The circuit in the image shows an AC voltage source that goes through several stages: a transformer, a full-wave rectifier, a filter, a 5V regulator, and a load. The key question is determining the voltage at point D, which is just after the filter.

Here's a breakdown of the circuit:

1. AC Input: The input is 120 Vrms (root mean square). This is the voltage entering the transformer.

2. Transformer: The transformer has a 10:1 step-down ratio, meaning the output at point B (secondary side of the transformer) will be 10 times lower than the input voltage.

   • The input is 120 Vrms, so the output at point B will be: $V_B = \frac{120 \text{ Vrms}}{10} = 12 \text{ Vrms}$ The transformer steps the voltage down to 12 Vrms.

3. Full-Wave Rectifier: After the transformer, the full-wave rectifier converts the AC voltage into pulsating DC. The rectified voltage (peak voltage, $V_{\text{peak}}$) can be calculated from the Vrms value. The relationship between Vrms and peak voltage is: $V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2}$ So, the peak voltage after rectification is: $V_{\text{peak}} = 12 \times \sqrt{2} \approx 16.97 \text{ V}$ Therefore, the peak rectified voltage is about 16.97 V.

4. Filter: The filter smooths out the pulsating DC to provide a near-constant DC voltage. For ideal filtering, the output voltage at point D would be close to the peak value from the rectifier, minus any small losses in the components. So, at point D, the voltage is approximately 16.97 V DC.

Thus, the voltage at point D is approximately 16.97 V DC.

Let me know if you need more details or explanations.

Follow-up Questions:

1. How does the full-wave rectifier work in converting AC to DC?
2. What role does the filter play in smoothing the DC voltage?
3. Why is the Vrms lower than the peak voltage?
4. What kind of filter is typically used in such circuits (e.g., capacitor filter)?
5. How would the output change if the transformer ratio was different?

Tip:

To accurately calculate DC voltage after rectification, always remember to account for the diode drops in a real-world circuit, typically around 0.7V per diode in silicon diodes.