Electrical – ambiguous component values (ex. U01 U47 M47 etc.)

First off, I'm going to presume that the TV is out of warranty. If it is still covered, your best bet is to call the warranty line and get it fixed that way.

It's hard to tell from your popped resistor photo, but is there a heat sink to the left of it in the image that is black? There might be something under there that blew out too.

As for DIYing the repair, you could try swapping out that through hole resistor that blew. However, while it's common for low quality capacitor to give out, resistors don't usually pop unless there is an over current situation, so there might be a larger issue at hand here. It also looks like there are 0 ohm and a 47K (? hard to read) ohm surface mount resistors in the damaged area on the back of the PCB that might need replacing.

If it's currently broken, there's no harm in attempting a repair. Find a friend who does electronics to help you source the parts, especially if you think you'll need to replace those surface mount resistors.

As for getting a replacement board, rather than having it replicated from scratch, you should be able to get a replacement board. The various boards in TVs are usually modular (power, tuner, etc), and that appears to be the power supply board. Try searching for "TVMAKE TVMODEL Replacement Power Supply Board" and see what comes up. I know the Vizio 50" plasmas have a nasty habit of blowing power supplies and thus they're pretty easy to find replacements for online.

Good luck and stay away from your boyfriend's books lest they burst into flames. ;)

The LED forward voltage drop will remain (roughly) the same, but the current can change, so the calculation becomes (same equation solving for I):

$$I_{LED} = {(V_s - V_f)\over{R}}$$

So for a 3V \${V_f}\$ and a 5V supply, the \$100\Omega\$ resistor would give \${(5V - 3V)\over{100 \Omega }} = 20 mA\$.

So if you know what current you want, just plug the values in, e.g. for 10mA:

$$R = {(5V - 3V)\over{0.01 A}} = {200 \Omega}$$

Basically, the fact that the supply and the LED forward voltage can be relied upon to be pretty static, means that whatever value resistor you put in will also have a static voltage across it (e.g ~2V in this case), so it just leaves you to find out that voltage and select a resistance value according to the current you want.

Below is the V-I curve of a diode (from the wiki LED page), notice the current sharply rises (exponentially) but voltage stays roughly the same when the "on" voltage is reached.

For more accurate control of the current you would use a constant current, which is what most LED driver ICs provide.