Three-phase bridge inverter with resistive star load A 3-phase bridge inverter is supplied by a 400 V DC battery. The load is star-connected with R = 10 Ω per phase (neutral isolated). Assuming square-wave operation, what is the peak value of phase current?

Introduction / Context:Inverters convert DC to AC. For a three-phase bridge with a star-connected resistive load, the line-to-line voltage steps between ±Vdc, and each phase sees a three-level line-neutral waveform whose maximum magnitude is tied to the DC link voltage. Estimating the peak current helps with device and protection sizing.

Given Data / Assumptions:

• Vdc = 400 V, square-wave inverter, ideal switches.
• Star-connected resistive load, Rph = 10 Ω, isolated neutral.
• Neglect device drops and switching transients for simplicity.

Concept / Approach:

In a standard three-phase bridge, each leg outputs ±Vdc/2 relative to the DC midpoint. The resulting line-neutral phase voltage is a stepped three-level waveform with peak magnitude approximately Vdc/2. For a pure resistor, instantaneous current follows i = v/R, so the peak current equals Vph,peak / Rph.

Step-by-Step Solution:

Verification / Alternative check:

Fourier fundamentals yield an RMS current of Ifund,rms = V1,rms / R, but peak instantaneous current is governed by the waveform peak, which is Vdc/2 for the phase voltage in a six-step inverter.

Why Other Options Are Wrong:

40 A assumes Vdc across a phase, which is not correct for a star load with isolated neutral. 10 A and 5 A underestimate the peak given Vdc = 400 V and R = 10 Ω.

Common Pitfalls:

Confusing line-line and line-neutral voltages; forgetting that maximum phase voltage magnitude is Vdc/2 in the six-step sequence.

Final Answer:

20 A