ANSWERS TO EVEN- NUMBERED PROBLEMS Chapter 1 1-2 Decreases by 4x 10 -I ' J . 1-4 2.5 A (0 < t < 6) ; -5 A (6 < t < 9) . Function is repetitive every 9 seconds . 1-6 p(t) = 180 - 180 sin tat W. 147 J . 1-8 P,, = 400 kW . Pa, = 20 kW . 1-10 (a) v, = v - VB, v2 = vB - v,.. (b) VB =- 19 V, v, = 34 V. 1-12 VP = VI ; V2 - (VI - V2) ; VR = VI - V2 + V3 ; VS = VI - V2 + V3 - V4- 1-14 (a) Delivers +400 W. (b) Delivers + 100 W. (c) Receives +550 W . (d) Receives +200 W. 1-16 No . 1 receives -700 W. No . 2 receives -650 W . No . 3 delivers +975 W . No . 4 delivers - 1575 W. No . 5 delivers -750 W. 1-18 75 MHz. 1-20 Nonlinear . 1-22 (a) [26 .4 +- 10 cos 10t - 25 sin 10t] A. (b) (- 16 + 4 cos 10t - 10 sin 10t) A. (c) ( - 22 - 5 .12 cos l0t + 12 .8 sin l0t) A. 1-24 v(t) = 0 when t < - 4 s ; -24e-3u+4I A when t > -4 s . 1-26 9 .02 x 10 - I' J .
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ANSWERS TO EVEN-NUMBERED PROBLEMS
Chapter 11-2
Decreases by 4 x 10-I ' J .1-4
2.5 A (0 < t < 6) ; -5 A (6 < t < 9) . Function is repetitive every9 seconds .
G, = (G, + GZ) ; it = (-i~ + i,, - i, . - id) . v = ( iTIG,) . i, = (G,IG,)i,-.i2 = (GZIG,-)i,- .
2-20
142.5 A .
2-22
(a) v, = -58.1 V. i � delivers 1046 W; ib delivers 988 W; v, . delivers 2896 W.(b) v, = -203 V. i� delivers 2640 W; ib delivers 3452 W ; v,. delivers 2424 W.2-24 (a) v. - ( i a + ib)l(G � + Gb + K) . Power = -Kva .(b) v, = -i.,l[G,(I + K) + G2]. Power = -KG,V2 .
2-26 v, = [i�(I + KG b) + ib]l(G, + G,, + KG,Gb) .2-28
(a) R, = 10 .3 fl . i,,. = 9.69 A . (b) R, = 8 .29 fl . i, = 12 .1 A.
2-30
Using the same subscripts as the resistors : (a) i, = 9.69 A, i2 = 5 .61 A, i3 = 4.08 A,is = 1 .71 A, i 5 = 1 .28 A, ib = i 7 = 1 .09 A. (b) i, = i2 = 1 .72 A; i3 = i 4 = 3 .45 A;i s = i6 = 3 .45 A; i 7 = 6.90 A; iR = 12 .1 A; i9 = 5 .17 A; i, )) = 4.31 A; i � = 62 = 0.862 A.
2-32
Using the same subscripts as the resistors : i, = 1 .12 A; i2 = 0.875 A; i 3 = 0.854 A;i4 = 0.250 A; i 5 = 0.521 A; i6 = 0.625 A; i7 = i9 = 0.333 A; ix = 0.146 A.2-34
(a) ( G2 + G,)( VI - V2) + G3( VI - V6) + G4( VI - V5) + ( G5 + G6)( VI - V4)+ G7V, = -la.
(b) (GI + G2 + G3)(VI - V2) + (G4 + G5 + G6) (VI - V3) = - la + ib - i, - id - i,-
3-10
v, = - 1 .45 V; V2 = 1 .39 V; V3 = 0.866 V. Starting with Ga and going in alphabeticalorder of the subscripts, the branch currents and power dissipated in the conductances are: 5 .68 A(16 .1 W), 8.52 A (24 .2 W), 5.8 A (8 .41 W), 0.261 A (0.136 W), 0 .174 A (0.150 W),'0.0866 A(0.0750 W), 5.56 A (7.73 W) . Power delivered by the three current sources : 34.8 W, 11 .6 W,10 .4 W.
3-12
New voltages : v, = 11 .6 V, V2 = 14.4 V, v 3 = 13.87 V, VREF = 13 V. Other quantitiesnot affected .
3-14 G2/[(G,G2 + G IG S + G,G4 + G2G3 + G2G4)] .
3-16
Starting with G, and going in numerical order of the subscripts, the branch currents are:6 .67 A, 13 .3 A, 0 A, 5 .77 A, 19.2 A.
