Question

ਲਾਸ : 5 . #1 #2

Problem #1 Solve for the Thevenin Equivalent for circuit 1 and the Norton Equivalent for circuit 2 with the following values for X.

a) X = 9 Ω resistor

b) X = 1 H inductor

c) X = ¼ F capacitor Problem

#2 Place a 4 Ω resistor across the + and – connecters in Problem #1 and solve for the value of voltage across and the current through the 4 Ω resistor for each value used for X.

Answer 1

Problem - 1 (a) X = 9.2 Resistor equivalent Thevenin circuit for #1 for the 2 91 1 Rok Path = (21|1 602) + 91 - 2X6 + 9 = + 9 /Rth = 10.50 Now for the er en 12 I = 2+6 DES = 9v Vy IX 6 = wO+ 10.52

X=1H for - inductor Circuit I There is source a DC W = 0 JX = jw = jxoxl zor ↑ for zu 0+ S 724 = Pon = 24 for uth pt I 360 12v With = 2x6 = 9V to I= = Thecenin equitalent with - O+ + for circuit -02 i in prima 7fuzon for Tal

GRAT U LALL for #2 Norton equivalent for RM or {gr Pan Rw = (2n) 11 (ar) II (91) = (2) ll en = (2) ll er = 6) 11 9 279 2 / +9 for In 3A 0 en zor şar => IN =3A Norton equivalent xo 3 Amp

DAAWAH Norton equivalent a supply wao - Iwc = 0 - open circuited for Rih Rth = 0 => Thevenin's equatent hot punible. for #2 Ru TT + {an a Trth 11111-! XA RE for = {r In Rom= 2x - 3A 0 En for =) IN = 3 Amp Norton equitabat JA ☺)