Question

Determine the peak-to-peak ripple voltage (in volts) when the peak outpot voltege of the bridge rectifier is 12 V.

Given that the input voltage has a frequency at the primary side of the transformer =51 Hz
And the FILTER at the outpot of the rectifier has a capacitor value of 56.8 μF and the resistor value of 228Ω.

Answer 1

The peak-to-peak ripple voltage in a bridge rectifier can be determined using the formula Vo = (1/(2πfRC)) x Vp

Step-by-step explanation:

The peak-to-peak ripple voltage in a bridge rectifier can be determined by calculating the voltage across the capacitor in the filter circuit. The formula for the peak-to-peak ripple voltage is given by:

Vo = (1/(2πfRC)) x Vp

Where:

• Vo is the peak-to-peak ripple voltage
• f is the frequency of the input voltage
• R is the resistance in the filter circuit
• C is the capacitance in the filter circuit
• Vp is the peak output voltage of the rectifier

Substituting the given values:

• f = 51 Hz
• R = 228 Ω
• C = 56.8 μF
• Vp = 12 V

Vo = (1/(2π(51 Hz)(228 Ω)(56.8 μF))) x 12 V = 9.5 V

Therefore, the peak-to-peak ripple voltage in this bridge rectifier circuit is 9.5 volts.