Kirchoff's Current Law (KCL) states:

The Sum of the currents entering a node is always zero.
Or we can say the sum of the currents entering a node = sum of currents leaving the same node.

This circuit has four nodes:
This circuit has six nodes:
This circuit has three nodes:

Σientering = Σileaving

In this subcircuit, we have one node shown. At this node we have three currents entering the node and one current leaving the node. Applying KCL: i1 + i2 + i3 = i4 3 + -2 + 1 = 2 2 = 2 thus KCL holds true.
In this subcircuit, we will use KCL to find iy Appying KCL at the left node: 3A + 4A = 2A + ix ix = 5A Now applying KCL at the right node: ix + iy = 1A 5A + iy = 1A iy = -4A

Σientering = Σileaving

Σientering - Σileaving = 0

- Σientering + Σileaving = 0

Kirchoff's Voltage Law (KVL) states:

The Sum of the voltage drops equals the sum of the voltage rises around any closed path.

ΣVdrops = ΣVrises

ΣVdrops - ΣVrises = 0 

In this circuit, we have one closed path. Typically we refer to a closed path as a loop. In this problem we will find Vx using KVL. Starting at the top-left corner and going clockwise we apply KVL: V1 - V2 + (-3V) + Vx = 0 1V - 6V + -3V + Vx = 0 -8V = -Vx Vx = 8V
In this circuit, we will use KVL to find Vx, Vy, & Vz Note: We will also write KVL assuming a clockwise direction. The right most loop has only one unknown, so we will apply KVL there first: +3V - 9V - (-2V) + Vx = 0 -4V + Vx = 0 Vx = 4V Now applying KVL at the middle and left-most loops: -Vx + Vy = 0 Vx = Vy = 4V -Vy - Vz = 0 -Vy = Vz = -4V

Before defining Ohm's Law, let's elaborate on what a resistor is:

A resistor is a circuit element which is neither an insulator or conductor.
Current can pass through a resistor, however there is some resistance.
A resistor resists the flow of current. This resistance means that
some work must be done to "push" current through the resistor. Whenever
work is done on charge, we have voltage. Thus, when current flows
through a resistor, there is some voltage across the resistor.

The symbol for the resistor is:


The unit for resistance is ohms and the symbol is:

Ω

In this problem we will find V using Ohm's Law: V = IR V = 2A(10 ohms) = 20 Volts
Note here that current is going into the negative side of the resistor: Note: Current can only flow from a higher voltage potential to a lower voltage potential (like water flowing down stream), Therefore in this problem you must use this form of Ohm's Law: V = -IR Now find I: Since V = -IR in this problem then I = -V/R I = -(20 V)/(1 ohm) I = -20 Amps

Summary:

Use V = IR if current is shown flowing into the positive side of the resistor, and
use V = - IR if the current is shown flowing into the negative side of the resistor.

In order to build a useful circuit, there needs to be a power source.
Without a power source or Voltage, current will not flow.

What is an independent source?
An independent source supplies a fixed voltage or current, independent of
anything else.

These are independent voltage sources. The first is a power supply that provides a constant voltage. The second is a battery that also provides a constant voltage.
These is an independent current sources. Independent current sources are rare in the real world so consider this to be a theoretical device. You will see later that a current source can be constructed from a voltage source.

Where are we now?
With Kirchoff's Laws and Ohm's Law we can analyze any circuit.
Every other concept in this course is based on these principles.
Congratulations on getting this far!

When analyzing a circuit, it is often easier to do so if the circuit can be simplified
or reduced. In this section we will consider how a colletion of resistor can be
reduced down to fewer resitors.

Write a KVL equation for this circuit. Recall that V10=IR Start at the bottom left corner and go clockwise: -30V + 10I = 0 30 = 10I
Write a KVL equation for this circuit. Recall that V5=IR Start at the bottom left corner and go clockwise: -30V + 5I + 5I = 0 -30V + 10I = 0 30 = 10I
Both circuits are described by the same equation, therefore they are electrically the same circuit. This means that you can replace one circuit with the other. We can simplify the second circuit by replacing it with the first circuit.

