Question

1. In the circuit shown in Figure 1, v.(t) = 10 cos(2t-30) V. Find (t). 22 1H vs 1F +v(1) Fig. 1 2. In the circuit shown in Figure 2, determine i(t). i(t) 20 2mH MM 100 5 cos 2000 V ) +50 uF 3200 Fig. 2 3. Find v (t) in the circuit of Figure 3 if the current i, = 2sin (100) A. 22 i 1.22 v S1222 + -j Fig. 3 4. Find I in the circuit of Figure 4. If o = 10 rad/sec, what is i, (t)? 42 20 -22 22 W 000 602-30° v (+) 319 W Fig. 4

Please solve it in the simplest possible way

Answer 1

0 Vs it) = 1000s (2t-30]V W-2 ond see XL = secol =j2x12524 te- jooc = -10.12 US Ct) = 10L-30 jos – Tuct) L VCH) = US « (-30-5) 2 +92-505 Vet) = 101-30 *(-30.5) 2+515 VGH = 2 4156-869 v Vita 2 cos 622-156-86)V scos 2004 T Soul 3 201 W22000 redlifee Xez fue 52000x50 kioo Rez -glown X = jwh 52000 AZMI? zjen By using nodal analysis

were there to busy) = { v (0.548+30.09038) = 2-5 p=4.5012-9-36v iz 560-4.0011736 Pa 0.96 25263 A 744) = 0.46 cos C2000t+52-63) A fx= 2 Sin(loot) A We 100 podl sec, fx=210 A wa jin Nr Via fm.(1-31) Va 2 6061-81) = 208284-450 P = L = 2520m P=10414 4135 is & px = 1.4146135°+260° is 1.2014-45 -Vst zitvco

Vs = 21tv - 26104194-45°+ 2.8281-655 us = 5.656245 lus = 56656 Sin Cloot-45) v - 1 godseca na LS 60136 ans - 60 -30 + 27,-j2 11+49, -63233 -4220- (6-21 -412 +213 -601-30 0 -41+ (+32) 12 -}2 13 20 70 320, – 32+2+4=32+32) 1370 32?i-/222 +41320 Qiz - j2ijtíz la 13 = -12 Pitizin 113.-jo-5P1+30:5720

the equation is subpiteering the equations 040 18 66-j2i - 4_12 +32 (-3089, 450-512) = 606-36 (6-13) -422 +-f2 P2 = 60L-36 {-}2} ; - 512 = 60L-36 → ☺ -49 +5+327 82-52(-jo-JP; +jo-siz) zo a & p; +041212 +31 +Pzzo :-571 + 6+52) 1220 ir = sp1 ao 6152 the equation is subriteiting the equation to 2-3), -5651) 2601-36 6+62 i [ 7as2 - 1) = 60 6.30 8, (325-50-75) 2006-30 i= 1798 [12" Foal, = 17.98 2170