GATE 2016 EE (EE2) – Question 37
For the ideal op-amp circuit, Vout in terms of V1,V2,V3 is

Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (D) $8V_1+2V_2-9V_3$
Explanation
No current enters the noninverting input, so its voltage is the conductance-weighted average $V_+=(V_1/1+V_2/4)/(1+1/4)=0.8V_1+0.2V_2$.
Negative feedback sets V−=V+. KCL there gives $(V_--V_3)/1+(V_--Vout)/9=0$, so Vout=10V−−9V3.
Substitution gives **8V1+2V2−9V3, D**.
More Operational amplifiers: characteristics and applications questions
- GATE 2015 EE (EE2) Q49
- GATE 2026 EE Q37
- GATE 2025 EE Q17
- GATE 2025 EE Q56
- GATE 2024 EE Q61
- GATE 2023 EE Q37
All Operational amplifiers: characteristics and applications questions →