A cell or a battery is essentially a charge "pump". Now, to help form an intuition for the answer to your question, fall back to the hydraulic analogy.

Two water pumps in parallel can produce twice the water flow of one (ideally).

Two water pumps in series can produce twice the pressure (or head) of one (ideally).
 I can see why it feels like black magic, since it seems like the electrons that leave the anode of one battery just go into the cathode of the next battery, and there seems to be no reason for the voltages to add up.

But let's look at it in a different perspective. Let's say you have a battery A with a voltage v0. Battery A is the only battery in a circuit with a resistor R also in the circuit. Resistor R's resistance is SO high, that battery A is hardly able to generate a tiny current through the circuit. v0 just isn't enough to produce a significant current through resistor R. 

But now, let's say you add battery B and battery C in series to battery A, with resistor R still in the circuit, too. Batteries B and C also have voltages of v0. 

In each battery, within a closed circuit, an electron really wants to leave the anode while another electron from the wire enters the cathode. When we just had battery A hooked up to resistor R, it's voltage v0 wasn't enough to make this happen, remember? But now, with all 3 batteries in series, not only does battery A's voltage want to naturally push an electron away from it's anode, but battery B's cathode also is exerting a pull on an electron from battery A's anode, too! And with just a voltage of v0, how is battery B able to give up an electron from its anode to so that it can receive one in its cathode from battery A? Because it also feels a pull from battery C's cathode, from battery C's v0! And last, you guessed it: battery C is willing to give up an electron from its anode to accept one in its cathode (with just the low, individual voltage v0) because it feels a pull from battery A's cathode as well! There's no first or last, it all happens at once, like a chain. Do you now see how the effects of each battery compound to produce a much higher voltage (aka a much higher motivation, or pull/push for electrons to flow) ?? Also, assume resistor R was between battery C and A. That part doesn't really matter.

Why do batteries in series add up voltage?

A device like the Power Integrations LinkSwitch-TN allows you to implement a simple buck converter that converts rectified mains voltage to a reasonably-well-regulated non-isolated low voltage with reasonably good efficiency.

If you need low ripple after the fact, you can put a linear regulator on the stepped-down output of the buck and minimize your resistive losses.

I've used these as primary-referenced control supplies and have had good success with them.

<img src="https://i.stack.imgur.com/XLOVg.png" alt="enter image description here">

Converting universal AC to 3.3v DC @ 0.1A

I'll just give a hint here - think about the voltage across R4, and how much current will flow through it. If you need to, remove R4 and work out the voltages at points 4 and 6.

If it wasn't quite as convenient (e.g. all resistor were not the same value), you can calculate the Vth and Rth (Thevenin) for R2, R5 and R3, R6 (i.e. points 4 and 6). Then work things out from there.
 Voltage Divider Basics and Analysis

I noticed on a later comment that you were really trying to understand how this could (potentially) be done using voltage divider theory. To start, let me remind you what a voltage divider really is:

Take this resistive voltage divider from Wikipedia:

<img src="https://i.stack.imgur.com/XcROL.png" alt="Resistive Voltage Divider">

The common equation to find the output is: V_out = R2 / (R1 + R2) * V_in

But what this really boils down to is this: V_out = R2 * I_R2

Of course, in this circuit, the current through R1 and R2 is equal, so finding the value is easily done by dividing the total voltage by the total current: I = V_in / (R1 + R2)

Because R2 is located between V_out and ground, the voltage across R2 is equal to V_out. However, what would happen if a third resistor were placed in parallel to R2? Consider this modified version of the above image:

<img src="https://i.stack.imgur.com/VBvqn.jpg" alt="Modified Resistive Voltage Divider">

The current relationship has now changed. In this circuit, I_R1 = I_R2 + I_R3. This makes the equation for the output a bit more complex: V_out = (R2 || R3) / (R1 + (R2 || R3)) * V_in 

Because R2 and R3 are in parallel: (R2 || R3) = (R2 * R3) / (R2 + R3)

But the voltage across R2 is still equal to V_out. Adding another series resistor changes this:

