Question

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier directly on the line voltage with Vf = 0.7V, what will be your output DC voltage in Volts (assuming a negligible ripple)?

Answer 1

The output DC voltage after using a full-wave rectifier on an AC source with a Vrms of 60V, considering a forward voltage drop of 0.7V and negligible ripple, is approximately 53.7V.

Step-by-step explanation:

To convert an AC voltage to DC voltage using a full-wave rectifier, you first need to understand the relationship between the root mean square (RMS) value of the AC voltage and its peak value. The RMS value is given as Vrms = 60V. For a full-wave rectified sine wave, the peak voltage (Vp) is Vrms × √2. Therefore, Vp = 60V × √2 ≈ 84.85V. However, because of the rectifier's forward voltage drop (Vf), we have to subtract this from the peak voltage to get the peak DC voltage (Vdc_peak). Therefore, Vdc_peak = Vp - Vf ≈ 84.85V - 0.7V ≈ 84.15V.

However, to find the average or output DC voltage (Vdc_avg) from a full-wave rectifier, you'd multiply the peak DC voltage by 2/π (since it's a sinusoidal wave), so Vdc_avg ≈ (2/π) × 84.15V ≈ 53.7V. Thus, the output DC voltage after using a full-wave rectifier directly on the line voltage, assuming negligible ripple, is approximately 53.7V.

Answer 2

`V_(dc)=84.15\ V`

Step-by-step explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

`V_(o)=√(2)\ V_(rms)`

Now by putting the values

`V_(o)=60√(2)\ V`

The output voltage of the DC current given as

`V_(dc)=V_(o)-V_f`

`V_(dc)=60√(2)-0.7\ V`

`V_(dc)=84.15\ V`

Therefore output voltage of the DC current is 84.15 V.