - x(q)De-At - t0B͞, t = t0 where x At0 B is the initial value and x (q) is the final value s1 = -0 + 0 22 - s2 = -0 - 0 22 - Overdamped response (i.e., > 1) x(t) = K1 e-A0 - 0 2 - 1Bt + K2 e-A0 + 0 2 Time constant of a first-order capacitive circuit = R Th C Critically damped response (i.e., = 1) x(t) = B1 e-0 t + B2 te-0 t Time constant of a first-order inductive circuit = L RTh Underdamped response (i.e., < 1) x(t) = e-t AA cos d t + A sin d tB, where = 0 , and d = 0 21 - 2 - 1Bt E–6 CHAPTER General form of a sinusoidal waveform x(t) = XM sin (t + ) where = Admittance 2 = 2f T Y = Conversion between sine and cosine functions Admittances of R, L, and C cos t = sin a t + b sin t = cos a t - I = Z V b YR = = G R YL = 1 = / 90° jL L Impedance YC = jC = C / 90° VM / v VM V = Z = = / v - i = Z / z I IM / i IM Admittances connected in series Impedances connected in series N 1 = a YS Y i=1 i N ZS = a Zi i=1 Admittances connected in parallel Impedances connected in parallel N Yp = a Yi i=1 N 1 = a Zp i = Zi Impedances of R, L, and C Passive element Impedance R Z = R L Z = jL = jXL = L / 90° , XL = L Z = C 1 = jXC = / 90° , XC = jC C C CHAPTER Average (real) power absorbed by an impedance (watts) P = VM IM cos Av - i B P = I2rms R = Maximum average power transfer theorem (When VOC and ZTh fixed, load ZL variable) ZL = RL + jXL = RTh - jXTh = Z*Th Z Th Voc Power factor (pf) pf = cos Av - i B = cos ZL S = P + jQ = Vrms / v Irms / -i + – V2rms R Complex power (volt-amperes) IL VL ac circuit Average power absorbed by a resistor ZL = Vrms Irms / v - i = I 2rms Z Voc & Z Th fixed, ZL variable Maximum average power transfer theorem (Special case: XL = 0) RL = 2R2Th + P = Re(S) = Vrms Irms cos Av - i B = I 2rms Re(Z) Reactive power (vars) Q = Im(S) = Vrms Irms sin Av - i B = I 2rms Im(Z) X2Th RMS value of a sinusoidal waveform Irms = E–7 Average (real) power (watts) Power triangle relationship tan Av - i B = tan Z = IM 12 Q P Im ▲ ▲ ▲ Average power absorbed in terms of rms values S P = Vrms Irms cos Av - i B +Q ▲ Re –Q ▲ S ▲ θv– θi P θv– θi ▲ ▲ CHAPTER 10 Magnetic flux, voltage and current relationships Voltage-current relationships for mutually coupled coils =N=Li webers v = d di = L dt dt i v1(t) = L1 di1(t) di2(t) + M dt dt v2(t) = M di1(t) di2(t) + L2 dt dt + i1(t) v N i2(t) + v1(t) – + M L1 v2(t) L2 – – Phasor voltage-current relationships for mutually coupled coils Ideal transformer equations v1 i1 N1 = - = v2 i2 N2 V1 = jL1 I1 + jMI2 V2 = jL2 I2 + jMI1 Energy stored in magnetically coupled inductors The coefficient of coupling i1(t) v1(t)+ – φ ▼ ▼ • A N2 N1 φ ▼ k = M 2L1 L2 where Յ k Յ Ideal transformer equations in phasor form V1 I2 N1 = = V2 I1 N2 The turns ratio of a transformer n = N2 N1 • i2(t) ▼ w(t) = 21 L1 Ci1(t)D + 21 