Semi-Converter Average Current Calculation

This solution calculates the average output current ($I_{avg}$) for a single-phase semi-converter with a resistive-inductive-EMF ($R-L-E$) load.

Given Parameters:

• Supply Voltage ($V_s$) = 200 V (RMS)
• Frequency ($f$) = 50 Hz
• Load Resistance ($R$) = $15 \ \Omega$
• Back EMF ($E$) = 80 V
• Firing Angle ($\alpha$) = $90^\circ$
• Load Inductance ($L$) is very large, ensuring ripple-free current.

Calculation Steps:

1. Calculate Peak Supply Voltage ($V_m$):

   The peak voltage is related to the RMS voltage by $V_m = V_s \sqrt{2}$.

   $V_m = 200 \ V \times \sqrt{2} \approx 282.84 \ V$

2. Calculate Average Output Voltage ($V_{avg}$):

   For a semi-converter with an RLE load, the average output voltage is given by:

   $V_{avg} = \frac{V_m}{\pi} (1 + \cos \alpha)$

   Substituting the values for $\alpha = 90^\circ$:

   $V_{avg} = \frac{282.84 \ V}{\pi} (1 + \cos 90^\circ)$

   $V_{avg} = \frac{282.84 \ V}{\pi} (1 + 0) \approx 90.05 \ V$

3. Calculate Average Output Current ($I_{avg}$):

   With a large inductance, the current is nearly constant and ripple-free. The average output current is determined by the average voltage available to drive the current through the resistance and back EMF.

   Using Ohm's law applied to the average values:

   $V_{avg} = I_{avg} \times R + E$

   Rearranging to find $I_{avg}$:

   $I_{avg} = \frac{V_{avg} - E}{R}$

   Substituting the calculated and given values:

   $I_{avg} = \frac{90.05 \ V - 80 \ V}{15 \ \Omega}$

   $I_{avg} = \frac{10.05 \ V}{15 \ \Omega} \approx 0.67 \ A$

The calculated average output current is approximately 0.67 A, which matches the closest option.