<img src="https://i.stack.imgur.com/7DHzw.jpg" alt="Modified Resistive Voltage Divider 2">

V_out = ((R2 || R3) + R4) / (R1 + (R2 || R3) + R4) * Vin

Now, the voltage across R2 is really equal to V_out - V_R4. Adding more resistors will only further skew the relationship between V_out and V_R2. Adding a resistor in series to R3 doesn't change things too much:

<img src="https://i.stack.imgur.com/8kCkw.jpg" alt="Modified Resistive Voltage Divider 3">

V_out = ((R2 || (R3 + R4)) + R5) / (R1 + (R2 || (R3 + R4)) + R5) * Vin

where R2 is in parallel with the series combination of R3 and R4... This is still a big resistive voltage divider with random parallel and series resistors. The voltage across R2 is equal to V_out - V_R5.

The real complication comes with the introduction of the bottom half of the Wheatstone Bridge formed by resistors 2 through 6. Your particular circuit is very simple because all of the resistors are of equal value (the bridge is balanced), but it is a bit harder to formulate equations for a generic circuit which might have a voltage across R4:

<img src="https://i.stack.imgur.com/BBRFT.jpg" alt="Modified Resistive Voltage Divider Circuit 4">

If you really want to keep using basic resistive relationships, the next best step is to simplify the circuit. One method is to do a Delta-Wye Transform, as you mentioned in the question. However, it makes a lot more sense to do it on the the bottom half or right side of the bridge where R2 is not involved. For example, converting the bottom half of the bridge yields:
<img src="https://i.stack.imgur.com/AwB4H.jpg" alt="Delta-Wye Transform">

The new resistor values are found as:

<ul>
<li>Ra = R4 * R5 / (R4 + R5 + R6)</li>
<li>Rb = R4 * R6 / (R4 + R5 + R6)</li>
<li>Rc = R5 * R6 / (R4 + R5 + R6)</li>
</ul>

But this still doesn't solve the problem because you need to know the voltage across R2 which is now equal to V_out - V_Na. This can still be solved using voltage dividers if you use a couple of different points as outputs:

<img src="https://i.stack.imgur.com/NjY4Q.jpg" alt="Modified Resistive Voltage Divider 5">

The final solution (the voltage across R2) can finally be found using a combination of various voltage dividers and the following equations:

<ul>
<li>Vx = [((R2 + Ra) || (R3 + Rb)) +Rc +R7] / [R1 + ((R2 + Ra) || (R3 + Rb)) +Rc +R7] * V_in</li>
<li>Vy = (Rc + R7) / [R1 + ((R2 + Ra) || (R3 + Rb)) + Rc + R7] * V_in</li>
<li>I_R2 = (Vx - Vy) / (R2 + Ra)</li>
<li>V_R2 = R2 * I_R2</li>
</ul>

Since you said you already found the answer using Mesh Analysis, I will go ahead and put the values you should get from the above equations:

<ul>
<li>Vx = 10V</li>
<li>V7 = 6 &amp; 2/3 V = 6.666...V</li>
<li>I_R2 = 0.025A</li>
<li>V_R2 = 2.5V</li>
</ul>

So it is possible to solve the circuit using nothing but voltage dividers, but we needed help from a resistor transformation. All of the basic circuit analysis techniques are based on Ohm's law: V = I*R, some of them just make more sense to use at times than others do.

Using Nodal Analysis

Considering all of your resistors are the same value, there are a few tricks to easily reduce the entire circuit down, but since that is pretty unrealistic in real life, I think it would be better to teach you a real analysis tool. There are many ways to analyse a circuit, but one of the easiest and best for many types of problems is Nodal Analysis. It is very easy to learn. 