L2 Ci2(t)D ; Mi1(t)i2(t) + v (t) –2 E–8 CHAPTER 11 Three-phase terminology Quantity Wye Delta Line current AIL B Ia , Ib , Ic Phase current AIp B Line-to-neutral voltage AVp B Van , Vbn , Vcn Phase voltage AVp B Line-to-line, phase-to-phase, line voltage AVL B Phase voltage AVp B Vab , Vbc , Vca Phase current AIp B Iab , Ibc , Ica Voltage, current, and impedance relationhips of Y and ⌬ configurations ⌬ Y Line voltage AVab or VAB B 13 Vp / + 30° VL / + 30° = VL / + 30° Line current IaA IL / IL / Phase voltage Vp / AVan or VAN B 13 Vp / + 30° Phase current IL / Load impedance ZY / - IL 13 / + 30° ZY / - E–8 CHAPTER 12 Resonant frequency of a series or parallel RLC circuit 0 = 1LC BW = HI - LO = Quality factor of a series RLC circuit Q = Bandwidth of a parallel RLC circuit where 0 L 1 L = = R 0 CR R BC Bandwidth of a series RLC circuit BW = HI - LO = 0 Q where LO = - 1 + + 2RC B (2RC) LC and HI = 1 + + LC 2RC B (2RC)2 Half-power (break) frequency of a first-order RC filter LO = 0 c HI = 0 c RC 1 + a b + 1d 2Q B 2Q 1 + a b + 1d 2Q B 2Q = 1 = RC Bandwidth of a series RLC bandpass filter BW = HI - LO = and R L 20 = LO HI L Quality factor of a parallel RLC circuit 0 C = R Q = BW AL V1 + C R Vo – E–10 CHAPTER 13 Laplace transforms of some special functions Laplace transform of a function f(t) LCf(t)D = F(s) = q 30 f(t) f(t)e-st dt The unit impulse function ␦At - t0 B = t Z t0 and F(s) ␦(t) u(t) s e-at s + a s2 t t0 + 3t0 - ␦At - t0 B dt = tn n! Sampling property of the unit impulse function f At B f(t)␦At - t0 B dt = b 0 3t1 t2 t1 t0 t2 t0 t1 , t0 t2 sn + 1 (s + a)2 te-at tne-at n! (s + a)n + sin bt b s + b2 cos bt s s + b2 The initial-value theorem lim f(t) = lim sF(s) t S0 SSq e-at sin bt b (s + a)2 + b2 e-at cos bt s + a (s + a)2 + b2 The final-value theorem lim f(t) = lim sF(s) tSq S S0 Some properties of Laplace transform Property Number f(t) F(s) Af(t) AF(s) f1(t) ; f2(t) F1(s) ; F2(s) f(at) s Fa b, a a a fAt - t0 BuAt - t0 B, t0 Ն e-t0 sF(s) f(t)uAt - t0 B e-t0 sLCfAt + t0 B D e-atf(t) F(s + a) n d f(t) dt n tf(t) f(t) t t 10 11 30 30 f() d t f1()f2(t - ) d snF(s) - sn - 1f(0) - sn - 2f1(0) p s0fn - 1(0) - dF(s) ds q F() d 3s F(s) s F1(s)F2(s) E–11 CHAPTER 14 The voltage-current relationship of a resistor in the s-domain The voltage-current relationship of a inductor in the s-domain V(s) = sLI(s) - Li(0) V(s) = RI(s) I(s) = The voltage-current relationship of a capacitor in the s-domain V(s) = I(s) v(0) + s sC V(s) i(0) + s sL Transfer or network function Yo(s) = H(s) Xi(s) I(s) = sCV(s) - Cv(0) CHAPTER 15 Trigonometric Fourier series of a periodic function f(t) Fourier transform pair F() = FCf(t)D = q f(t) = a0 + a Dn cos An0 t + n B q 3-q f(t)e-jt dt n=1 f(t) = F-1 CF()D = q = a0 + a an cos n0 t + bn sin n0 t n=1 q