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Topic 4: Thévenin and Norton Exercise 4.2: Thévenin and Norton Equivalent Circuits 1 Find the Thévenin and Norton equivalents

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Voccer Uth ORG) using mesh analysis, Let mesh currents be es, es, & i respectively, from mesh o, i = -6A 0. y KVL in mesh ③ -by solving eqs Q &@ we get @ *; 6siz-zeig = 9 9 x ) = 1 + x + 1 ) = -4-5) : 6572-32ig = 9 tsin + () = add (+) -32i3+ (67465)finding Rih : (08) RN s.cxcurrent source o.c village source. . 33 S t . Rth (or) RN OB 321 T 103 € RH Loß. AL QA = [4+ 60123]finding (In ker se Using mesh analysis, from mesh Ó 2 by KvL in mesh @ se –9+ 33 iz + 32 (12-13) =0 6512 -3zig = 9 → ... by KIN (68) Ise = iz IN (OV) Isc = :-0.9665 A j Thevinin es? Circuit 1. Rth = 7.12754 Vth =-3656 i Norton eq? ckt مونه ERN = 7.12

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