Hello there,
There are a lot of questions in those files, so we'll have to take them one at a time. We'll start with scan1.pdf Question number 1.
You basically have two laws to follow:
1. The sum of currents entering and leaving a node is zero.
2. The sum of voltage drops around a closed path is zero.
To start with question 1, you would first assign nodes to each junction,
then write simultaneous equations relating the currents flowing into and out of the
nodes, then solve those equations for the unknowns.
If we call the junction of 6v and 2 ohm resistor node v1, the junction of the three resistors
node v2, and the junction of the 9v and 4 ohm resistor node v3, we can start writing equations.
We will assume current flowing left to right in the 2 ohm resistor, and top to bottom
in the 4 ohm and 3 ohm resistors. That means that the current flowing through the 2
ohm resistor and into node v2 has to equal the current flowing through the other two
resistors out of node v2 using Law #1 above.
Starting with node v1:
(v1-v2)/2=(v2-0)/3+(v2-v3)/4
Note that all we did here was take the difference between the voltages at three pairs of nodes and divide by the resistance between them. Because each pair divided by a resistance is equal to a current, we could rewrite that equation as I1=I2+I3 if we wanted to show the currents entering and leaving node v2.
Now since we already know the voltage for v1 and v3 we can fill those in:
(6-v2)/2=(v2-0)/3+(v2-9)/4
and get rid of the zero:
(6-v2)/2=(v2)/3+(v2-9)/4
and now we have only one unknown so we only have to solve for v2:
v2=63/13
To calculate the current through the 3 ohm resistor now we only have to divide v2/3 to get the current.
We can double check this result by using the result for v2 and calculate the current in the other two resistors
and make sure that Law #1 is obeyed. If not, we made a mistake.
This type of analysis is also known as Nodal Analysis.