1) According to Kirchhoff's current law (KCL), the current going into any one node (point in the circuit) must be equal to the current coming out of any one node.
So divide the circuit up into individual currents flowing from one node to another. The direction of the current flow is entirely up to you, it will just affect the polarity of the resulting current:

<img src="https://i.stack.imgur.com/KchBP.jpg" alt="KCL">

2) Form relationships between the currents:

<ul>
<li>I1 = I2 + I3 </li>
<li>I2 = I4 + I5 </li>
<li>I6 = I3 + I4 </li>
<li>I7 = I5 + I6 </li>
<li>I7 = I1</li>
</ul>

3) Express the currents as voltages/resistances

<ul>
<li>I1 = (15V - V3) / R1</li>
<li>I2 = (V3 - V4) / R2</li>
<li>I3 = (V3 - V6) / R3</li>
<li>I4 = (V4 - V6) / R4</li>
<li>I5 = (V4 - V5) / R5</li>
<li>I6 = (V6 - V5) / R6</li>
<li>I7 = (V5 - 0V) / R7</li>
</ul>

4) Using Algebra, combine and reduce the equations to find the values of the unknown voltages. To answer your specific question, the voltage across R2 would be V3 - V6 which is equal to R2 * I2.

When use Voltage divider in circuit analysis

Depending on the MOSFET package, mica washers are an option. 

A common through-hole package for MOSFETs is the TO-220, where the metal tab looks like this from the back:

<img src="https://i.stack.imgur.com/A7GwL.png" alt="TO-220"> (source: Wikipedia)

Edit: OP clarifies that their MOSFETs are TO-247 packages, for which the TO-220 insulators can be used - or TO-247 ones possibly, though those don't seem to be sold as a distinct product much.

For these, mica insulator kits are available at extremely low prices, containing the washer itself, a nut, bolt, spring washer, and an insulating sleeve to isolate the bolt from the metal tab:

<img src="https://i.stack.imgur.com/zs1wy.png" alt="Mica Insulator Kit"> (source)

Insulation is preferably done for each of the MOSFETs you are attaching to a heat-sink, to avoid undesired short circuits. 

You should also use a bit of heat-sink compound between the mica and the metal tab, as well as between the mica and the heat sink itself.
 The modern alternative to mica and grease is the Sil-Pad products from Bergquist and others. These are not very expensive and they're a lot more convenient to work with.
 For very good thermal performance and isolation, go with Bergquist Hi-Flow.

It's phase change material that goes liquid-ish above 55C, filling whatever voids are necessary, plus it has a Kapton insulator in the middle to give you good dielectric strength. It works best with clip-type connectors that hold constant pressure, otherwise you need to pre-bake the heatsink and torque down screws a second time once the material flows.

What material should I use to insulate my MOSFET from a heatsink?

Test this:
Open a terminal (press <kbd>Ctrl</kbd>+<kbd>Alt</kbd>+<kbd>T</kbd>).
Run this:
<pre><code>sudo -i

apt-get update
apt-get autoremove
apt-get clean
 
UNUSCONF=$(dpkg -l|grep "^rc" | awk '{print $2}')
apt-get remove --purge $UNUSCONF

NEWKERNEL=$(uname -r|sed 's/-*[a-z]//g'|sed 's/-386//g')

ADDKERNEL="linux-(image|headers|ubuntu-modules|restricted-modules)"

METAKERNEL="linux-(image|headers|restricted-modules)-(generic|i386|server|common|rt|xen)"

UNUSKERNELS=$(dpkg -l | awk '{print $2}' | grep -E $ADDKERNEL | grep -vE $METAKERNEL | grep -v $NEWKERNEL)

apt-get remove --purge $UNUSKERNELS

update-grub
</code></pre>
 I believe the "Remove obsolete packages" actually just runs a <code>sudo apt-get autoremove</code>. Try it, see if it helps.
 I find that this answer of an unrelated question might provide a utility and command that seems to remove more of the unused stuff:

<ol>
<li>Install the "deborphan" package.</li>
<li><code>sudo deborphan | xargs sudo apt-get -y remove --purge</code></li>
</ol>

How to remove obsolete packages after failed release upgrade via do-release-upgrade on Ubuntu?

As Scott Weldon said, there doesn't seem to be any way to do this other than disabling sharing of the friend whose library is in use, or using offline mode as Jaxir said to prevent the library from being in use in the first place. You can change the order in which your friends share libraries with you by having them unshare and reshare the library with you, in order to play your games with the library of whichever friend is least likely to be using it, but for a general solution to choose which friend's library to use, you'll need sfs-select. 

