TiernanW said:
However in one video I seen for Kirchoff's Laws (Sorry can't link to it forgot what its called), the guy took the bottom 4 sided network (in this case with the 2 bulbs) and the current went round in a square all in the same direction, so that would mean that it flows to the right through the first bulb. That confused me a little bit. What is the correct way?
It's impossible to speculate what the man in that unknown video was doing but perhaps this will help...
In general when you apply Kirchoff's Laws to solve a problem the first thing you do is mark up the drawing with arrows showing the direction of +ve current. It's not always obvious which direction current is flowing but that doesn't matter.
You can mark up the circuit with current arrows in an arbitrary way just as long as long as you are consistent when you come to mark up the voltage drops. The next step is usually to write and solve a bunch of simultaneous equations to get values for all the voltage drops and currents. If one of the currents turns out to be -ve that means your earlier assumption about the direction of the current was wrong (wrong isn't quite the right word but it will do).
For example consider this circuit with two batteries and a resistor.
The left hand battery is 12V and the right hand 9V so it would be reasonable to assume that current would flow from left to right through the resistor (eg clockwise).
However for the moment I will pretended that it's not obvious which direction current flows and will define +ve current as anti-clockwise as shown below. To be consistent with my assumption that current flows anti clockwise I also have to define the direction of the voltage drop across the resistor V
R as shown on the drawing. That also "looks wrong" but it must be consistent with the direction I assumed the current was flowing.
Now we apply KVL...
I'll start at point A and go around clockwise (again it's an arbitrary choice).
+12V +V
R + (-9V) = 0
Solve to give
V
R = -3V
Oh look, the voltage is negative which means the voltage across the resistor is actually the other way to my original assumption.
Then applying Ohms law
I = V
R/R = -3/3 = -1A
Oh look the current is negative which means the current is actually flowing in the opposite direction to my original assumption.
The point is that it doesn't matter which way you assume current is flowing as long as you are consistent. In the end it all falls out in the wash.