F()ejt d 2 3-q where t1 + T0 an = f(t) cos n0 t dt T0 3t1 bn = T0 3t1 Fourier transform of some special functions f(t) F() t1 + T0 f(t) sin n0 t dt and Dn / n = an - jbn Exponential Fourier series of a periodic function f(t) q f(t) = a cn ejn0 t n = -q ␦(t - a) e-ja A 2A␦() ej0 t 2␦A - 0 B cos 0 t ␦A - 0 B + ␦A + 0 B sin 0 t j␦A + 0 B - j␦A - 0 B e-at u(t), a a + j e-␣∑t∑, a 2a a + 2 where t1 + T0 cn = f(t)e-jn0 t dt T0 3t1 Relationships between various Fourier series coefficients Dn / n = 2cn = an - jbn e-at cos 0 tu(t), a e-at sin 0 tu(t), a j + a (j + a)2 + 20 0 (j + a)2 + 20 E–12 Some properties of Fourier transform f(t) F() Property Af(t) AF() f1(t) ; f2(t) F1() ; F2() f(at) Fa b, a a a Time-scaling fAt - t0 B e-jt0 F() Time-shifting ejt0 f(t) FA - 0 B Modulation dn f(t) (j)n F() dtn (j)n tnf(t) q 3-q Linearity f1(x)f2(t - x) dx dn F() Differentiation dn F1()F2() Convolution q F (x)F2( - x) dx 2 3-q f1(t)f2(t) Convolution property of the Fourier transform Vo() = H()Vi() CHAPTER 16 Two-port network admittance equations Two-port network impedance equations I y y V B R = B 11 12 R B R I2 y21 y22 V2 I1 V1 B Impedance parameters I2 Linear network V1 z z I R = B 11 12 R B R V2 z21 z22 I2 V2 z11 = V1 I1 I2 = z12 = V1 I2 I1 = z21 = V2 I1 I2 = z22 = V2 I2 I1 = Admittance parameters I1 y11 = V1 V2 = y21 = I2 V1 V2 = y12 I1 = V2 V1 = y22 = I2 V2 V1 = Two-port network hybrid equations B V1 h h12 I R = B 11 RB 1R I2 h21 h22 V2 Hybrid parameters h11 = V1 I1 V2 = h12 = V1 V2 I1 = A = V1 V2 I2 = h21 = I2 I1 V2 = h22 = I2 V2 I1 = B = V1 -I2 V2 = C = I1 V2 I2 = D = I1 -I2 V2 = Two-port network transmission equations B E–13 Transmission parameters V1 A B V R = B R B 2R C D I1 -I2 Two-port network parameter conversions z z B 11 12 R z21 z22 y22 ¢ D Y -y21 ¢Y z22 ¢ D Z -z21 ¢Z -z12 ¢Z T z11 ¢Z B z11 z21 D z21 ¢Z z21 T z22 z21 -y22 y D 21 - ¢Y y21 ¢Z z D 22 -z21 z22 z12 z22 T z22 y11 D y21 y11 -y12 ¢Y T y11 ¢Y y11 y12 R y21 y22 -1 y21 T -y11 y21 -y12 y11 T ¢Y y11 A C D C D B D B A c C B D D D ¢T C T D C ¢H h22 D -h21 h22 - ¢T B T A B h11 D h21 h11 B d D ¢T D T C D - ¢H h D 21 -h22 h21 B h12 h22 T h22 -h12 h11 T ¢H h11 -h11 h21 T -1 h21 h11 h12 R h21 h22 ... a short circuit since the voltage across a short circuit is zero In addition, in order to zero a current source, we replace the current source with an open circuit since there is no current in... This Student Problem Companion is designed to be used in conjunction with Basic Engineering Circuit Analysis, 8e, authored by J David Irwin and R Mark Nelms and published by John Wiley & Sons,... that R R = and = R2 R1 Therefore if + vo - 30 R = 24kΩ (two 12kΩ resistors in series) R1 = 12kΩ R2 = 48kΩ (four 12kΩ resistors in series) then the design conditions are satisfied 31 CHAPTER PROBLEMS