GATE 2022 EC – Question 58

Correct answer: -0.51 to -0.49

Explanation

Let the ladder nodes be $V_a,V_b,V_c,V_d$ (left to right, $R=1\ \text{k}\Omega$). The inverting input is a virtual ground, so the $2R$ at the end of the ladder carries $V_d/2R$. KCL at $d$: $V_c-V_d=V_d$, so $V_c=2V_d$. At $c$: $2(V_b-V_c)=2(V_c-V_d)+(V_c-1.6)$, giving $V_b=4V_d-0.8$. At $b$: $2(V_a-V_b)=2(V_b-V_c)+V_b$, giving $V_a=8V_d-2$. At $a$: $V_a+(V_a-1.6)+2(V_a-V_b)=0$, which gives $24V_d=8$ and $V_d=\frac13$ V. The inverting amplifier has gain $-\frac{3R}{2R}=-1.5$, so $V_0=-1.5\times\frac13=-0.5$ V.