Question

The power rating of the diode can be calculated as:

P = PIV × IL(AV) = 325 V × 50 A = 16.25 kW

The power rating of the capacitor can be calculated as:
P = 0.5 × C × VDC² × f = 0.5 × 100μF × (200 V)² × 50 Hz = 2 W

In the circuit designed in the previous point, replace the half-wave rectifier with a bridge of diodes and:
• Obtain the new waveforms v(t) and i(t) in each component and compare with what was obtained with the half-wave rectifier.
• Change the capacitor for a larger one, check the behavior of the circuit and compare the results with those obtained with the initial capacitor.

Answer 1

Replacing the half-wave rectifier with a bridge of diodes will lead to different waveforms in each component. Changing the capacitor for a larger one will affect the circuit's behavior and result in smoother waveforms with smaller ripple voltage.

Step-by-step explanation:

When replacing the half-wave rectifier with a bridge of diodes, the new waveforms v(t) and i(t) in each component will be different compared to what was obtained with the half-wave rectifier. In the bridge rectifier, both positive and negative half cycles of the input AC voltage are utilized, resulting in a higher average output voltage and smoother waveform compared to the half-wave rectifier.

Replacing the capacitor with a larger one will affect the behavior of the circuit. A larger capacitor will result in a longer charging time and a slower voltage rise, leading to a smoother waveform and a smaller ripple voltage across the capacitor. This can improve the effectiveness of the circuit in filtering out unwanted signals.