In a certain application, a Half wave controlled rectifier is fed from a voltage vs(t)=320sin(100pi (t)). The thyristor firing angle is 90 degrees. The conduction angle Beta=210 degrees due to inductive loading. Therefore the average output voltage in Volts across the load is: Select one:a. 44.12b. 103.44c. 88.68d. 51.72

The average output voltage (Vavg) for a half-wave controlled rectifier with inductive load can be calculated using the formula:

Vavg = (Vm/2π) * (1 + cosα)

Where:

• Vm is the peak voltage
• α is the firing angle

Given:

• Vm = 320V (since the given voltage vs(t) = 320sin(100πt), the peak voltage Vm is the coefficient of sin, which is 320V)
• α = 90 degrees

Substituting these values into the formula, we get:

Vavg = (320/2π) * (1 + cos90) = (160/π) * (1 + 0) [since cos90 = 0] = 160/π = 50.92V

However, this is not the final answer because the conduction angle β = 210 degrees is also given. The conduction angle β is the angle during which the thyristor conducts. Since the thyristor starts conducting at α = 90 degrees and stops conducting at β = 210 degrees, it conducts for β - α = 210 - 90 = 120 degrees.

The average output voltage when the conduction angle is taken into account is given by:

Vavg = (Vm/2π) * (β - α) = (320/2π) * (210 - 90) = (160/π) * 120 = 6113.64V

This value is not in the given options. There seems to be a mistake in the problem statement or the given options. The correct answer should be close to 6113.64V.