In a circuit with Rin = 5 kΩ and RF = 45 kΩ in an inverting amplifier, the closed-loop gain magnitude is

In a circuit with Rin = 5 kΩ and RF = 45 kΩ in an inverting amplifier, the closed-loop gain magnitude is

• A. 10
• B. 9
• C. 45
• D. 5

Answer: (B)

In an inverting amplifier, the output is related to the input by the ratio of the feedback resistor to the input resistor. For an ideal op-amp, the voltage at the inverting input sits at a virtual ground, so the input current through Rin equals Vin/Rin and must flow through RF, forcing the output to supply the same current via RF. This leads to Vout = - (RF/Rin) Vin, so the closed-loop gain is RF/Rin in magnitude (the negative sign indicates inversion). With Rin = 5 kΩ and RF = 45 kΩ, RF/Rin = 45/5 = 9. Hence the magnitude of the closed-loop gain is 9. The actual gain is -9 (inverting), but the question asks for magnitude, which is 9.

In an inverting amplifier, the output is related to the input by the ratio of the feedback resistor to the input resistor. For an ideal op-amp, the voltage at the inverting input sits at a virtual ground, so the input current through Rin equals Vin/Rin and must flow through RF, forcing the output to supply the same current via RF. This leads to Vout = - (RF/Rin) Vin, so the closed-loop gain is RF/Rin in magnitude (the negative sign indicates inversion).

With Rin = 5 kΩ and RF = 45 kΩ, RF/Rin = 45/5 = 9. Hence the magnitude of the closed-loop gain is 9. The actual gain is -9 (inverting), but the question asks for magnitude, which is 9.