Full-bridge inverter fed from 30 V DC battery If the inverter produces a 50% duty rectangular AC output of amplitude ±Vdc across a load, what is the output voltage RMS value?

Introduction / Context:
A single-phase full-bridge inverter connected to a DC source ideally produces a square-wave AC output that toggles between +Vdc and −Vdc. For many introductory problems, the interest is in the RMS magnitude of this square wave.

Given Data / Assumptions:

• DC input Vdc = 30 V.
• Ideal switches, 50% duty symmetric square wave.
• Load is such that the inverter maintains ±Vdc across it (e.g., stiff DC link, appropriate gating).

Concept / Approach:
The RMS of a bipolar square wave that alternates between +A and −A with equal durations is simply A. Unlike sine waves, there is no √2 relation here; that relation applies to sinusoidal waveforms (V_rms = V_peak/√2). For square waves of amplitude Vdc, V_rms = Vdc.

Step-by-Step Solution:
Define waveform: v(t) = +Vdc for half cycle, −Vdc for half cycle.Compute RMS: V_rms = √[(1/T) ∫_0^T v^2(t) dt] = √[(1/T)(Vdc^2 * T)] = Vdc.Insert Vdc = 30 V → V_rms = 30 V.

Verification / Alternative check:
Fourier series shows a fundamental of 4Vdc/π, but RMS of the full square wave remains Vdc because higher harmonics restore total mean-square value.

Why Other Options Are Wrong:

• 15 V and 21.2 V apply to different duty or sinusoidal assumptions.
• 152 V and 302 V are nonsensical for a 30 V DC link.

Common Pitfalls:

• Using sinusoidal conversion V_rms = V_peak/√2 incorrectly for a square wave.

Final Answer:
30 V