lushf seLJHsf leksjfhs, Lecture notes of Electrical Circuit Analysis

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February 5, 2006 CHAPTER 10 P.P.10.1 10sin(2t) ——> 1020°, w=2 2H — > joL=j4 1 0.2F —» —~=-}25 jac Hence, the circuit in the frequency domain is as shown below. Vv; Ws Q V> 4 > >—|-—> 1“ \M\y + A 1020° A (t) 2Q Vx 3 j42 > 3Vx Atnodel, 10=~++ 2 node |, => : 2 - j2.5 100 = (5+ j4) V, —j4V, (1) Vv, V,-V, 3V,-V. At node 2, — —1_=+ , —*__* where V, = V, j4 - j2.5 4 - j2.5V, = j4(V, — V,)+2.5BV, -V,) 0=-(7.54+ j)V, +(2.5+4 jl.5)V, (2) Put (1) and (2) in matrix form. [ 5+j4 -j4 Tv, | [100] |_7.5+ j4) 2.54 stv, | lo | where A = (5+ j4)(2.54 j.15)—(-j4)(-(7.5+ j4)) = 22.5— jl2.5 = 25.74 Z - 29.05° 2.5+ 51.5 j4 V, lasea sa 100 ¥- 22.5 j12.5 ia v= 2.54 1-5 4 gg) __2-915230.96° 1) 11 39 60,019 22.5— j12.5 25.742 - 29.05° 7.54 j4 8.5 228.07° == (L100) = = (100) = 33.02 457.1 2° > 22.5 — j12.5‘ ) 25.742 - 29,059 ) docsity.com