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Problem 6.1. i= ct = 5(2e" -6 +e )=10(1 - 3t}eFA p= vi=10(1-3t)e**- 2 e?* = 2001 - 3)" W 0-160 Problem 6.3 i= ct = 40x10? 2 = 480 mA a(t ; ¢ " Supe b 5) Pay . TF this covrent waveform is applted toa 2OM! capacitor, “pew find the voltage v(t) across the ca paciton. Assume Y(0)-0. t y(t)2 ) dt + vlo) +t u A sec 064 <1 242 at anys — + 2 cm YW) Zoy0-® [ars 40 =/0 AV ° ‘Z ; Vit) = 0,000V 2/0 V TAKE bs G4 anys £ 5 z = Ube So xto-” feat) it 410 = /6 (44 -4°-3) +10 V 9(2) = & OPV 2c4 L>=O amps re £ i es tayo = 6xre V vA) = Zoyro~” j oe ¥ Supp 6.12) 4A voltage of GOcos yme V appeavs across the fermi nals ofe 4 mt CApacrtow, Caleclale the covreat thracgh the capacctos and the energy stored init from #20 to 20.1265. “£2C iy = 3x10 %Go (anys art) = -0.22 A singnt Aoys - * ¢t t t..125 W af prs) it f or = =92.60 sin qntcosutrt dt = 57230) sn pt dt ° ° 0 OFS > 2167 — od FAT Problem 6.13 Under de conditions, the circuit becomes that shown below: i, OQ 1, 530Q + 20 Oy = 309 ‘1 V2 / 60v ’ Vi = 30i2 = 30V, v2 = 60-201, = 40V in = 0, i = 60/(304+104+20) = 1A Thus, vy; =30V, v2=40V Problem 6:19 Find the equivalent capacitance between terminals a and & in the circuit of Fig. 6.53, All capacitances are in uF. a 0 «o—} —_ fe IN 4 20 ot 3 ae yerO ty 20 . 10 bo We combine 10-, 20-, and 30- F capacitors in parallel to get 60 w F. The 60 - uF capacitor in series with another 60- 4 F capacitor gives 30 uF. 30+ 50=80uF, 80+40=120 wF 12 120 | je re 12 80 120- 4: F capacitor in series with 80 yz F gives (80x120)/200 = 48 48+ 12=60 60- 4 F capacitor in series with 12 w F gives (60x1272 = 10 4 F Supp 6-27) Three Capacitors , C= SaF, Caz 10m and Cz220uf ave Connected tn sevies with a 200V sovrce, Compete : (a) the totel capacitance (b) the charge on tach capacitor (C) the tutel energy stored in the series combsnetion 1 7 1 - *) c-h+h9 S> sare + pur’ + dono’ PCF 2.86 F », eer nce = bere tls 208 ACV AAW HQ hb = in ey z je peoy" B+ B70 @( ded rh)7200 8+ 6:70 Ge bnky 6) w= FCGVs dt 2.86 x00°3 go0t> oT = S2mT Prvhlenn 6-25’) “= ab «3 so ~dv 4 -4e 4 dy 2 i . cL +Z =O (éndadt 4 3 (ood tte > we" 0°? & -IS4 4 HPhg - 0%, 7° 14, its b(igiserd aud > a 4 oe 8. Bats ot +5, ‘ bo te * 30°? -Ga -2hg tHley 20 ‘hel (deg’ bg ec lt 48 fq _ 4b es. @ LG eset jitrh [a 7? + po +i? 40 -( “4 L = O 47 -20 dv dy og th 432 \ei7-4a _ 108 ae el ig | 6 (69-49) CuCuas-7—)+b DGorsm) SA I 47? ~20 -G@ -2 tt 5224 ay =p Cag? aad al 20tmA, O= > L=10mH Problem 6.5} - spa = L 60 20 30 10 L,, =10(25+10) = = 7.778 mH 10x35 45 di_ iy Problem 6.59 (a) v,=(L, +1) a L, +L, di di ‘1 aba V2 elie di di v, =v, =L,—t=L,—2 ) vyev, gla g i, =i, +i, di, _ diy diy vy vy Gi+L,) dt dt dt L, L, LL, jet vat = LL, LiL, diy - L, : L, L, °L, +L, dt L, +L, inay, fete fp diy ee L, , L, +L, dt L, +L, aloe vlo) = 0 6of/40> 24m STR {2 singe mV Zz tomH + Gomi vy SH4omts G 20+ 24 4/6 -G0 mH _fFTH = lo wl + 4 id al a0 vit) erie) = doyo”? ji sindt Ko dt +0 < +m atlt= g (ease “safe 4 = cofi-cos ut) @?? aqan OE & = send sons? o- UC smut) P ‘ = 4.8 x09 sit) = AGA At mV Serp 6.66) Find 4 and Vv, assuming thet Problem 6.7/_ By combining a summer with an integrator, we have the circuit below: Ri “a 1 —W— —— Rz V2 AM ae ®B | [TP 3M 1 1 1 =- dt - dt -— dt vor" Jv R,c vs R,C vs For the given problem, C = 2uF, R,\C=1 —* R=1(OC= 100°/(2) = 500 kQ RoC = 144) ——» Ro = 14C) = 500k0/(4) = 125 kQ. R3C = 1/110) ——-» R;=1/(10C)=50kQ Problem 6.73 Consider the op amp as shown below: Let va=vyp=V 0-v At node a, —— = aR Vi-V_ VOY, ao R AW R b vi mam R dt Vv; =2v~v, +R (2) dt At node b, Combining (1) and (2), or showing that the circuit is a noninverting integrator. Problem 6.93 Since two 10pF capacitors in series gives 5p1F, rated at 600V, it requires 8 groups in parallel with cach group consisting of two capacitors in series, as shown TTT « TETTTTTT -TTPELETT Answer: 8 groups in parallel with each group made up of 2 capacitors in series, \ Ft aN Jt ae) \ a JI \ Supp G, #4) Whea a ca pacifon is connected toa de Sovree, hs voltage SES from 20V to 26V im tus with an averege charging corvent of 0.64. Det evn ine the valve of the capacitance. _ d s Av £=C = D> 4°90 . ~ at] 4X0 C= Lye Bu 7 96 ZG-Z6 = (SK? F 2 1G we 
