The solution to this problem requires the use of AC circuit analysis techniques to determine the voltage and current waveforms in the circuit. We can start by calculating the impedance of the load:

Z = R + jωL
where R is the resistance, L is the inductance, ω is the angular frequency (2πf) and j is the imaginary unit. Plugging in the values, we get:

Z = 5 + j(2π×50×16×10^-3) = 5 + j5.03

The voltage across the load is given by the product of the supply voltage and the firing angle of the SCR, which determines the portion of the half-cycle during which the SCR conducts:

Vload = Vsupply × cos(α)

where α is the firing angle. At full conduction (α = 0), the voltage across the load is equal to the supply voltage. At zero conduction (α = 90°), the voltage across the load is zero. For intermediate values of α, we can use trigonometric functions to calculate the voltage. For example, at α = 30°, we have:

Vload = 230 × cos(30°) = 199.9 V

The current through the load is given by Ohm's law:

Iload = Vload / Z

Using the value of Vload from above, we get:

Iload = 199.9 / (5 + j5.03) = 28.5 − j28.5 A

The real part of this complex number represents the average DC current flowing through the load, while the imaginary part represents the AC component due to the inductance. The total power delivered to the load is given by the product of the voltage and current:

Pload = Vload × Iload

Using the values from above, we get:

Pload = 199.9 × 28.5 = 5704.15 W

Note that this is the apparent power, which includes both the real power (due to the resistance) and the reactive power (due to the inductance). To calculate the real power, we need to take the real part of the complex power:

Preal = Re(Pload) = Re(Vload × Iload) = Re(28.5 − j28.5 × 199.9) = 4025.9 W

This is the power that is actually dissipated in the load, and represents the average power over one half-cycle of the AC waveform. The reactive power is equal to the imaginary part of the complex power:

Q = Im(Pload) = Im(Vload × Iload) = Im(28.5 − j28.5 × 199.9) = 4025.9 var

This is the power that is stored and released by the inductance, and represents the energy that is transferred back and forth between the load and the source during each half-cycle.

In summary, the load consisting of a 5 ohm resistance and a 16 mH inductance is fed by a single-phase SCR-based AC regulator with a supply voltage of 230 V. The voltage across the load varies with the firing angle of the SCR, while the current through the load has both DC and AC components due to the inductance. The power delivered to the load is a combination of real and reactive power, with the