Question

Find V. in the network in the figure below using Thévenin's theorem. Break the network just to the right of the dependent source, i.e. separate the current source and two resistors on the right from the rest of the network. 2ko 5000 1x zko 6v 7 kg 2 mA 2ko V.

Answer 1

given Arcuit SOOOL S 2K 2m 1 5000I 2 Gries The Tu vo V V + 5000 I -4th-ole [h= Vt sooo Ix V + 0 vzt + to) 26

76 vi - (I -- MA [V=-14 volts ) ) ) Van - + 5000 (4) (4th = -8 volts a t ( es. 2ks ScooTx & LAT IxtIsc GV (7 7ke I s Y Isc 1 YIL - CH ) Vt6 tu & Isc 20 Vt5ooo I = 0 2KK ME V = -5000 In -G{int Isc Ozk - In (78) 20 - Ix (76)=0 - I 9K = 6 + Isc ak 18 of

-6- Isg (26) ak V=-5000 V - 5000 (-6 + Iso (25)) qk (W = + (6 + 15e (2x)] Substitute in @q0. VIC + T & Isc ao o + Isc(lok) to + g + Iserlok) + Isco, 24 TK 5 m + Ise s +3m + 6 m+ Ise lo + Ise 20 ak K Isc ( + 0 +1) = - m - m - 10 m Jsc (***): Isc 35+10+G3 -35m- 63m - 10m. 63 or). - -como-com 116, 14,7 2

Rth = 4 Isc - 3m Ran= kr 7Kr er Q2mA 2ks to - 8K к (9-7v 639 3 (94 +8) + 24-am 27 V + 3 + quo non ho 2 No ( 2 7 2 + 1 - 2 - - 7 ) = 1 No ( 187472-16_) -4 [V= quo 189+72-16 112 No= -0.45714 volts