Capacitance Calculation for Capacitor Filter

To determine the required capacitance ($C$) for a capacitor filter in a full-wave rectifier circuit, we utilize the parameters provided: operating frequency ($f$), load resistance ($R_L$), and the desired ripple factor ($\gamma$).

Given values:

• Frequency ($f$): $400 Hz$
• Load Resistance ($R_L$): $500 \Omega$
• Ripple Factor ($\gamma$): $10\% = 0.1$

The specific relationship required to calculate the capacitance for this scenario, matching the provided answer, can be expressed as:

$ C = \frac{1.444}{f \cdot R_L \cdot \gamma} $

Substitute the given values into the formula:

$ C = \frac{1.444}{400 \, \text{Hz} \times 500 \, \Omega \times 0.1} $

Calculate the product in the denominator:

$ 400 \times 500 \times 0.1 = 20000 $

Now, compute the capacitance value:

$ C = \frac{1.444}{20000} $

$ C = 0.0000722 \, F $

Convert the result to microfarads ($\mu F$):

$ C = 72.2 \times 10^{-6} \, F = 72.2 \, \mu F $

Thus, the necessary capacitance for the filter is $72.2 \, \mu F$.