Question R1 R2 R5 R6 in2 Vb (R1) (R2) (R5) (R6) (Vb) U1 U2 O R3 in1 Va (R3 4 (R4) (Va) For the circuit shown assume the OPAMP is ideal In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, R6, write the equation for the node voltage at X, due to Va only Preview In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, and R6, write the equation for the node voltage at X, due to Vb only R1 R2 R5 R6 in2 Vb (R1) (R2) (R5) (R6) (Vb) U1 U2 O R3 in1 Va (R3 4 (R4) (Va) For the circuit shown assume the OPAMP is ideal In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, R6, write the equation for the node voltage at X, due to Va only Preview In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, and R6, write the equation for the node voltage at X, due to Vb only Preview In terms of Vx, R5, and R6 Write the equation for the node voltage at out Preview In terms of Va, Vb, R1, R2, R3, R4, R5, and Rç, write the equation for the node out, due to Va and Vb Preview

ZSTLP5 The Asker · Electrical Engineering

please help! this is one problem that ive been stuck on

 

 

Transcribed Image Text: R1 R2 R5 R6 in2 Vb (R1) (R2) (R5) (R6) (Vb) U1 U2 O R3 in1 Va (R3 4 (R4) (Va) For the circuit shown assume the OPAMP is ideal In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, R6, write the equation for the node voltage at X, due to Va only Preview In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, and R6, write the equation for the node voltage at X, due to Vb only Preview In terms of Vx, R5, and R6 Write the equation for the node voltage at out Preview In terms of Va, Vb, R1, R2, R3, R4, R5, and Rç, write the equation for the node out, due to Va and Vb Preview
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Transcribed Image Text: R1 R2 R5 R6 in2 Vb (R1) (R2) (R5) (R6) (Vb) U1 U2 O R3 in1 Va (R3 4 (R4) (Va) For the circuit shown assume the OPAMP is ideal In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, R6, write the equation for the node voltage at X, due to Va only Preview In terms of Va, Vb, Vx, R1, R2, R3, R4, R5, and R6, write the equation for the node voltage at X, due to Vb only Preview In terms of Vx, R5, and R6 Write the equation for the node voltage at out Preview In terms of Va, Vb, R1, R2, R3, R4, R5, and Rç, write the equation for the node out, due to Va and Vb Preview
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(1)Hooln: Given that !a) When v_(b)=0,v_(x)0v_(a)^(')==> from virtual grand concept, the potential at (2) is eguat to potential at (0., Now, Appy, Kcl(2)(2)=>quadnu_(2)=(v_(a)R_(4))/(R_(3)+R_(y))Now, Apply, iscl 20{:[=>(V_(1)-0)/(R_(1))+(V_(1)-V_(x))/(R_(2))=0],[=>quad(v_(1))/(R_(1))+(v_(1)-v_(x))/(R_(2))=0],[=>quad(V_(1)R_(2)+V_(1)R_(2)-V_(2)R_(1))/(R_(1)R_(2))=0],[=>quadV_(1)(R_(1)+R_(2))=V_(x)R_(1)],[=>nu_(1)=((nu_(x)R_(1))/(R_(1)+R_(2)))],[" But "nu_(1)=v_(2)quad(:'" virtual Ground ")" concept. "],[=>(V_(x)R_(1))/(R_(1)+R_(2))=(V_(a)R_(y))/(R_(3)+R_(y))],[=>quadnu_(x)=[(R_(1)+R_(2))/(R_(3)+R_(4))]((R_(4))/(R_(1)))*V_(a)quadv_(01 r)]:}(b) When V_(a)=0,quadV_(x) due to V_(a)!={:[=>(V_(2)-0)/(R_(3))+(V_(2)=0)/(R_(4))=0],[=>(V_(2)R_(3)+V_(2)R_(3))/(R_(3)R_(4))=0],[=>V_(2)(R_(3)+R_(4))=0],[=>quadV_(2)=0]:}lly Apply kcl@ (1),{:[quadquad(v_(1)-v_(b))/(R_(1))+(v_(1)-v_(x))/(R_(2))=0], ... See the full answer