You can do this a number of ways. I usually drive transistors in common-emitter mode when using them as switches. The emitter of the NPN transistor goes to the common (ground) rail, the collector to the cathode of the LED and then a current limiting resistor to B+. The transistor will turn on when the base voltage is approximately 0.7V above the emitter voltage. Use a resistor on the base of the transistor (to the I/O pin) since BJTs are current controlled devices.
When the I/O line goes high (say +5V), you will have 5V across the resistor and the B-E junction of the transistor. The transistor B-E junction will drop 0.7V with the remainder (4.3V) across the resistor. Say you use 1k for the base resistor. This will give you 4.3/1000 or 4.3mA of base drive current. This is TONS of current, and since the B-E junction is forward-biased it will amplify the base current by the gain of the transistor (usually between 50 and 200) and limit the C-E current to this level. (When using transistors as switches you usually don't care about the actual collector current, you just want to make sure the transistor is allowing much more current than you will actually draw.) The end result is that the transistor is fully on and your LED will light, limited by the series resistor you chose to limit the LED current to something like 10-20mA.
When the I/O line is low, the B-E junction has no appreciable voltage across it and the transistor is considered "off." I like to include a 4.7k-ish resistor between the base and emitter to make sure that the transistor doesn't accidentally switch on due to noise or a floating I/O pin.
This basic NPN switch circuit works well as long as your C-E voltage is under about 30V and your load is mainly resistive and relatively low current (under a few Amps). When you're trying to control higher voltages you have to start looking at either more specialized (high voltage) transistors or more complex drive circuits. When driving inductive loads (relays, motors, etc.) you need to protect the transistor from the inductive kickback that occurs when current stops flowing through an inductor. When driving high current loads you may have to again look at specialized transistors or more complex drivers to ensure the transistor remains fully switched-on.
If your load must be connected to common (instead of the transistor emitter) then you can use a PNP transistor. Emitter to B+, collector to the LED anode, LED cathode to a current limit resistor and the other end of the resistor to common. Now a logic '1' should turn the LED off and a logic '0' should turn it on, but you've got a problem. The problem is that your I/O line cannot turn the transistor off, because the highest voltage it can reach is 5V (in our example). This would maintain 1V across the B-E junction and the transistor would remain on, even only if partially on. In this case I like to "cheat". You can turn the I/O line into an input to turn the transistor off (remember I like to have a 4.7kish resistor between the base and emitter). This is not ideal because it slows down your turn-off (which may or may not be a problem) but also because you now have (in this example) 6V going to an I/O line. It may not be able to withstand this kind of voltage and you can damage the line or the internal protection circuitry on the input. What I do to mitigate the problem is to use an NPN transistor to turn on the PNP transistor. This doesn't solve the turn-off problem but for most general cases it's nothing to worry about.
Your alternator has a non-zero impedance, which is higher than the impedance of the partly-charged battery in this case. That's what causes the voltage to drop when the battery is connected.
Or, in other words: The alternator cannot deliver as much current as would be required to put the battery's terminals to 14V, so it delivers as much as it can and the voltage drops. - With higher rpm the picture (voltage) will likely change. Besides, note that an alternator -as the name suggests- produces (rectified) alternating current (AC) where peak voltage and RMS may differ, so be sure what you are measuring. (This is, by the way, one of the reasons why you should never run your car's engine without connected battery; the battery, like a giant capacitor, smoothes the system's voltage ripple and spikes away which would otherwise put significant stress on other electric components.)
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To corectly measure power on a three phase system (without neutral) you need two wattmeters. For a purely reactive load, one wattmeter will read positive X watts and the other will read negative X watts meaning the power consumed is X-X watts or 0 watts (as one would expect).
You have to remember that the line voltage and line/phase current on a purely reactive load will not be 90º apart because it is three phase. Neither will the phase angle be 0º for a purely resistive load. This is because phase voltage is not in-phase with line voltage.
On a resistive circuit the phase voltage and phase current will be exactly in phase BUT, phase voltage leads line voltage by 30º and in the OPs question the wattmeter is connected across two lines (i.e. it is measuring line voltage) hence the purely reactive load has a phase angle of 60º instead of 90º as you would get on a single phase circuit: -
If you added a 2nd wattmeter in the conventional way that two wattmeters are connected for three-wire, 3-phase power measurements you'd find that the 2nd wattmeter reads a negative power balancing completely the 1st wattmeter reading: -
Shown is a single wattmeter method that works if the load is balanced. The 2 wattmeter method works with balanced or unbalanced loads whether thy be delta or star connected loads.