A Full Wave Rectifier (FWR) is an electronic circuit that converts AC power to DC power. The output of an FWR has a ripple voltage, which is an AC voltage superimposed on the DC voltage. The ripple voltage should be minimized to achieve a stable DC voltage. This response will explain how to calculate the minimum capacitance required to achieve a ripple voltage of less than 5% in a FWR.


The formula for calculating the ripple voltage in a FWR is:

Vr = (I/(2*f*C))

Where:
Vr = Ripple voltage
I = Load current
f = Frequency of AC power (50 Hz in this case)
C = Capacitance


To calculate the minimum capacitance required, we need to rearrange the formula for capacitance:

C = (I/(2*f*Vr))

We know that the load current (I) is 1kΩ, and the ripple voltage (Vr) should be less than 5% of the output voltage. Let's assume the output voltage is 12V.

Vr = 0.05*12 = 0.6V

Now we can calculate the minimum capacitance required:

C = (1/(2*50*0.6)) = 0.0167F or 16.7μF

Therefore, the minimum capacitance required to achieve a ripple voltage of less than 5% in a FWR with a load of 1kΩ and an output voltage of 12V is 16.7μF.


In conclusion, the calculation of the minimum capacitance required in a FWR involves using the formula for ripple voltage and rearranging it for capacitance. It is important to ensure that the ripple voltage is minimized to achieve a stable DC voltage. In this case, a capacitance of 16.7μF is required to achieve a ripple voltage of less than 5%.