Question

1) The source vokage of the circuit shown in Figure 4 is v. = 20 cos(39) V. Determine ico) an 1.) z2(t). The source voltage of the circuit shown in Figure 4 is vs = 20 cos(3t) V. Determine i(t) and ། { [t AA ཆ ༽ ཐ 3༽ ། 1 ད་ ༢༥༣ 'A 3རིན་

Answer 1

M=3H 10r i tule 834 The given a wils are opposing each other as the current is flowing inside the dotted teminal in one oil and bearing the dotted terminal from another coil. 1(M) is obie IL coils are in this 2 coils are in this 2 opposing each each other. aiding winery other, its of Equivalent circuit can be written as its & Equivalent can be written as circuit ₂rM LM m Latm mm M3

T Equivalent. Replacing the circuit by its 10 LitM L2+M Mhmmm - 3 ? gam { 3 LitM = 3+3=64 L₂+M= 8+3 =114 M = -34

Its Equivalent Reactance is given by. jx+1= jw C4, +M) =j3x6 = jion was rad (given) jXL z = jw (L2 +M) =j3x11 =7332 jxm = jw (-M) =-; 3x3 = -jar

By using source transformation, Isa 2018 3t = 260934. 200l 36 - 2 cos 3t. Jy - m zloist C 10 and 3 are connected in parallel. its Equivalent is a a cox3 lot3 = 30 = 2.3072 13 Again by source transformation, = vs C Vs= Is R= 2063tx 2.307 = 46615063t.

2.30tr i 18 i iz - ME j33 Vs-4,615 loj3t. To find i we werd need to find Equivalent impedance seen by source. 2.307 ) 18 j33 - M [-ja) and (3 +j33) are in parallee, Z,= (-; 9) (3+j33) -; 277 297 -jq+3+j33 -j9+3+j33 298.22 L-sela 24018 (82.87 Zi = 0.416 - 12.32; Z = 12.32 /–88.06

zeg 2.307 +218 + Z, = 2.307 +378 +0.416 -12.321 = 2.723 +j 5.68 = 6.298 [ 64.38 vs = t new value. zeg 4.615 cost 60298/6438 ? 4.615 10 6.298164.38 = 0.732-64.38 A. = 0.316 – 0.663 2 can be found out by urrent division sule. = ii (opposite branch impedance) sum of the impedance. 12= ii (-jq) (0.316 - 0-66j) (-39) ja+3 tj 33 3+j24 = -0.147 + j0.229 z = 0,272 /122.73 A.