Electronics Engineering (ELEX) Board Practice Exam

Given a full wave rectified voltage of 18 Vp across a 500 µF capacitor filter, what is the ripple if the load current is 100 mA at a supply frequency of 50 Hz?

1. 0.025

2. 0.032

3. 0.040

4. 0.020

The correct answer is: 0.032

To determine the ripple voltage in a full-wave rectified circuit, the formula used is derived from the capacitance, load current, and supply frequency. The ripple voltage (Vr) can be calculated using the formula: \[ V_r = \frac{I}{f \cdot C} \] where: - \( I \) is the load current (in amperes), - \( f \) is the frequency of the ripple (which is twice the supply frequency for full-wave rectification), - \( C \) is the capacitance (in farads). In this scenario: - The load current \( I \) is 100 mA or 0.1 A, - The supply frequency \( f \) is 50 Hz. Since it is a full-wave rectifier, the ripple frequency will be \( 2 \times 50 = 100 \) Hz, - The capacitance \( C \) is 500 µF, which is equal to \( 500 \times 10^{-6} \) F or \( 0.0005 \) F. Now, we can substitute these values into the ripple voltage formula: \[ V_r = \frac{0.1 \, A}{100 \