3-18
31 .3 A.
3-20
v� source : 1276 W; Vb source : 1253 W; i, source : 666 W.
3-22
Same as in (3-20) above.
3-24
v, = 6 .97 V; V2 = 4 .76 V; V3 = 6 .18 V . Starting with G. and proceeding in alphabeticalorder of the subscripts, the branch currents are: 5.58 A, 4.42 A, 0.618 A, 5.68 A, 3.81 A .
Use Eq . (3-92) with v, = (R2R3Vi, + RIR3Vi2 + R,R 2 Vi3 )/(R,R 2 + R,R3+ R2R 3 ) and G4 instead of G, .
Answers to Even-Numbered Problems
81 5
Chapter 4
4-2
(a) i, = -0 .106 A ; i2 = 0.417 A ; i3 = - 1 .298 A . Currents in the other branches are:1 . 19 A in R 2 , 1 .61 A in R5 , and 1 .72 A in v� . 20.6 W delivered by v� , 21 .4 W delivered by v,, .VAB = 18 .8 V . (b) is = 0.636 A, ib = i, . = 0.454 A . Currents in the other branches are: 0 .182Ain R c , 1 .09 A in v� , and 0.636 in R� . 10 .9 W delivered by v.. v u, = 3 .62 V .
4-4
(volvbe) = [1 + (R2/R,)] .4-6
Same result as Eq . (3-93) in Chapter 3 .
4-8
v� delivers 1084 W; i b delivers -288 W . 520 W in R,, 22.7 W in R2, 163 W in R 3 ,and 90.8 W in R4 .
4-10 i� delivers (vbia) W ; vb delivers v b [(vblRL) - ia] W; R L consumes (v~Rc)W . vb receivespositive power (or delivers negative power) when RL > (vblia) .
4-12
i� ib i, . i,, i,, it i,
4-14
R,R . , = R 2R 3 . R, andR�, have no effect on the condition for balance .4-16 Open circuit : i, = 2v,/(R� + Rb) ; VAB = v,(Rb - R.)l(Rb + R�) .Short circuit : i, = v,(R. + R b)l2R,Rb . i s, = v,(Rb - R �)l2R �Rb .4-20
R, = Ru R bl(Ra + R b + R c ) ; R 2 = R o R,J(R o + R b + Rc) ;R 3 = Rb R,J(Ra + R b + R,.) .R� = (R,R 2 + R,R 3 + RZR 3 )/R 3 ; R b = (R,R2 + R,R 3 + RZR3)/R2 ;Rb = (R,R2 + R,R 3 + RZR3)/R, .
4-22 v 3 = - KZRZR3V,/{(R, + K 2R,)(R, + R 2 + R3)
4-24
v� delivers vaic W. i, delivers (R bic - v .) i, W ; Rb consumes (Rbi,. 2)W .i,. receives positive power when R b < ( VQ/ic) .
4-26
v � = R[(2/3) i~ + ib + (4/3) ic ] .4-32
Elements of the matrix are : (R, + R 2 + R3 + R4), -R 4 , -(R 2 + R 3), -R 4 ,(R 4 + R6 + R, + R s ), - (R6 + R,), - (R2 + R3), - (R6 + R,), (R2 + R3 + R6 + R2 + R5) .i, = 4.57 A, i2 = 2.57 A, i 3 = 2.86 A .4-34
Elements of the matrix are : (R, + R 2 + R 3 ), -(R 3 + KR 5 ), KR5 , -R 3 ,(R3 + R 4 + R, (1 + K)], -R5 (1 + K), 0, -R 5 , (R 5 + R6) .
4-36.
4.57 x 10-2 S, 2.57 x 10 -2 S, 2.86 x 10 -2 S .
4-38
Elements of the matrix are: (R, + R 2), -R2 , (K - R2), (R2 + R3) .
(PLIP.,) = RLI(RL + R,) . (PUP S ) = 0 when RL = 0 and approaches 1 as R L ~ X. P, ismaximum when R L = 0 ; PL is maximum when R L = R, ; (PLIP.J is maximum when RL = X-
v,.(t) _ [V� (I - k)12][1 - e -" I where ? = (K + 3)RC/2 .
