Question

please show full working and check calculations/formulas again before submitting
2. (50 pts) In this problem, assume the operational amplifier is ideal and its open-loop gain A=c0. The current sources and voltage sources are ideal. Find 1) V.; 2) the current I flowing out of or into the output of the op amp and its direction (e.g. out of or into); 3) the currents flowing in the voltage sources I and 12 as denoted, and their directions; 4) the voltage at nodes Vi and V2. 2k12 I 2k92 =ov Vo 1 MAY 2 U 10kg 36k2 VT 31k2 3k92 2mA

Answer 1

2L • 2K12 - GV FVV I imp ТСе 5k 1 +5V - VOA 5mA Tokel amit CV 36к. + 2 56lka 3KL IMA -4.5mA سهم 5mA IMA - 3y + tekur 2. m A 3V WEE . In: 5 - 5mA and Lom=0 A ( of IK and Ilokn = 0 A (i opamp dont dorow current) 45mA So I = 5mA 5kn ê 5KR X IMA = 5v Vzkr = 2kaXIMA=2v V-2V 36V 9) V- = 8V By Virtual short v = V+=8V -> V+-5 - I3kX3K1 = 0 =) 8-5-I34X3452 = 0 - Isk a = 1 mA = I3km ] 5V T 1km Ysmo 3k2 5mA - IMA Skyrima

2) I Gkr = IMA 5) Vakn = 6kn XIMA = 6 V V XA 3kt 31+ vo va RmA =) V2 -3 -3 V=0 - VL : 6 V y 3) V = 6-6=OV 2ks- 2k1, Ima 2mA 43 nA GV 4MA Lov EVOV TIMA Gu} Skr 4 Izka - 3 MA * output current is 4mA into the op- Amp ak 2mn I = 2 MA Hann 1=20A - GVT 5V + 5kr |_= 8v V MA IMA IMA Ovi IMA => V, -5=8V +1v,-13V 0 Answers Vorov ; 1:4mA into Op-Amp; I,-2MA ; I=5MA V = 13V, V2 =6V