Question

The diameter of high power diode is proportional to its maximum forward current.

a). Calculate the ac ripple voltage of 1-ph full-wave rectifier supplying a resistive load if input ac voltage source is 100V. Use two decimal accuracy.

Answer 1

The AC ripple voltage of a 1-ph full-wave rectifier cannot be calculated with the information provided. Additional details such as the capacitance, load resistance, and frequency of the AC supply are necessary. However, electric power dissipation in a resistor can be explained by the formula P = I²R, indicating that power is proportional to the square of the current multiplied by the resistance.

Step-by-step explanation:

To calculate the AC ripple voltage of a 1-ph full-wave rectifier supplying a resistive load with an input AC voltage source of 100V, we need to understand that the AC ripple voltage (Vripple) is the fluctuation in voltage within the output of a rectifier. In a full-wave rectifier circuit, this ripple voltage is typically represented by the equation Vripple = (Vpeak - Vload)/fCRL, where Vpeak is the peak voltage of the AC input, Vload is the voltage across the load, f is the frequency of the AC supply, C is the capacitance of the filter capacitor, and RL is the load resistance.

Unfortunately, with the provided information, we cannot calculate Vripple without knowing the values for the capacitance of the filter capacitor, the frequency of the AC supply, and the load resistance. Additionally, this calculation would require further information such as the characteristics of the full-wave rectifier in terms of voltage drop across diodes, which is essential to deduce the Vload.

Answering your initial statement, power dissipation in resistors is explained by the formula P = I²R, which means that electric power is proportional to the square of the current through the resistor multiplied by the resistance. This insight comes from the formula P = V²/R, which implies that for a lower resistance connected to a given voltage source, the greater the power delivered.

Answer 2

The AC ripple voltage of a 1-ph full-wave rectifier with an input AC voltage of 100V is calculated as 51.40V. This is derived by converting the AC input to peak voltage, estimating the average DC output, and then subtracting the average from the peak.

Step-by-step explanation:

To calculate the AC ripple voltage of a 1-ph full-wave rectifier supplying a resistive load with an input AC voltage of 100V, we must understand that the ripple voltage is the residual periodic variation of the DC voltage within a power supply which has been derived from an alternating current (AC) source. The formula for ripple voltage (Vr) in a full-wave rectifier is given by Vr = Vm - Vdc, where Vm is the peak value of the input AC voltage and Vdc is the average (DC) output voltage. For a full-wave rectifier, Vdc is approximately 0.636 of the peak voltage Vm. Since the input AC voltage here is given as 100V rms, we convert this to peak voltage Vm by multiplying it by √2, which gives us the peak voltage of 141.42V. Now, the average voltage Vdc becomes 0.636 × 141.42V, which is approximately 90.02V. To find the ripple voltage Vr, we subtract Vdc from Vm: Vr = Vm - Vdc = 141.42V - 90.02V, which is 51.40V. Thus, the ripple voltage is 51.40V.