6-20
(a) v,. increases to 9.93 V at t = 50 ms, decays to zero and increases again to 9 .93 V att = 200 ms and decays to zero . (b) v,. increases to 3.94 V at t = 50 ms, decays to 1 .45 V att = 150 ms, increases to 4.81 V at t = 200 ms, and decays to zero .
i, .(t) = 0.03e -1 .33, A . (c) v,.(t) = (45 - 20e-1 .331 ) V; i,.(t) = 0.0133e-1 .33' A .-3.13x 104,
-3.13x I Oar6-36
(a) iL(t) = 0.03e
A; VL(t) _
- 14.1 e
V.(b) iL(t) = (0 .2 - 0.2e-3.131104x) A ; VL(t) = 94e -3 .131104 f V .(c) iL = (0 .2 - 0.17e-3 .13x104,) A; vL(t) = 79.9e-3.13x1041 V .
6-40
R = 7 .36 fl; ; L = 92 mH; C = 6793 N.F .
Answers to Even-Numbered Problems
81 7
6-22 v, I -- 80e -s ' V. v,.2 = 100 + 20e-s ' V .6-24 R = 2M. L = 39 .2 mH .
6-26 iL(t) = - 170 mA (t < 0) ; (-42.6 - 127 e -108x 1 `4) mA (t > 0) .
6-30 iL = (V�,IR)(I - e - ") (0 < t < tp); (V,�IR)(1 - e-'PI)e-"-'n" (t > t") .
6-32 R=4M.L= 111 H .
6-34 (a) v,.(t) = 25e - '-"' V ; i,(t) = -0.0167e-1'33' A. (b) v,.(t) = 45(1 - e -1 .33') V;
Chapter 7
(c) (-7.24 cos 50t + 49.0 sin 50t) V.7-8
(a) 36.0 cos (1001 + 56.3°) V. (b) 28 .0 cos (200t + 105°) V . (c) 10 cos 3001 V.7-10
(a) Amplitude = 2A cos ((~/2) . (b) Phase = (4)/2) . (c) Maximum when
= 0andminimumwhen ~ = 180° .
7-12
i(t) = 4.47 cos 105t A . v(t) = 112 cos 1051 . V . Pa, = 250 W.7-14
i(t)' _ -7 .2 x 10-2 sin (5 x 106t) A . p(t) = -0.432 sin 10 7t W.7-16
v(t) = 17 .8 cos (1 .26 x 104t) V. i(t) = -5.61 sin (1 .26 x 104t) A.p(t) = -50 sin (2.52 x 1041) W .
7-18
v (t) =
-30 sin 500t V . p(t) =
-60 sin 1000t W.7-20
i(t) = 20 cos 1000t A. v(t) _ -50 sin 10001 V. L = 2.5 mH .p (t) = -500 sin 2000t W.7-22
(a) Peak value increases by a factor K,2 and frequency is not affected for all three elements .Constant component for the resistor changes by a factor of K,2, but no change for the inductor orcapacitor . (b) Peak value unaffected for the resistor, decreases by K2 for the inductor, and increasesby K2 for the capacitor . Frequency increases by a factor K2 for all three elements . Constantcomponent not affected for all three elements . (c) No effect on any of the items .7-24
i(t) = 10 cos (2000t - 2 .86°) A.7-26
v(t) _ [I~l
G2 + 1/w2L2] cos (wt + arc tan 1/wLG) .7-28 Im = V�,/VR2 + w2L2 . tan
_ -(wLIR).7-30
G = 0.313 S . C = 1 .56 x 10-2 F .
7-32
(a) 1168 r/s . (b) 316 r/s . (c) 184 r/s .
7-34
p(t) = [19.9 + 42 .3 cos (5000t - 61 .9°)] W. Pave = 19.9 W-7-36
(a) Im = 4.62 A. p(t) = [500 + 577 cos (2wt - 30°)]W . (b) Im = 4.62 A .p(t) = 500 + 577 cos (2wt + 30°) W. (c) I�, = 4 A. p(t) = 500 + 500 cos 2wt.7-38
(a) 0.577 Vm. (b) 0 .745 I,� .
Chapter 8
8-2
(a) v1(t) = 10 cos (200t - 135°) V; (b) v2(t) = 10 sin (2001 - 75°) V;(c) v3(t) = 10 cos (200t - 180°) V.
8-4
(a) f,(t) = 2N/p-2+ q2 cos [bt + arc tan (qlp)] . (b) f2(t) = 2V/-n'-+a2 cos bt .(c) f(t) = 2
p2 + q2 cos [bt + r + arc tan (qlp)] . (d) f4(t) =
(a) R = 10 f, X = 4 .12 dl, G = 0.0856 S, B = -0 .0353 S . (b) R = 5 .95 dl,-4.91 f, G = 0.1 S, B = 0 .0824 S . (c) R = 0.684 dl, X = 1 .880 11, G = 0.171 S,-0.470 S . (d)R=0.12 fl, X= -0.16 fl, G=3 S,B=4S .