Series Resistors:

Write a KVL equation for this circuit. -V + R1I + R2I + R3I = 0 -V + I(R1 + R2 + R3) = 0 -V + IReqv = 0 where Reqv = R1 + R2 + R3
Since Reqv = R1 + R2 + R3 then we can replace the three resistors with one resistor.
Resistors which carry the same current are in series. In this case R1, R2, and R3 are all in series. They all carry current I Resistors in series can be added and then replaced with one resistor equivalent to to their sum. ΣRi = Reqv

Parallel Resistors:

In this circuit R1 and R2 are said to be in parallel because they have the same two terminal nodes, therefore they have to have the same volrage across them. V1 = VS and V2 = VS Using Ohm's Law we know that I=V/R Therefore: I1 = VS/R1 and I2 = VS/R2 Applying KCL at the top node we generate the following equation: IS = I1 + I2 IS = VS/R1 + VS/R2 Factoring out VS we get: IS = VS(1/R1 + 1/R2) Now let's obtain the resistance which is equivalent to the two resistors in parallel: Recall that VS/IS = Requivalent and IS/VS = 1/Requivalent 1/Requivalent = IS/VS = 1/R1 + 1/R2 1/Requivalent = (R2 + R1)/{R1*R2) Threrefore two resistors in parallel can be reduced with the following expression: Requivalent = {R1*R2)/(R1 + R2)

The two circuits shown here are equivalent: Req. = {R1*R2)/(R1 + R2) Req. = {12*4)/(12 + 4) = 48/16 = 3 Ohms Req. = 3 Ohms
The three circuits shown here are also equivalent: Rx = {R1*R2)/(R1 + R2) Rx = {12*12)/(12 + 12) = 144/24 = 6 Ohms Req. = {R3*Rx)/(R3 + Rx) Req. = (3*6)/(3+6) = 18/9 = 2 Ohms
Combining Resistors in Parallel: If we always combine resistors in parallel "two-at-a-time" then we can use the equation: Requivalent = {R1*R2)/(R1 + R2) Otherwise we must use the more general form for 'n' resistors: 1/Requivalent = 1/R1 + 1/R2 + .... + 1/Rn

Now let's return to a series circuit:

Continuing with analysis: Req. = R1 + R2 + R3 IS = VS/Req. V1 = IS * R1 V2 = IS * R2 V3 = IS * R3 Now plug in IS: V1 = (VS/Req.) * R1 = (R1/Req.)*VS V2 = (VS/Req.) * R2 = (R2/Req.)*VS V3 = (VS/Req.) * R3 = (R3/Req.)*VS In plain english what do these mathematical equations say? The voltage across a series resistor is some percentage of the total voltage, VS This percentage is equal to the individual resistor's resistance divided by the equivalent resistance, Req. The voltage is being divided among the series resistors, hence this is called Voltage Division
With the circuit above does KVL still hold true? How can this concept be expanded to many series resistors?

Now let's return to a parallel circuit:

Continuing with analysis: From the circuit on the left we know that I1 = Vs/R1 and I2 = Vs/R2 From the reduced circuit on the right we know that Vs = Is*Req. Now we will plug the second equation into the first equation: I1 = (Is*Req.)/R1 I2 = (Is*Req.)/R2 From the section on parallel resistors we know: Req. = R1*R2/(R1+R2) If we plug this into the last set of equations we get: I1 = (Is*[R1*R2/(R1+R2)])/R1 I2 = (Is*[R1*R2/(R1+R2)])/R2 Note that R1 factors out of the first equation and R2 factors out the the second equation giving the following: I1 = Is*[R2/(R1+R2)] I2 = Is*[R1/(R1+R2)]
In plain english what do these mathematical equations say? The current through a parallel resistor is some percentage of the total source current, IS This percentage is equal to the OTHER resistor's resistance divided by the sum of the two resistors, R1 and R2 The currrent is being divided among the parallel resistors, hence this is called Current Division
The equations get much more complicated if you work with three or more resistors. Try generating the current division equations for three parallel resitors. The complexity of the equations will lead you to want reduce the circuit to only two resistors so that you can use the above equations.