Question E7G07:
What voltage gain can be expected from the circuit in Figure E7-3 when R1 is 10 ohms and RF is 470 ohms?  

Choices:  
A: 0.21  
B: 4700  
C: 47  
D: 24    

Correct Answer is  C: 47  

To find the voltage gain of the op-amp in Figure E7-3 (an inverting amplifier), use:
$$A_v = -\frac{R_F}{R_1}$$

Compute it:
$$A_v = -\frac{470\ \Omega}{10\ \Omega} = -47$$
Magnitude of gain is 47, which matches choice C.

Why C (47) is correct:
- In an inverting op-amp, the closed-loop voltage gain is set by the ratio of the feedback resistor to the input resistor. Here, $R_F/R_1 = 470/10 = 47$.

Why the others are incorrect:
- A: 0.21 — This resembles the inverse ratio, but the true inverse is $10/470 \approx 0.021$, not 0.21. Wrong operation and decimal.
- B: 4700 — Likely from multiplying $470 \times 10$ or confusing units. Gain is a ratio, not a product; it should be 47.
- D: 24 — Could come from a mistaken halving or misusing a voltage-divider idea. The correct formula does not yield ~24.

Useful formulas and tips:
- Inverting op-amp: $$A_v = -\frac{R_F}{R_{in}}$$
- Non-inverting op-amp (for contrast): $$A_v = 1 + \frac{R_F}{R_{G}}$$
- The negative sign in the inverting case indicates phase inversion; exams often ask for magnitude only.