Question

4. Find I using Norton's theorem 1 kn 1 kΩ V + 2V 2 k2 12 V Answer: = 4mA 5. Use Thèvenin's theorem to find Va. 1000 6 kn 1 kn + 2 kΩS 2 kΩ 3 V 1 mA V Answer: V = V 6, Use Norton's theorem to find V Š1 k2 4000 2 kn 3 kn V3 kn 1 mA Answer: V= =1.4516V

Answer 1

1Xn 1KA +V 12v akn 1K 12 24m 1K IK M M wel 17 Noc NMA 1K TK (12 12-V-- Vb 1a-39v 2 So x 1060 4 Noc 36m Novtn equralnt 4MA 2.25k 4m

100 DM k ML AK Im+ V2 K 3K taking bommon +3-G av, 4VtaN30 Dn * On Sob V, in 4(a)2 (32-3 mA 14 14K 3 v 14 J4 NO 14 o

61C V Y43 2K IK Voc VI Q64 Kn M/ NO -13 3 K 18K3 3 mA 3 m al mA al N

AY 3K 71K 4K 1m G 2-V - Tmt 4K 0 3k 2 phe K 4K 3K - 3 4 t N +12-D +12-D 12 N2- 1RV3+12 Sub m $2

A V+12 6. 5 A4V - 15V3-12-70 14-26 12 12+12 14 RVAR0 132-5 V E1-2SA +1S3 V132 -21 12t 26 8(64,13V4) 13 V132 12-3913V132 14R)12 V Yom 2Im A

くK Vp Im A ay-a24 4 0 -1m t V- 2K 4K .Vh N2- 4K L3222-124 2 3K 1K 3K N-23-121 2 N-C-30 2 from 3 3No-t 31

IS 48 V]1 31 IL 21 45 21 4 Kn 31 IM 45.K7m4 31