1. Alex Kearns Richard Kerr Rocio Rodriguez

2.  An electric circuit is a collection of electrical devices connected by conductors, usually wires, to a power source, which supplies a flow of electrons.  Electricity is defined as this flow of electrons originating from and returning to the power source.

3.  Voltage: potential difference in charge between two points in an electrical circuit  represented by the uppercase letter V  Unit of voltage is the Volt (joule/coulomb)  Potential difference is directly proportional to the force that pushes electrons/current from one point to the other  Electrical potential difference can be thought of as the ability to move electrical charge through a resistance  Resistance: The opposition that a substance offers to the flow of electric current.  represented by the uppercase letter R  unit of resistance is the ohm (volt*meter)  Current: A flow of electrical charge carriers, usually electrons.  represented by the uppercase letter I  unit of current is the ampere (coulomb/second)  one ampere of current represents one coulomb of electrical charge (6.24 x 1018 charge carriers) moving past a specific point in one second.

4. Any voltage V, current I or resistance R in an electrical circuit can be determined without actually measuring it if the two others values are known.  This law can be used to determine the amount of current I flowing in the circuit when voltage V is applied to resistance R. Ohm's law is:

5.  Current = Voltage / Resistance.  In the following circuit, assume that resistance R is 2 and voltage V that is applied to it is 12 V. Then, current I flowing in the circuit can be determined as follows:

6.  Series Circuit:

7.  An open in the circuit will disable the entire circuit.  The voltage divides (shared) between the loads.  The current flow is the same throughout the circuit.  The resistance of each electrical device can be different.

9.  The total resistance RO in this circuit is equal to the sum of individual resistance R1 and R2.  In other words: The total resistance(RO) is equal to the sum of all resistances (R1 + R2 + R3 +.......)  Therefore, the strength of current (I) flowing in the circuit can be found as follows:

11.  Parallel Circuit:

12.  A Parallel Circuit has multiple paths or branches to ground. Therefore:  In the event of an open in the circuit in one of the branches, current will continue to flow through the remaining.  Each branch receives source voltage.  Current flow through each branch can be different.  The resistance of each branch can be different.

14.  Resistance R0 (a combination of resistances R1 and R2) in a parallel connection can be determined as follows:  From the above, the total current I flowing in this circuit can be determined from Ohm's law as follows:  The total current I is also equal to the sum of currents I 1 and I 2 flowing through individual resistances R1 and R2

17.  Kirchhoff’s Node Law:  the sum of all the currents entering a node is equal to the sum of all the currents leaving the node.

18.  Kirchhoff’s Voltage Law:  The directed algebraic sum of the voltages and voltage drops around a loop must be zero.  In other words, the sum of the voltage drops will always equal the voltage provided by the power source.

24.  Assign currents to each part of the circuit between the node points. We have two node points Which will give us three different currents. Lets assume that the currents are in clockwise direction.

25.  So the current on the segment EFAB is I 1, on the segment BCDE is I 3 and on the segment EB is I 2  Using the Kirchhoff's current Law for the node B yields the equation  I 1 + I 2 = I 3. For the node E we will get the same equation.  Then we use Kirchhoff's voltage law  -4 I 1 + (-30) -5 I 1 - 10I 1 +60 +10I 2 =0  When through the battery from (-) to (+), on the segment EF, potential difference is -30, and on segment FA moving through the resistor of 5W will result in the potential difference of -5 I 1 and in a similar way we can find the potential differences on the other segment of the loop EFAB.  In the loop BCDE, Kirchhoff's voltage law will yield the following equation:  -30 I 3 + 120-10I 2 +60 =0

26.  Now we have three equations with three unknowns:  I 1 + I 2 - I 3 =0 -19 I 1 +10 I 2 = - 30 -10 I 2 -30 I 3 = -180  The system above has the following solution:  I 1 =2.8302 I 2 =2.3774 I 3 =5.2075  This linear system could have been solved using linear algebra with much less complicated math. Linear Algebra is more useful when the network is very complicated and the number of the unknowns is large.

27.  Write down Kirchhoff’s law for each loop. The result, after simplification, is a system of n linear equations in the n unknown loop currents in this form: where R 11, R 12,..., R mm and V 1, V 2,..., V m are constants.  Solve the system of equations for the m loop currents I 1, I 2,..., I m using Gaussian Elimination.

29. The number of loop currents required is 3. Let’s choose the loop currents shown.

30.  Write down Kirchhoff's Voltage Law for each loop. The result is the following system of equations:

31.  Put these equations in a matrix and solve using Gaussian Elimination:

32.  Solving the system of equations using Gaussian elimination gives: I 1 = - 4.57, I 2 = 13.7 and I 3 = - 1.05