For the ideal AC-DC rectifier circuit shown in the figure below, the load current magnitude is Idc I_dc=15 A and is ripple free. The thyristors are fired with a delay angle of 45°. The amplitude of the fundamental component of the source current, in amperes, is ______. (round off to two decimal places)

The correct answer is 17.65.

By analyzing the given circuit diagram:

We can identify that the upper two devices of the bridge are thyristors (indicated by the control gate terminal), and the lower two devices are diodes. This circuit is a single-phase half-controlled bridge rectifier (also known as a semi-converter). The load contains a series resistor and inductor, carrying a ripple-free load current $I_{dc}=15\text{A}$.

For a single-phase semi-converter, the source current $i_s$ is a quasi-square wave with a pulse width of $\pi -\alpha$ per half-cycle, where $\alpha$ is the firing delay angle. The fundamental component amplitude (maximum/peak value) of this source current waveform is derived using Fourier series analysis as follows:
$I_{s1m}=\frac{4I_{dc}}{\pi }\cos \left(\frac{\alpha }{2}\right)$

We are given the following values:
Load current magnitude, $I_{dc}=15\text{A}$
Firing delay angle, $\alpha =45°$

Substituting these values into the fundamental current amplitude equation:
$I_{s1m}=\frac{4\times 15}{\pi }\cos \left(\frac{45°}{2}\right)$
$I_{s1m}=\frac{60}{\pi }\cos (22.5°)$
Using $\cos (22.5°)\approx 0.92388$ and $\pi \approx 3.14159$:
$I_{s1m}\approx \frac{60}{3.14159}\times 0.92388\approx 19.0986\times 0.92388\approx 17.645\text{A}$

Rounding off to two decimal places, the amplitude of the fundamental component of the source current is $17.65\text{A}$.