GATE EC 2021 Set 1 — Question 19

The circuit shown is an ideal OPAMP configured as a summing amplifier with the non-inverting input connected to $V_{REF}$.
For an ideal OPAMP, the voltage at the non-inverting input ($V_+$) is equal to the voltage at the inverting input ($V_-$) due to the virtual short concept. Thus, $V_- = V_+ = V_{REF}$.Applying Kirchhoff's Current Law (KCL) at the inverting input node ($V_-$):
$\frac{V_{IN} - V_-}{R_{IN}} + \frac{0 - V_-}{R_1} + \frac{V_{REF} - V_-}{R_2} + \frac{V_{OUT} - V_-}{R_F} = 0$Substitute $V_- = V_{REF}$ into the KCL equation:
$\frac{V_{IN} - V_{REF}}{R_{IN}} + \frac{0 - V_{REF}}{R_1} + \frac{V_{REF} - V_{REF}}{R_2} + \frac{V_{OUT} - V_{REF}}{R_F} = 0$The term $\frac{V_{REF} - V_{REF}}{R_2}$ becomes 0, meaning resistor $R_2$ is effectively bypassed or irrelevant in this configuration.The KCL equation simplifies to:
$\frac{V_{IN} - V_{REF}}{R_{IN}} - \frac{V_{REF}}{R_1} + \frac{V_{OUT} - V_{REF}}{R_F} = 0$Rearranging the equation to solve for $V_{OUT}$:
$\frac{V_{OUT} - V_{REF}}{R_F} = - \frac{V_{IN} - V_{REF}}{R_{IN}} + \frac{V_{REF}}{R_1}$
$V_{OUT} - V_{REF} = - \frac{R_F}{R_{IN}} (V_{IN} - V_{REF}) + \frac{R_F}{R_1} V_{REF}$
$V_{OUT} = V_{REF} - \frac{R_F}{R_{IN}} V_{IN} + \frac{R_F}{R_{IN}} V_{REF} + \frac{R_F}{R_1} V_{REF}$
$V_{OUT} = V_{REF} \left(1 + \frac{R_F}{R_{IN}} + \frac{R_F}{R_1}\right) - \frac{R_F}{R_{IN}} V_{IN}$This equation is in the linear form $V_{OUT} = K_c - K_g V_{IN}$, where:
$K_c = V_{REF} \left(1 + \frac{R_F}{R_{IN}} + \frac{R_F}{R_1}\right)$
$K_g = \frac{R_F}{R_{IN}}$We are given two conditions: $1 = K_c - K_g (0.1)$ (Equation 1) $6 = K_c - K_g (1)$ (Equation 2)Subtract Equation 1 from Equation 2:
$(6 - 1) = (K_c - K_g) - (K_c - 0.1 K_g)$
$5 = -K_g + 0.1 K_g$
$5 = -0.9 K_g$
$K_g = -\frac{5}{0.9} = -\frac{50}{9} \approx -5.555$Since $K_g = \frac{R_F}{R_{IN}}$, and resistances ($R_F$, $R_{IN}$) are positive values, their ratio must also be positive. The calculated negative value for $K_g$ indicates an inconsistency in the problem statement or the given values. This type of inconsistency often leads to the question being marked as "Marks to All" (MTA) in competitive exams, as indicated in the answer key for this question.However, if the question implicitly asks for the magnitude of the gain factor associated with $V_{IN}$, then $|K_g| = |-\frac{50}{9}| = \frac{50}{9} \approx 5.555$.
This value matches option (D).Final Answer is based on the magnitude of the calculated gain factor, assuming the problem intended a positive ratio for $R_F/R_{IN}$.The final answer is $\boxed{\text{5.555}}$