Question

File Edit View Tools HUD - View only - Isrc = 3.79 mA, lur2 = 1.84 mA Vc = 4.00 V, Ri= 3.3 k2, R2 = 2.2 k 2. R3 = 1.0 k2, R = 4.7 k92 Use Node-voltage method to calculate the node voltages in this circuit. Verify your results in MultiSim Question 1. Question 2 Question 3. Question 4. V V, V V TV IV TV IV

Answer 1

VM-2m- {f} 13 O VSE Isrc, = 379m A; I steg = 1.84mA V src=40; R,= 3.3 ks; &q=22ks R3= 1.ok sa; R4 = 4,7k2. Apply Nodal Analysis at Node v » 0.303 ♡ - v2 mt 3.799A + 0.454 (U,G) MAIO (0.7574,-0.30342 -0.45444 +3.79) mA = Apply Nodal Analyfif at Node Vol 0 Ve-vsve very t o

(vg - 4 +0.212 % - 0.21 244 +0.30] -0.303 V Jm. 1.515V 0.21.244,-0.334,=4 --> 0 => 0. 757V, 20.303 vg -0.454 44 = -3.79 19) Apply Nodal Analysig at Node 'Y RO -1.84M AT V VG-Vg 4.7km 1 (4.1ke) 2.2k2 - 1.84mA +0.21244 -0.212V 2 - 0.45 444-0.4594 met -0.242V4-0-21213-0.4541,= (8 + from the given Circuit we can say V3 =vsrc=4v from ② v= -0.212v-o.qu2v4-18 v=-0.466qve -0.5731_3.96 Put & in & @ => 1.515v2-0.21 244 -0.30] (-0.466942-051744-1985

(2) > 1.515 V2 - 0212 44 to 1414 Vg to 161 V4 +1.199 =0 :.1.656 Vg - 0.05104 + 1199 = 0 157(-0.46690,-0.63344 -3.96 -0.3o3V2 – 0.454 V +3.79=0 -0.3534% -0.403402997 - 0.303 V 0.45444 +379- 7 - o. 6064 V - 0.857444 + 0.793=000 Solve 6 o goby + 6 => V4 = 101.4058 1.651 +0.3534 4 =1.4058 v Y3 -0.030V4 + 0.72-0 -Y2 - 1.041 Vy + 1.30 -0 =-0.6807 :v = -0.4669 Vq -0.5v, - 1.400 +2.02=0 -3.96 4= - 2002 =-0.4669 (-0.6807) -1.44 -0.533 (1.4058)-3.96 y = 1.402. 1.4059 = 0.317-0.749-3.96 V = -4.392 .:V2 = -0.68o7

Note V3=Vsrc=4v;

Please give ? if u understood the concept.