When \(S_1\) is closed for time \(t_1\), a current flows through the load.

Upon opening \(S_1\), a path must be provided for the current in the inductive load to prevent high voltage induction.

To mitigate the induction effects, a diode \(D_m\), commonly known as a freewheeling diode, is typically connected.

Diode \(D_m\) facilitates a path for the inductive load current to avoid voltage spikes.

Diode \(D_1\) is in series with the switch to prevent negative current flow in the presence of AC input.

In DC supply scenarios, \(D_1\) is unnecessary.

\(S_1\) in conjunction with \(D_1\), mimics the behavior of an electronic switch.

At \(t = 0^{+}\) , the switch is closed, and the current is zero.

In the absence of \(L\), the current would rise instantaneously.

Due to \(L\), current rises exponentially with an initial slope of \(V_s/L\).

The circuit operation is divided into two modes:

Mode 1: Begins when the switch is closed at \(t = 0\).

Mode 2: Begins when the switch is opened.

\(i_1\) and \(i_2\) are the instantaneous currents for Mode 1 and Mode 2.

\(t_1\) and \(t_2\) represent the durations of these modes.

**Mode-1:**At the end of this mode: \[I_1=i_1(t=t_1)=\frac{V_s}{R}\left(1-e^{-tR/L}\right)\]

If \(t_1\) is sufficiently long, \(i_1\) reaches a steady-state value.

The steady-state current, \(I_s = V_s/R\), flows through the load.

**Mode-2:**Switch is opened, allowing the load current to flow through \(D_m\).

The current through the \(D_m\) can be determined: \[\begin{aligned} 0 & =L\frac{di_2}{dt}+Ri_2\\ i_2(t) & =I_1e^{-tR/L} \qquad \because i_2(t=0)=I_1 \end{aligned}\]

At \(t = t_2\), this current starts to decay exponentially towards zero.

This decay occurs if \(t_2 \gg \frac{L}{R}\).

Given \(V_s=220~\mathrm{V}\), \(R=0\), and \(L=220~\mu H\).

Draw the load current waveform if the switch is closed for a time \(t_1 = 100~\mu s\) and is then opened

Determine the final energy stored in the load inductor.

\[\begin{aligned} i\left(t\right) & =\frac{V_{s}}{L}t\\ \mathrm{at~}t=t_1, \quad I_0 &=V_st_1/L\\ &=200\times100/220=100~\mathrm{A}\\ W & = 0.5LI_0^2 = 1.1~\mathrm{J} \end{aligned}\]