Question

Chapter 5, Problem 5.074 2 Your answer is partially correct. Try again. Find 1, in the network in the figure below using Thévenin's theorem. Break the network at the 2k Ohm resistor containing lo. 2 ΚΩ 3 ΚΩ 7 kΩ 8 kΩ 5 ν Vx5 ΚΩ Οιν ντο -2.2573 ' - 2.209 1ο.536

Answer 1

JE 261 -W 38 IV SK Vini & VAL viwit Wr MAN SVI Node A vouage - Vse Now Nodal oenalaysis at Node 4 Vxe . Vx+Vx-5 2 ) - VX Vase I op Vol

3 Vet love of the - Ve se 3 1914 - cument ID Protein breito apping av appling kVL :- p yker - Vth + VATIE VG / The - Uth & VATTEO Vith VaprI oth a 42.Ti T With 2.25T Now, for Rth a deactivate all independent current source and voltage source. Note :: we cannot deactivală dependent sources.

mechodi Assume are one cerrent source of IA and 'V' vollage across it ki > Shoot cruit r S.C. 1 " , 2+ + + Rtha om { sur VW trite I current flows from але адаме 7 kn also iz ve

Loop & KUL + 3 (Inieci) a Va t xao 3 (1-2) + 3 ii f g veco 3(1-2) + 3 Ve + 2 VX 20 Cache) 301-) = -( 13ya) zi-13 Vxe = 3 from loopa KVL 8 (1-) - Ve - Yi oo 8-Vxaisi ao 15 substitucline value of i in eq @ :foom eq 0 3(8-V) - 13V2 33 Preto - 13 V = 3 8-14V2 = 15. ve = -7/14 = %

Now, Vx=-12 from eg, @ 8tY2 2.0.566 0.5 15 Now, appey ing kul in accross source loop. current A Spa 1 37h +v+Verico ra Ti-Ve = y*** 4.466 15. v 8 latha ta 6.466 notre So, now we got wth = 2.257 V Rth = 4.466k5 a : 2ks 18 ndes Io = ?

from thevenin circuit Rth mu uth T & R YI. - 4.466 kr Šak2 2.257 1 - oo I, 2.25 4.46672 Io = 0.349 m A