Question:
A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in \( 180^\circ \) conduction mode. The fundamental frequency of the output voltage is \( 50 \, \text{Hz} \). If the DC input voltage of the inverter is \( 100 \, \text{V} \) and the value of the load inductance is \( 20 \, \text{mH} \), the peak-to-peak load current in amperes is (rounded off to the nearest integer).
A single-phase full bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in \( 180^\circ \) conduction mode. The fundamental frequency of the output voltage is \( 50 \, \text{Hz} \). If the DC input voltage of the inverter is \( 100 \, \text{V} \) and the value of the load inductance is \( 20 \, \text{mH} \), the peak-to-peak load current in amperes is (rounded off to the nearest integer).
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For VSI circuits with inductive loads, the peak-to-peak current depends on the fundamental voltage and load reactance.
Updated On: Oct 11, 2025
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Solution and Explanation
Step 1: Calculate the fundamental voltage.
The fundamental voltage amplitude is:
\[
V_1 = \frac{4 V_{DC}}{\pi}.
\]
Step 2: Calculate the load current.
For a purely inductive load:
\[
I_{pp} = \frac{V_1}{\omega L},
\]
where \( \omega = 2 \pi f \). Substituting values:
\[
I_{pp} = 50 \, \text{A}.
\]
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