Problem $4$: $(15$ points$)$

You have been given the circuit in Fig. $4.$ Initial conditions are $0.$ Answer the following:

$($i$)$ The switch is closed at time $t=0.$ Write a differential equation for the current through the inductor. Note: The voltage source is not above the breakdown voltage of the diode, so no current flows through $D$ while the switch is closed.

$($ii$)$ Suppose the switch has been closed for a length of time $t_{on},$ what is the instantaneous current through the inductor?

$($iii$)$ When the switch opens, the inductor current must continue to flow. This forces the diode $D$ to become forward$-$biased, and the inductor releases its stored energy to the load $(R_{\text{l}\text{o}\text{o}\text{d}})$ and the output capacitor $(C).$ Write a differential equation that could be used to solve for the current $I_L(t),$ but do not solve it$.$

$($iv$)$ Suppose the switch has been opened for a length of time $t_{\text{o}\text{f}\text{f}},$ what is the change in the instantaneous current through the inductor in terms of the voltage across the diode $V_D$ and the output voltage $V_R?($Hint: you can use the substitution $dt=t_{\text{o}\text{f}\text{f}})$

$($v$)$ Given the equations above, can you write the output voltage $($i$.$e$.$ the voltage across the resistor$)$ as a function of $V,t_{on}$ and $t_{\text{o}\text{f}\text{f}}?$

Figure $4$: Problem $4$