Solutions to Problems Chapter l 2.6 1.1 (a)M-IC3T3/2;(b)M-IL-3y4J2; (c)ML2 r 2r 1 1.3 (a) 102.5 W; (b) 11.8 V; (c) 5900 W; (d) 6000 w 1.4 Two in series connected in parallel with two others. The combination connected in series with the fifth 1.5 6 V; 16 W 1.6 2 A; 32 W; 8 V 1.7 ~ D.;!b n 1.8 in;~ n 1.9 67.5 A, 82.5 A; No.1, 237.3 V; No.2, 236.15 V; No. 3, 235.4 V; No. 4, 235.65 V; No.5, 236.7 V Chapter 2 2.1 3il + 2i2 = 0; Constraint equation, 15vl - 6v2- 5v3- 4v4 = 0 i 1 - i2 = 5; i 1 = 2 A; i2 = - 3 A -6v1 + l6v2- v3- 2v4 = 0 2.2 va\+va!=-5;va=2i2;va=-3il; -5vl- v2 + 17v3- 3v4 = 0 -4vl - 2vz- 3v3 + l8v4 = 0 i1 = 2 A;i2 = -3 A 2.7 (a) 1 V ±and 2.4 .!?.; (b) 10 V +and 2.3 v13 + v2 2 = 0 (from supernode 1 + 2); 10 n; (c) 12 V ±and 4 n constraint equation v1- v2 = 5; v1 = 2 V; v2=-3V 2.8 3470 w 2.5 2.9 384 000 w 2.10 (a) 1 .!?., lSi W; (b) Ra = 0 (negative values of Ra not considered), 162 W 2.11 (a) 8 .!?.; (b),~ W; (c) 4 .!1. 2.12 ! n. By the principle of superposition if 1 A is fed into any junction and taken out at infinity then the currents in the four branches adjacent to the junction must, by symmetry, be i A. Similarly, feeding in 1 A at infinity and extracting it at a junction causes currents of! A in branches adjacent to the junction. If both these operations are 90. carried out simultaneously at adjacent P.2.5 junctions, the current in the branch joining The network is unique the junctions is ! A and the potential across 200 SOLUTIONS TO PROBLEMS 201 it is! V. Thus I A is passed between Junctions Otapter 4 which have a potential difference of 2 V and 4.I 36J;VtF=V2F=4V;24J the effective resistance is! n 2.I3 ! n. By similar reasoning to the above 4.2 At t = 0: ie3 = ics = ~ A; vc3 = vcs = 0 V; iL2 = iL7 = 0; VL2 = ~ V; 2.14 (a)Rt=R2=R3=~n. vL7 = ~ V; at t = ~: vL2 =vL7 = ic3 = ics = 0; (b) Gt = G2 = G3 = i S iL2 = iL7 = 5 A; vc3 = 20 V; vcs =50 V; 2.I5 .!,} n t 2.I6 (a)! M and i S; (b) 4 A~ and i S; node a: (va- Vb) +! J(va- vc) dt +Jbva = 10; (c) 3 At and! S 0 2.I7 2.1mA- d 2.18 0.67mA ~ node b: (vb - va) + 3-(vb - vct) = 0; dt 2.19 0.513 A~ 2.20 ru = ~ n; '12 = r21 =' n; r22 = ~ n; t Rt = J,fi 51; R3 =' 51; R2 = ~ H node c: ! J (vc - va) dt + ~(vc - vct) 2.21 gu = tf S;gt2 = g2t =- fl1 S;g22 = !¥ S; 0 dvc G1 = f¥ S; G3 = WS; G2 = fl S +5-=0· dt ' Otapter 3 t 3.I (a) 2~ F; (b) 3! F; (c)~ F node d: 3-d (vd- vb) +-I f(vd- ve) dt 3.2 (a) j H; (b)~ H; (c)~ H dt 7 3.3 133 pF + 556 pF = 689 pF total; the 0 same value of 689 pF 1 + -(vd- vc) = 0; 3.4 Eight possible connections: (a) coil one 4 alone-4 H; (b) coil two alone-2 H; (c) series opposition-4 H; (d) series addition-S H; t (el parallel addition-;{ H; (f) parallel reversed node e: -ve + -1 f (ve - vct) dt = 0 -g H (g) coil two short-circuited H; 6 7 -1 0 (h) coil one short-circuited-;{ H 3.5 (a) 11 H; (b) 4~ H 3.6 4.3 (a) At t = 0: iL4 = iL6 = vc2 = vc3 = 0; VL4 = 80 V; VL6 = 40 V; ic2 = ic3 = 4 A; 3H 3H 3H (b) at t = ~: ic2 = ic3 = VL4 = VL6 = 0; iL4 = iL6 = 5 A; vc2 ~50 V; vc3 = 75 V; t (c) 5io: + 4 -(io:-d i(3) +-1 J(ia- i-y) dt = 100; $r"· dt 3 FJ:E(a) 0 t 10i(3 + 1- f (i(3) dt + IO(i(3- i-y) 0 d +4-(i(3-ia)=O; dt di lO(i-y - i(3) + 6 ~ + 5i-y (b) t dt P.3.6 + t J (i-y - ia) dt = 0 0 202 AN INTRODUCTION TO ELECTRICAL CIRCUIT THEORY 4.4 Four meshes; three topological and t one pseudo mesh; node A:.!:. I (va - vb) dt + (va - vc) 40 d t mesh a: 10(i0 - i0 ) + 2 -(i0 - i(3) dt + ~ 1(va - vc) dt = 0; t t + t J(iQ- i-y) dt = 0; 1 [ (vb - 20) 0 node B: 4 J (vb - va) dt + 2 - 0; t 0 d d I [ vd) 0; mesh{3: 2-(i(J-i0 }+i(3+- J {i(J-i-y)dt node C: (vc- va) + 2 -(vc- = dt 3 0 dt t + 4(i(3- i")') = 0; node D: 2 -(vd-d vc) +-1 f vd dt = 0; dt 2 t 0 1 . t mesh -y: S j (i-y - i0 ) · dt + 4(i-y - i(3) node E: (ve - va) dt + (ve + 3)3 = 0 0 t I 0 di 13 8 + '::'::1.+ 6i = 4.6 vc 1 = 4(1 - e- 1 ); 36 J; 24 J; 12 J 7 dt "Y o·, 4.7 (a) v = 4(1 - e-t/4 ); (b) v = 4 e-t constraint equation; io = 10 4.8 (a} 0; (b) 8 V; (c) i = 1.6 e-t/6; (d) 3.2 V 4.5 4.9 (a)for 0 <:. t <:. 1 v = 5 (1 - e-2t); fort;;;. 1 v = 4.32 e-2(t-I); (b) 4.32 V; 0.588 V; 4.32 x e-18 V"" 0 4.10 (a} v = 6(1 - e-t/3); (b) v = 8(1 - e-t/4 ) 4.11 (a) i = 2(e-t/4 - e-t/2); (b) 2.77 s; (c) 0.5 A; (d) 1 F 4.12 (a) 0; (b) 0; (c) 0.5 V; (d) Vba = 2(e-tl2 - e-t) 4.13 v = 3 + e-t/S 4.14 (a) v = 1.03 e-t/4 sin (0.97t); (b) 2; (c) 4 S; 4.1,5 i 1 = 6t + 0.6(1 - e-2t); i2 = 1.2(1- e-2f); (a.) at t= O;i, = i2 = O;at t = oo;i2 = 1.2,i1 = oo Chapter 5 (b) { w1o-1 3 - 10-3 w }) P.4.5 x cos + tan -I (wt 0.5 SOLUTIONS TO PROBLEMS 203 (d) w 0 = 1000 rad s-1 ; Q = 2; B = 500 rad s-1. 5.17 (a) 5 mH; (b) 20 nand 5 mH; For iL = 0.707(iL)m8 x, w = 1167 or (c) 10.6 n 625 rad s-1 5.18 G = 1.72 S; C = 6.9 mF; 0.25 W 5.3 w = 4640 rad s-1 and IL(max) = 1.75 Is 5.4 (a) 8 W; 0.895; 1.414 A; 6.32 V; 5.19 iL=2.5sin (2r-i}P=9~W (b) 25 mW; 0.707; 5 rnA; 7.07 V; (c) 5 mW; 5.20 0.414 1; 70.7 J.LA; 70.7 V w 5.5 (a) 0 V mean; 7.07 V r.m.s.; (b) 0 V mean; 5.78 V r.m.s.; (c) 0 V mean; 10 V r.m.s.; Chapter 6 (d) 3.19 V mean; 5 V r.m.s.; (e) 6.38 V mean; 6.1 (a) z 11 = z 22 = !<za + Zb); 7.07 V r.m.s. z12 = Zzt = '(Zb - Za); 5.6 Io=1LO;[L=l.59L-90°; (b)Z1 =Z2 =Z3 ;Z3 = hzb- Za) Ic = 3.14 L +90°; I= 1.85 L 51.2° 6.2 (a) Yu = Y:z:z = _!_(_!_+ ..!__); 5.7 (a) Z = 30 - j31.8 = 43.8 L- 47°; 2 Za Zb Y= 0.015 + j0.016 = 0.022 L +41° (b)Z = 15.6 + j15.1 = 21.7 L +44°; Y12 = Y21 = _!_(.!_- __!_\ Y= 0.033- j0.0318 = 0.046 L -44° 2 Zb Za/ (c) Z = 30- j0.4 = 30 L -46'; (b) Yt = Yz = Y3 = Y = 0.033 + j0.0004 = 0.033 L +46' ..!__; _!_(_!_- ..!__\_ Zb 2 Za Zb/ (d) Z = 30 + j0.4 = 30 L +46'; Y= 0.033- j0.0004 = 0.033 L -46' 6.3 The amplitude phasors are: I 1 = 7 + j4; h = 3.5 + j2; Ic = 1 + jl.5;fL = 2.5 + j0.5; 5.8 (a) 16000 L 51.6°; (b) 85 L -32°; V1 = 6- j4; V2 = 12- j8; VL= 2 -j10. (c) 21.6 L -90° 6.4 Is= 4 + j2; VL = 8 + j4; the amplitude 5.9 (a) Y = 0.25 L +2°; G = 0.25 S; phasors are V1 = 6- j2, V2 = 12- j4, B = +0.0097 S; (b) Y = 0.2 L -36.8°; VL= 8 +j4,/s =4 + j2,/2 = 2 + j1, G = 0.16 S; B = -0.12 S Ic = 1 + j3, 1L = 1 - j2. 5.10 (a) 30 sin (2t + 0.927); (b) 30 sin 6.5 18.3:1 (2t - 0.927); (c) 36 sin 2r; (d) 25 sin 2t 6.6 (a) R 5 = 20 n; Ls = 0.4 H; (b) Rp = 100 n; 5.11 (a) 5 L 53.1°; (b) 2.83 L 45°; (c) 4 L 0° Lp = 0.5 H.