SFS-select is a python tool to enable and disable friends' shared libraries by modifying the local cache remembering which accounts have authorized you. It's a bit of a clunky workaround, but it lets you pick exactly which friend's library you would like to use for a shared game, without needing your friends to log in to your computer again, or changing the actual permissions registered on the Steam server at all. It includes both graphical and command-line versions. You can download it here or read about it on the Steam forum thread.
 I had the same problem. Unfortunately, this seems to be by design. I don't have a solution, but I do have a workaround: temporarily disable the sharing of user A's library with you, and you will be able to play user B's copy of the game.

This is rather inconvenient though, especially if this happens frequently (or if you have several shared libraries with the same game), so hopefully Valve gets this fixed.

Steam Family sharing - Overlapping shared libraries

I use Xee. It is so much better than preview for images (IMHO); especially since you can traverse images within the same folder by clicking the arrow buttons. All of the button short cuts are reprogrammable. 

It supports lossless cropping and rotation. Just see the webpage. 

If you like it, buy milksnake to help the developer out - I did. 
 This command worked for me:

<pre><code>jpegtran -crop WxH+X+Y infile.jpg outfile.jpg
</code></pre>

While perhaps not lossless I found FFmpeg to also do quite well:

<pre><code>ffmpeg -i infile.jpg -filter crop=W:H:X:Y -qmax 1 outfile.jpg
</code></pre>
 Based on this answer, I can suggest using the command-line tool <code>jpegtran</code>, the GUI-based Jpegcrop Windows program, or one of the many programs based on IJG jpegtran code (although not all of them support cropping).

OS X: How to perform a lossless crop of a JPEG image?

In Mac OS X Lion 1.07, Terminal now supports 256 colors, Background Color Erase (BCE) and some other features that make it more compatible with modern xterm, and the default value for $TERM is now xterm-256color instead of xterm-color (which represents an older branch of xterm with less functionality).

If you upgrade to Lion, any settings profiles that used xterm-color will use the new default. (All "Declare terminal as" does is set the value of $TERM, by the way.)

The error you're seeing indicates that the appropriate terminfo files for xterm-256color are either missing or inaccessible. You might try installing Lion again. Also, some people have reported that installing Xcode 4 on Lion can cause this problem.

You can test the validity of xterm-256color with the following shell command:

<pre><code>infocmp xterm-256color
</code></pre>

This will either display an error or the appropriate terminfo values for xterm-256color, and, if found, it displays the pathname to the terminfo file it used.
 xterm-256color should be present in <code>/usr/share/terminfo/78</code>.

Run this command and double check if it's there:

<pre><code>ls -laR /usr/share/terminfo
</code></pre>

In the past, there were issues with Xcode messing up permissions to a huge degree. Worst case scenario, use disk utility and repair your permissions. You will need to boot from an install DVD and verify/repair your boot partition.
 Looks like your <code>TERM</code> value isn't recognized. You are probably setting the TERM value in your <code>.profile</code> (or similar) or you've got missing/corrupt files or invalid permissions in <code>/usr/share/terminfo/</code>.

I can't clear the terminal in Mac OS X Lion - "terminals database is inaccessible"

You can use another variable, though this is not particularly elegant.

<pre><code>wire[31:0] y;

assign y = func(A) ^ func(B);
assign x = y[15:0];
</code></pre>

A better approach would be to use a function. 

<pre><code>function [15:0] trunc_32_to_16(input [31:0] val32);
 trunc_32_to_16 = val32[15:0];
endfunction

assign x = trunc_32_to_16(func(A) ^ func(B));
</code></pre>
 In your example, you are implicitly truncating bits.

Making the truncation explicit can often remove the warnings in simulation/lint/synthesis.

One way to do this in-line is to use a cast operator, e.g.:

<pre><code>typedef logic [15:0] HALF_WORD;
assign x = HALF_WORD'((func(A) ^ func(B));
</code></pre>

This approach may make sense if it is obvious from the context that all the bits being dropped are 0's.

