Understanding Ripple Voltage in Full-Wave Rectifiers

In electronics, a rectifier converts alternating current (AC) to direct current (DC). However, the output DC is not perfectly smooth; it contains fluctuations called ripple voltage. A full-wave rectifier utilizes both halves of the AC input cycle. A filter capacitor is often added to reduce this ripple voltage, smoothing the output DC.

The question asks for the ripple voltage of a full-wave rectifier circuit with specific components: a 100-µF filter capacitor and a load drawing a current of 50 mA.

Calculating Ripple Voltage Using the Filter Capacitor

The approximate peak-to-peak ripple voltage ($V_r$) for a full-wave rectifier with a capacitive filter can be calculated using the formula:

$ V_r \approx \frac{I_L}{2 f C} $

Where:

• $I_L$ is the DC load current.
• $f$ is the frequency of the AC input voltage.
• $C$ is the capacitance of the filter capacitor.

Given Parameters:

• Load Current ($I_L$): 50 mA
• Filter Capacitance ($C$): 100 µF
• Rectifier Type: Full-Wave

Assumption for Frequency:

The frequency ($f$) of the AC input is not explicitly given. In many standard power supply applications, the AC mains frequency is either 50 Hz or 60 Hz. We will assume a frequency of 60 Hz for this calculation, as it leads to a result consistent with the options provided.

Step-by-Step Calculation:

1. Convert units to standard base units:
   • Load Current: $I_L = 50 \text{ mA} = 50 \times 10^{-3} \text{ A}$
   • Capacitance: $C = 100 \text{ µF} = 100 \times 10^{-6} \text{ F}$
   • Frequency: $f = 60 \text{ Hz}$
2. Substitute the values into the ripple voltage formula:

   $ V_r \approx \frac{50 \times 10^{-3} \text{ A}}{2 \times 60 \text{ Hz} \times 100 \times 10^{-6} \text{ F}} $

3. Calculate the denominator:

   Denominator $= 2 \times 60 \times 100 \times 10^{-6} = 12000 \times 10^{-6} = 0.012 \text{ A/V (or C/s)} \times \text{s} = 0.012 \text{ Farads (Incorrect unit derivation, focus on numerical calculation)}$

   Denominator $= 2 \times 60 \times (100 \times 10^{-6}) = 120 \times 100 \times 10^{-6} = 12000 \times 10^{-6} = 0.012$

4. Calculate the ripple voltage ($V_r$):

   $ V_r \approx \frac{0.05 \text{ A}}{0.012} \approx 4.167 \text{ V} $

Conclusion:

The calculated ripple voltage is approximately 4.17 V. Comparing this value to the given options:

• 3 V
• 0 V
• 1.2 V
• 4 V

The calculated value of 4.17 V is closest to 4 V.

Therefore, the ripple voltage of the full-wave rectifier under the given conditions is approximately 4 V.