Discussion
The mean output voltage of a single-phase fully controlled bridge with firing angle \( \alpha \) is
Answer: B
Correct answer: \( (2V_m/\pi)\cos\alpha \)
Integrating the supply waveform over the conduction interval of a fully controlled bridge gives
\[ V_{dc} = \frac{2V_m}{\pi}\cos\alpha \]The cosine is what makes the bridge so useful: at \( \alpha = 0 \) the output is the full \( 2V_m/\pi \), at 90 degrees it is zero, and beyond 90 degrees it is negative, so the same bridge inverts and returns power to the supply. The form with \( (1 + \cos\alpha) \) belongs to a half-controlled bridge, which cannot go negative.
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