Consider the single-phase voltage source inverter circuit feeding an inductive load (L). The inductor current (i_L) in the steady state is triangular with a peak value of 5 A as shown. The rms value of the current through the switch S₁ is:

The correct answer is 2.04 A.

Step-by-Step Explanation:

1. Understanding the Waveform and Circuit Operation
From the given waveform of the inductor current $i_L\left(t\right)$, we can observe the following:

• The peak current value is $I_p=5\text{A}$.
• The total time period of the load current is $T=20\mu \text{s}$.
• During the first half-cycle ($0<t<10\mu \text{s}$), a positive voltage is applied across the inductive load, causing the current to increase linearly from $-5\text{A}$ to $5\text{A}$.

2. Conduction Interval of Switch S₁
During the positive voltage half-cycle ($0<t<10\mu \text{s}$):

• For $0<t<5\mu \text{s}$, the load current $i_L$ is negative. In this interval, the current flows back to the source through the feedback/anti-parallel diodes $D_1$ and $D_2$.
• For $5<t<10\mu \text{s}$, the load current $i_L$ is positive. This current flows through the active switches $S_1$ and $S_2$.

3. Mathematical Derivation of the RMS Current through S₁
Let us define a local time variable $\mathrm{t\text{'}}=t-5$ for the conduction interval of $S_1$, where $\mathrm{t\text{'}}$ ranges from $0$ to $5\mu \text{s}$. During this period, the current through $S_1$ increases linearly from $0$ to $5\text{A}$:
$$i_{S_1}\left(\mathrm{t\text{'}}\right)=\frac{5}{5}\mathrm{t\text{'}}=\mathrm{t\text{'}}$$
Calculating the RMS value of the current over the active half-cycle period ($T_h=10\mu \text{s}$):
$$I_{S_1,\text{rms}}=\sqrt{\frac{1}{T_h}{\int }_0^5{\left(\mathrm{t\text{'}}\right)}^{2}d\mathrm{t\text{'}}}$$
Substitute the values and evaluate the integral:
$$I_{S_1,\text{rms}}=\sqrt{\frac{1}{10}{\left[\frac{{\left(\mathrm{t\text{'}}\right)}^3}{3}\right]}_0^5}=\sqrt{\frac{1}{10}·\frac{125}{3}}=\sqrt{\frac{125}{30}}=\sqrt{\frac{25}{6}}$$
Evaluating the square root:
$$I_{S_1,\text{rms}}=\frac{5}{\sqrt{6}}\approx 2.0412\text{A}$$
Thus, the RMS current through the switch $S_1$ is 2.04 A.