GATE 2020 EC Q.18: solution and answer

Official answer: 6Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Assumptions: the op-amp is ideal (infinite input resistance and infinite open-loop gain), so no current flows into either input terminal. With negative feedback present, the inputs are at the same potential as long as the output stays within the $\pm 5\text{V}$ supply rails.

2. Virtual short: $V_{-}=V_{+}=V_i=2\text{V}$.

3. Current in the 1 k$\Omega$ resistor from the inverting input to ground: $I_1=\frac{2\text{V}}{1\text{k}\Omega }=2\text{mA}$. No current enters the inverting input, so the same $2\text{mA}$ must come from the output through the 1 k$\Omega$ feedback resistor: $I_f=2\text{mA}$.

4. Output voltage (non-inverting amplifier): $V_o=V_i\left(1+\frac{R_f}{R_1}\right)=2\left(1+\frac{1\text{k}\Omega }{1\text{k}\Omega }\right)=4\text{V}$. This is inside the $\pm 5\text{V}$ rails, so the op-amp is not saturated and the virtual-short assumption holds. Check: $I_f=\frac{4-2}{1\text{k}\Omega }=2\text{mA}$.

5. Load current: $I_L=\frac{V_o}{1\text{k}\Omega }=\frac{4\text{V}}{1\text{k}\Omega }=4\text{mA}$.

6. KCL at the output node: the op-amp output feeds both the load and the feedback path, so $$I_o=I_L+I_f=4\text{mA}+2\text{mA}=6\text{mA}$$