A single-phase bridge voltage source inverter (VSI) feeds a purely inductive load. The inverter output voltage is a square wave in 180° conduction mode. The fundamental frequency of the output voltage is 50 Hz. If the DC input voltage of the inverter is 100 V and the value of the load inductance is 20 mH, the peak-to-peak load current in amperes is _____ (rounded off to the nearest integer).

The correct answer is 50.

Let us analyze the operation of a single-phase bridge voltage source inverter (VSI) feeding a purely inductive load. The inverter operates in 180° conduction mode, which produces a square-wave output voltage.

For a single-phase bridge VSI, the output voltage waveform $v_0\left(t\right)$ across the load is a square wave with a peak value equal to the DC input voltage $V_{dc}$. Specifically:
$v_0\left(t\right)=V_{dc}$ for $0<t<T/2$
$v_0\left(t\right)=-V_{dc}$ for $T/2<t<T$
where $T$ is the time period of the output voltage waveform.

The fundamental frequency $f$ is given as 50 Hz. Therefore, the time period $T$ is:
$T=\frac{1}{f}=\frac{1}{50}=0.02\text{s}=20\text{ms}$

The load is purely inductive with inductance $L=20\text{mH}=20\times {10}^{-3}\text{H}$. The relationship between the voltage across the inductor and the current through it is given by:
$v_0\left(t\right)=L\frac{d{i}_0\left(t\right)}{dt}$

During the positive half-cycle, i.e., from $t=0$ to $t=T/2$, the voltage is constant at $v_0\left(t\right)=V_{dc}=100\text{V}$. During this period, the current increases linearly from its minimum value (peak negative current, $-I_{peak}$) to its maximum value (peak positive current, $I_{peak}$).

We can integrate the inductor equation over the positive half-cycle:
${\int }_{-I_{peak}}^{I_{peak}}d{i}_0={\int }_0^{T/2}\frac{V_{dc}}{L}dt$

This simplifies to:
$2I_{peak}=\frac{V_{dc}}{L}·\frac{T}{2}$

The term $2I_{peak}$ represents the peak-to-peak load current ($I_{p-p}$). Therefore:
$I_{p-p}=\frac{V_{dc}·T}{2L}$

Substituting the given values into the equation:
$V_{dc}=100\text{V}$
$T=0.02\text{s}$
$L=20\times {10}^{-3}\text{H}$

We get:
$I_{p-p}=\frac{100\times 0.02}{2\times 20\times {10}^{-3}}$
$I_{p-p}=\frac{2}{40\times {10}^{-3}}=\frac{2}{0.04}=50\text{A}$

Thus, the peak-to-peak load current is 50 A.