The image shows a circuit with an AC input of 120 Vrms passing through several components: a transformer, full-wave rectifier, filter, 5V regulator, and a load. The question is asking for the voltage at point C, which is after the full-wave rectifier.

Here's a breakdown:

1. Input voltage at A (AC input) is 120 Vrms.
2. The transformer has a 10:1 ratio, meaning it steps down the voltage. The secondary side at B would have: $V_{B} = \frac{120 \text{ Vrms}}{10} = 12 \text{ Vrms}$
3. After the full-wave rectifier at point C, the peak voltage $V_p$ can be calculated from the RMS voltage: $V_p = V_{rms} \times \sqrt{2} = 12 \times 1.414 \approx 16.97 \text{ V}$
4. However, the full-wave rectifier introduces a diode drop (about 2 diode voltage drops in total for a full-wave rectifier using silicon diodes), which subtracts around 1.4V (0.7V per diode). This results in: $V_p \approx 16.97 - 1.4 = 15.57 \text{ V}$

Therefore, the voltage at point C is approximately 15.6 Vp.

The correct option is 15.6 Vp.

Do you want more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the difference between peak and RMS voltage?
2. How does a transformer step down voltage?
3. Why do rectifiers drop voltage, and how is the drop calculated?
4. What role does a filter play in this circuit?
5. What is the purpose of the 5V regulator in this circuit?

Tip: In full-wave rectification, both halves of the AC signal are used, which improves efficiency compared to half-wave rectification.