(a) w = (G/C) . V = (0.707/-45°)(Is/G) (b) 2.64(G/C) . V = (0.353/-69 .3°)(I,/G) .8-38
9(w2 + 1)/(w4 + 9w2) .9(jw) = 7r + arc tan w + arc tan (3/w)
9-38
(V2/V,) = 1/[(1 - 0.125w 2 + j0 . 125w] . Amplitude response starts at I at w = 0, reachesa peak value of 2.86 at 2.74 r/s, and approaches zero as w ---> -. Phase response starts at 0° atw = 0, becomes - 45° at 2.37 r/s, and approaches zero as w -> x.9-42 (V2/V,) = (1 - K)l(2 + jwRC) . 0.186(1 - K)/ - 68.2° at w = (5/RC) ;0.354(1 - K) -45° at w = (2/RC) ; 0.447(1 - K)L26.6° at w = RC;0.485(1 - K)/- 14 ° at w = (1/2RC) ; 0 .498(1 - K)//--5 .7 ° at w = (1/5RC) .
Z;~ = R(1 - a)2/(1 + ja 2wRC), where a = (N,INz) .
10-22
(a) Vub = 398L° V . (b) V�,. = 398/-30° V . (c) Vb� = 230/-120° V .(d) V, .u = 398° V. (e) V,b = 398L° V.
10-24
(a) 1. = 5 .02/41 .6 ° A . I b = 5 .02/-78 .4° A . I, . = 5 .02L° A.(b) 252 W. (c) 756 W.
10-26 Z2 = 3Z, .
10-28
(a) 1� = 3 .26° A. Ib = 11 .4/-75° A . I,. = 8.97° A.(b) Z�b consumes 722 W, Zn ,. 1083 W, and Z, .a 468 W. (c) 2273 W.
10-30
Phase sequence a-b-c : (a) I, = 13 .9 ° A ; IZ = 10 .4L:: 75° A ; 1 3 = 13 .9 / - 168° A .(b) 1� = 27 .8/15 .2' A ; I b = 17 .8/-126° A ; 1, = 17 .8° A. (c) Z, : 483 W. Z z : 809 W.Z 3 : 1451 W. (d) 2741 W . Phase sequence a-c-b : (a) 1, = 13 .9° A. I z = 10.4 165° A .1 3 = 13 .9/-48.4 ° A . (b) IQ = 15 .3L°A . Ib = 23 .3/- 176° A . 1, . = 23 .2 -34 .3° A .Answers to (c) and (d) are not affected by the phase sequence .
10-32
(a) Y = 4 .17 x 10-Z/-60° S . 48 .1 0 in parallel with 88 .2 mH . (b) C = 56 .7(c) Same average power as before . Apparent power = 20 kVa .
W, reads 20 kW . Wz reads 25 kW . Total average power = 45 kW . Total reactive power- 5 kVAR (current lagging) . Total apparent power = 45.3 kVA .
Chapter 11
11-2
(a) L = 0.507 p,H . (b) I R = l OL° A . I L = 62 .8/ - 90° A . Ic = 62 .8L°A .
11-4
iG = 25 cos 1000t + 20 cos 2000t + 15 cos 3000t mA . iL = 50 cos (1000t - 90°)+ 20 cos (2000t - 90°) + 10 cos (3000t - 90 ° ) mA . is = 12 .5 cos (1000t + 90°)+ 20 cos (2000t + 90°) + 22 .5 cos (3000t + 90°) mA . is = 45 .1 cos (1000t - 56.3°)+ 20 cos (2000t) + 19 .5 cos (3000t + 39.8° ) mA .
11-6
(a)G=0.01 S.C=50p,F.L=0.2mH .(b) W, . = 50 x 10-6 [l + cos (2 x 1041)] J . WL = 50 x IO-'[I - cos (2 x 10 4 t)] J .(C) Wmax = 10-4 J .
11-8
(b) w = 1/
(LC - G2Lz/2) .
11-10
(a) Co = 1/wo2L . (b) C = (1/o2L) ± (N/3G/w~) .
11-12
(a) G = 0.0223 S . L = 4.56 VH . C = 88.9 nF . (b) iG = 3.34 cos (1 .57 x 10°t) A .iL -- 20 .9 cos(1 .57 x 1Obt - 90 °) A . is = 20 .9 cos (1 .57 x 1Obt + 90°) A .(c) wL = 5 x 10'[1 - cos (3 .14 x 1O b t)] J . W,. = 5 x 10'[1 + cos (3 .14 x 1O bt)] J .(d) wR = 10-3 J/cycle .
