Question

(5pts) 8. For the circuits shown below, the measured voltage Vx is close to 8, 6V b. 7.5V c, 9V d. none of the above 1kΩ 1kΩ Λ 10kΩ 10kΩ Λ 5V 1kΩ 10ν the vacitance

Answer 1

Given ckt: IK12 IKA 13 TOKR TOKIL MMM 11-13)>(12-13) + Kn V (D. 5V toy * (11-12) We have to find Vx. There are 3 meshes, Let I current flows in mesh i Let 12 current flows in mesh (2) Let Iz current flows in mesh (3) Net current flowing through common vesitor 10 Kv2 · (between mesh 0 and 3) I - I3 Net current Howing through common resistor loks (lying beton mesh (2) and mesh)] = 12-13 Net currunt flowing though common resistor 162 (lying betovech mesh (1) and mush(21] = 1, -12 Now, Applying kvh in meshús, Ikr -5 + 10 (1,-13) + 1411-12) - 0 10KR & KR lokh = (1 2 = 13) -5+10.1, +1,-12-1013 = 0 + 5V (t Ik2&vx A lov T 101, +1, 12-10 13=5 111, -12 -1013 = 5 (11-12) i) (say?

& IKS Tokr Tokn Applying KVL in mesh (2), --1(1,-1) +10 (12-13) +10 = 0 (4-13) -1,+12+1012-1013 +10 = 0 SvD o 1K2 V IL 1, - 1112+10 13 = 10 ----YA(z) (soy) (12-13) o lov lov Applying KVL in mesh (3) 1 (13) + 1-(1) - 10 (12-13) -10 (1,-13) = 0 213-1012+1013-101 +1063=0 - 101,- 1012 + 22 13 = 0 22 13 = 101,+1012 13 = 10(1; +12) 22 13 = 5 (11+12) equ(3) (say) Substituting 's from equabon 3) in equahon ; we get 111, - 12 -10 (4+1)] = 5 111, -12- 501, - 50.19 = 5 17 12171 - 1112-501,- 5012 = 55 711,-6112 - 55 epa (4) (say)

Substituting 13 fwm equabon ® in equabon Günzurget 1,- 1112 +100 (1,7 12,7 = 10 I - 11 I2 + 50 11 + 50 = 10 1111 - 12/12 +501,+501, = 110 6111 - 71 12 = = 110 equahon 5 say) Now, solving equation (4) and equabion() weget TIL - 67.19 = 55 from eju (4) X 71 612, – 7/1, - 110 -fromegn (5) x 61 on multiplying aquahin en with 71 and equebim © with 61, We get 50411 - 4331 12 = 3905 3721 1 - 433112 6710 - + 1320 I, to -2805 2805 771 1326 (kon : 2.125 MA - equ (6) Subsiduting 4 - - 2.125 mA in equation (4), Wéget 71 (-2.125) - 611,= 55 150.875 - 61 1 = 55

- 150.$75 - 6/12 = 55 6112 = -150.875 -55 6 1 12 = -205.875 I2 = -3.375 MA Here, value of t which is - 2.125 mA & value of tz which is – 3.375 mA comes out to be negative, it means our assumed direction of I and I are opposite to that of actual detection. Now, from circuit, Vx = (1K2) (1,-12) Ve= (1K1) (-2.125-6-3.375)]mA V2= 1. (-2.125 +3.375) V. Va= 1.25 V .: Therefore Mequired value of Vx is 1.28 y. 80, option d None of the above.