Solution

The single-phase full bridge VSI provides a square wave output voltage ($+V_{dc}$ or $-V_{dc}$) to a purely inductive load $L$. The frequency is $f$. We have $V_{dc} = 100$ V, $L = 20 \times 10^{-3}$ H, and $f = 50$ Hz.

Peak Load Current Calculation

During each half-cycle, the voltage across the inductor is constant ($V_{dc}$ or $-V_{dc}$), causing the current to change linearly. The peak current ($I_{peak}$) is reached at the midpoint of the half-cycle ($t = T/2 = 1/(2f)$). The magnitude of this peak current is calculated as:

$ I_{peak} = \frac{V_{dc}}{L} \times \frac{T}{2} = \frac{V_{dc}}{2Lf} $

Substituting the given values:

$ I_{peak} = \frac{100 \text{ V}}{2 \times (20 \times 10^{-3} \text{ H}) \times 50 \text{ Hz}} $

$ I_{peak} = \frac{100}{1} = 50 \text{ A} $

Interpreting the Result

The calculated peak load current is $50$ A. The peak-to-peak current is $I_{pp} = 2 \times I_{peak} = 100$ A. Given the answer hint suggests the value is $50$ A, it is interpreted that the question requires the peak current value.

• Peak Load Current: $50$ A

Rounded to the nearest integer, the result is 50.