If some of the bits may be nonzero then I would suggest still using an intermediate net like @dwikle suggested in a previous answer, since it makes it more clear that you are actually throwing away bits. Here it is again for reference.

<pre><code>wire[31:0] y;

assign y = func(A) ^ func(B);
assign x = y[15:0];
</code></pre>
 I think that this might help keep the line count down.

<pre><code>wire [15:0] not_used ;

assign {not_used, x} = (func(A) ^ func(B));
</code></pre>

Not sure if that is valid with assigns though.

How to truncate an expression bit width in Verilog?

Wow! What country &amp; area are you in? 
It would be interesting to know. 

<blockquote>
 Is this probably even a three phase system? 
 Are they 120 degrees apart? 
 Could each wire be just one of three independent single phase systems since the power company has no control over people unbalancing the loads?
</blockquote>

You can easily [tm] find out. 
See section "Method" below.

3 phase power when 1 or 2 phases is missing is not very good for 3 phase equipment so that does not sound like a very useful choice. 

You didn't ask, but you could easily make an "auto-swapper" using relays that would deliver power to the output from one of whatever phases were available. Don't tell your neighbours - if everyone had these they would probably dance together as the power company struggled to balance the system :-).

An auto-swapper could be built with one relay with 3 changeover contacts plus one relay with one changeover contact.

Vout = A + B/A + C/A/B

If A is alive, use it via an A operated contact. 
If A is dead the A operated contact blocks A. 
If A is dead use B if B is alive. (B contact and /A contact) 
If A is dead and B is dead use C. (C contact + /B contact.)

If /A and /B and /C, go for a walk.

C could also have a C operated contact if you don't want C connected when dead.

Some equipment may not like power being disconnected and connected again at relay-operate speeds. 

<hr>

METHOD:

<h2>WARNING</h2>

A meter used for the following test MUST be rated for 400V mains if you are on a 230 VAC or similar system. NOT JUST "has a 400V capable range" .

Usually people will say "CAT 3" or "CAT 4" but even that is misleading. 
This paper The myths of instruments and safety addresses up a few comminly held and potentially dangerous ideas. 

For now, understand that the measurement I am about to suggest involes 400 VAC in a 230 VAC system and 200 VAC in a 110 VAC system. Your meter and leads and anything electrically connected must withstand mains PLUS any spikes that may be present. Given the situation you describe, spikes and worse could be quite likely. Be aware that arc over in an under-rated multimeter can kill you and people have died in such circumstances, even measuring 230 VAC!. 

You could also use a resistive divider. They too can have issues.

If proceeding ... :-)

Name you phases A B C. 
Measure phase to phase to phase in all combinations. AB BC CA.

If two readings are about 1.7 x Vmains (200V or 400V for 110VAC/230VAC systems) you have two phases of a 3 phase system.

If 3 readings are ~= 1.7 x Vmains you have 3 phases. 

If readings are ~= 0V or ~= 2 x Vmains (220 VAC or 460 VAC you have independent single phases from separate transformers. 
If they are in phase you get about 0V. This would be usual. 
If you get about 2 x maians they are out of phase. This would be unusual due to construction methods.

I said "independant separate phases" meaning that they may have two transformers with 3 phases R G Y phases. If you have R from transformer 1 on your phase A, and R from transformer 2 on your phase B then you will probably get 0V between them.

They COULD even feed you R G Y phases from 3 seperate transformers. You could use that as 3 phase power but the current flows would be "interesting".
 I expect you don't really have a 3 phase system. If each line were a phase from a 3 phase transformer somewhere, the power company would have a difficult time turning off individual phases. However, what you describe doesn't make much sense either, so I can't really guess.

You can find out whether the phases are in sync, 180° out of phase, or three phase. Measure the voltage with a portable AC voltmeter of each of the three combinations of hot leads. If any two are 180° out of phase, you will be twice the hot-to-neutral voltage. If they are in phase you will get zero. If they are 120° apart, as they would be in a 3 phase system, you will get 1.7 times the hot-to-neutral voltage.

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