11-16
(a) BW = 200 r/s . Half-power at 124 r/s and 324 r/s . (b) Y = 0 .01 S. (c) 0.0141L° S .
Answers to Even-Numbered Problems
821
11-18
(a) G = 2 x 10-4 S. L = 0 .319 mH. C = 1 .27 nF . (b) Half-power at (1 .495 x 10 6)r/s and (1 .652 x 10 6) r/s . (c) wT = 3.96 x 10-9 J. (d) wR = 2.5 x 10 -9 Rcycle .
11-22
(a) BW = 1 .25 x 105 r/s . Half-power at 6.262 x 106 r/s and 6.387 x 10 6 r/s .(b) R = I M SZ .
11-24
(a) 6324 r/s . (b) G = 2 x 10 -4 S; L = 50 mH; C = 500 nF . (c) 1, . = 790° mA.I, �;, = 792/ -86 .4° mA . (d) BW = 400 r/s . Half-power at : 6128 r/s and 6528 r/s .
11-26
RL = 0.305 f. L = 4.86 wH . C = 3 .18 nF .
11-32
(a) L = 0 .507 I,LH . (b) VR = IOL° V. VL = 1 .59
° V. Vc = 1 .59/-90 ° V.
11-34
(a) R = 100 S2 . L = 0.5 H. C = 20 nF . (b) wL = 1 .25 x 10`11 + cos(2 x 10°t)] J . w, = 1 .25 x 10-3 [1 - cos (2 x 1040] J. (c) WT = 2.5 x 10' J.11-36 (b) w =
N/-1/LC - RZ/2Lz11-38
(a) R = 44 .7 fl . L = 0.178 mH. C = 2.28 nF . (b) VR = 15010° V.VL = 93810° V. V, . = 938/-90° V . (c) wT = 5 x 10 -°[1 + cos (1 .57 x IO6t)] J.w, . = 5 x 10'[1 - cos (1 .57 x 1O6t)] J. (d) wR = 1.01 x 10-3 J/cycle .
-0.781, - 6 .30; poles at s = 0, -1, - 16 .(b) Zeros at s =
±jO.495, ± j2.33; poles at s = 0, ±j2.08. (c) Zero at s =
- I ;pole at s = 0 .(d) Zero at s = - 1/RC(4 - 3KR) ; pole at s = -1/3RC .
12-10
(a) Zeros at s = 0, - 1, - 16; poles at s =
-0.781, - 6 .30 . (b) Zero at s = 0; polesat s = ±jO.495, ±j2.33. (c) no critical frequencies . (d) pole at s = - 1/(4 - 3KR)RC .12-12
Series : Any pole of Z, or Zz is a pole of the total impedance. Any zero common to bothZ, and Z2 is a zero of the total impedance. Parallel: Any zero of Z, or Z: is a zero of the totalimpedance. Any pole common to both Z, and Zz is a pole of the total impedance.12-16 (V�lV,) = (AIR,C,)(s + 1/R2C2)/(s + 1/R,C,)[s + (1 + A)/R,C=] .
Answers to Even-Numbered Problems
5+j377t) .
12-18
The asymptotic plots of the functions are as follows . (a) Horizontal at -40 dB up to 100r/s and downward at -20 dB/decade after that . (b) Upward at a slope of + 20 dB/decade from w= 0 to 100 r/s and levels off at a constant value of 0 dB after that . (c) Horizontal at -80 dB upto 100 r/s and downward at -40 dB/decade after that . (d) Upward at a slope of +40 dB/decadeup to 100 r/s and horizontal at 0 dB after that .
12-20
The asymptotic plots of the functions are as follows . (a) Horizontal at 40 dB to 0.01 r/s,upward at 20 dB/decade from 0.01 to 0.1 r/s, horizontal at 60 dB from 0.1 r/s to I r/s, and upwardat 20 dB/decade from 1 to 10 r/s, and horizontal at 80 db after that . (b) Upward at 20 dB/decadeto 10 r/s, horizontal at 20 dB from 10 r/s to 100 r/s, and downward at -20 dB/decade after that .12-22
(a) Single-stage asymptotic plot : Upward at 20 dB/decade to 2000 r/s, horizontal at 80 dB
from 2000 r/s to 104 r/s, and downward at -20 dB/decade after that . (b) For the three-stageamplifier, upward at 60 dB/decade to 2000 r/s, horizontal at 240 dB from 2000 to 104 r/s, anddownward at -60 dB/decade after that . (c) 240 dB . (d) Bandwidth = 1000 r/s (approx) .12-24