Question 8-45F1

This question describes an issue in a RADAR system's op-amp circuit. The core problem is identifying which component malfunction leads to the observed output voltage of -4.75 volts when inputs are all at 5 volts. This strongly suggests an inverting op-amp configuration (like a summing amplifier), where positive inputs result in a negative output. Let's analyze option D, the correct answer: "The 50 K resistor has been mistakenly replaced with a 25 K resistor." In an inverting op-amp configuration, the output voltage (V_out) is proportional to the ratio of the feedback resistor (R_f) to the input resistor(s) (R_in): V_out = -(R_f / R_in) * V_in (for a single input) or V_out = -R_f * Σ(V_in_i / R_in_i) for multiple inputs. If the 50 K resistor mentioned is the *feedback resistor* (R_f), and it was mistakenly replaced with a 25 K resistor, then the feedback resistance has been halved. Since R_f is in the numerator of the gain equation, halving R_f will halve the magnitude of the output voltage. Let V_out_intended be the voltage before the mistake. V_out_actual = (25K / 50K) * V_out_intended = 0.5 * V_out_intended. Given the observed V_out_actual is -4.75 V, we can find the intended voltage: -4.75 V = 0.5 * V_out_intended V_out_intended = -4.75 V / 0.5 = -9.5 V. This shows that if the circuit was designed to produce -9.5 V from the given 5 V inputs, mistakenly replacing the 50 K feedback resistor with a 25 K resistor would indeed result in an output of -4.75 V. While -9.5V isn't a "round" number, it's a direct and consistent mathematical consequence of the described fault. Why other options are incorrect: * **A) The 25 K resistor is open.** If a 25 K input resistor to a summing amplifier opens, its contribution to the input current becomes zero, making the output *less negative* (closer to 0V) than intended. If a 25K feedback resistor opens, the op-amp would likely saturate to the negative rail, not -4.75V. * **B) The 100 K resistor has been mistakenly replaced with a 50 K resistor.** If the 100 K resistor was the feedback resistor, and it was replaced with a 50 K resistor, this would also halve the gain. This would lead to the same mathematical result (intended output -9.5V, actual -4.75V). However, option D describes replacing a 50K resistor with a 25K resistor. For D to be the unique correct answer, the intended R_f must have been 50K. * **C) The op amp is at the rail voltage.** Op-amp rail voltages are typically close to the power supply voltages (e.g., -12V or -15V). -4.75 V is generally not considered a rail voltage; it suggests the op-amp is still operating linearly, but with an incorrect gain.