Question

The RMS output voltage at fundamental frequency of a single-phase, full-bridge inverter with input voltage of 48V DC, feeding a load of 2.4 \( \Omega \) is:

• A. \( \frac{4 \times 48}{\sqrt{2} \pi} \) V
• B. \( \frac{48}{2\sqrt{2} \pi} \) V
• C. \( \frac{\sqrt{2} \times 48}{\pi} \) V
• D. \( \frac{4 \times 48}{\pi} \) V

Hint:

The fundamental RMS output voltage of a single-phase full-bridge inverter is given by \( V_{1,\text{rms}} = \frac{4V_{\text{dc}}}{\sqrt{2} \pi} \), derived from the Fourier series of a square wave.

Answer 1

The Correct Option is A

Step 1: The output voltage waveform of a single-phase full-bridge inverter is a square wave. 
Step 2: The fundamental RMS output voltage (\( V_{1,\text{rms}} \)) is given by: \[ V_{1,\text{rms}} = \frac{4V_{\text{dc}}}{\sqrt{2} \pi} \] where:
- \( V_{\text{dc}} = 48V \) (DC input voltage)
- \( \pi \) appears due to the Fourier series expansion of a square wave. 
Step 3: Substituting the given values: \[ V_{1,\text{rms}} = \frac{4 \times 48}{\sqrt{2} \pi} \text{ V} \] 
Step 4: Thus, the correct answer is \( \frac{4 \times 48}{\sqrt{2} \pi} \) V.