Half-wave rectifier with resistor and inductor load

Well, first of all the answer of @SpehroPefhany is good. But I will show some math involving this circuit. I will assume an ideal diode.

The current provided by the source can be found using:

$$text{I}_text{in}left(tright)=maxleft(0,frac{text{V}_text{m}left(omegatext{L}left(expleft(-frac{text{R}t}{text{L}}right)-cosleft(omega tright)right)+text{R}sinleft(omega tright)right)}{text{R}^2+left(omegatext{L}right)^2}right)tag1$$

If you wonder where that formula is coming from, it is the same as:

$$text{I}_text{in}left(tright)=maxleft(0,mathcal{L}_text{s}^{-1}left[frac{text{v}_text{in}left(text{s}right)}{text{R}+text{sL}}right]_{left(tright)}right)tag2$$

Where $mathcal{L}_text{s}^{-1}left[cdotright]_{left(tright)}$ is the inverse Laplace transform.

The voltage across the components can be found using:

• Resistor (Ohm's law): $$text{V}_text{R}left(tright)=text{I}_text{R}left(tright)cdottext{R}tag3$$
• Inductor: $$text{V}_text{L}left(tright)=text{I}_text{L}'left(tright)cdottext{L}tag4$$

---

If you're a user of Mathematica I provided some code that you can use to make a picture of all the important values.

Clear["Global`*"];
Vm =;
[Omega] =;
R =;
L =;
Vin = Vm*Sin[[Omega]*t];
Iin = Max[0, 
   InverseLaplaceTransform[((LaplaceTransform[Vin, t, s])/(R + s*L)), 
    s, t]];
VR = Iin*R;
VL = D[Iin, t]*L;
Plot[{Vin, Iin, VR, VL}, {t, 0, 4*((2 Pi)/[Omega])}, 
 AxesLabel -> {HoldForm[Time[s]], HoldForm[Voltage[V] & Current[A]]}, 
 PlotLabel -> None, LabelStyle -> {GrayLevel[0]}, ImageSize -> Large]

When for example $text{V}_text{m}=omega=text{R}=text{L}=1$, we get:

The diode conducts whenever the anode voltage is greater than the cathode voltage (ideally).

In this case the anode voltage is zero, but the cathode voltage is less than zero so the diode continues to conduct.

Unfortunately, this image does not explicitly show the reference ground, which is the (-) terminal of Vs.