This circuit has four nodes:  
This circuit has six nodes:  
This circuit has three nodes: 
Σi_{entering} = Σi_{leaving}
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: i_{1} + i_{2} + i_{3} = i_{4} 3 + 2 + 1 = 2 2 = 2 thus KCL holds true. 

In this subcircuit, we will use KCL to find i_{y} Appying KCL at the left node: 3A + 4A = 2A + i_{x} i_{x} = 5A Now applying KCL at the right node: i_{x} + i_{y} = 1A 5A + i_{y} = 1A i_{y} = 4A 
Σi_{entering} = Σi_{leaving}
Σi_{entering}  Σi_{leaving} = 0
 Σi_{entering} + Σi_{leaving} = 0
ΣV_{drops} = ΣV_{rises}OR
ΣV_{drops}  ΣV_{rises} = 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 topleft corner and going clockwise we apply KVL: V_{1}  V_{2} + (3V) + V_{x} = 0 1V  6V + 3V + Vx = 0 8V = Vx Vx = 8V 

In this circuit, we will use KVL to find V_{x}, V_{y}, & V_{z} 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) + V_{x} = 0 4V + V_{x} = 0 V_{x} = 4V Now applying KVL at the middle and leftmost loops: V_{x} + V_{y} = 0 V_{x} = V_{y} = 4V V_{y}  V_{z} = 0 V_{y} = V_{z} = 4V 
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: Now find I: Since V = IR in this problem then I = V/R I = (20 V)/(1 ohm) I = 20 Amps 
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. 
Write a KVL equation for this circuit. Recall that V_{10}=IR Start at the bottom left corner and go clockwise: 30V + 10I = 0 30 = 10I 

Write a KVL equation for this circuit. Recall that V_{5}=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. 
Write a KVL equation for this circuit. V + R_{1}I + R_{2}I + R_{3}I = 0 V + I(R_{1} + R_{2} + R_{3}) = 0 V + IR_{eqv} = 0 where R_{eqv} = R_{1} + R_{2} + R_{3} 

Since R_{eqv} = R_{1} + R_{2} + R_{3} then we can replace the three resistors with one resistor. 

Resistors which carry the same current are in series. In this case R_{1}, R_{2}, and R_{3} 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. ΣR_{i} = R_{eqv} 
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. V_{1} = V_{S} and V_{2} = V_{S} Using Ohm's Law we know that I=V/R Therefore: I_{1} = V_{S}/R_{1} and I_{2} = V_{S}/R_{2} Applying KCL at the top node we generate the following equation: I_{S} = I_{1} + I_{2} I_{S} = V_{S}/R_{1} + V_{S}/R_{2} Factoring out V_{S} we get: I_{S} = V_{S}(1/R_{1} + 1/R_{2}) Now let's obtain the resistance which is equivalent to the two resistors in parallel: Recall that V_{S}/I_{S} = R_{equivalent} and I_{S}/V_{S} = 1/R_{equivalent} 1/R_{equivalent} = I_{S}/V_{S} = 1/R_{1} + 1/R_{2} 1/R_{equivalent} = (R_{2} + R_{1})/{R_{1}*R_{2}) Threrefore two resistors in parallel can be reduced with the following expression: R_{equivalent} = {R_{1}*R_{2})/(R_{1} + R_{2}) 
The two circuits shown here are equivalent: R_{eq.} = {R_{1}*R_{2})/(R_{1} + R_{2}) R_{eq.} = {12*4)/(12 + 4) = 48/16 = 3 Ohms R_{eq.} = 3 Ohms 

The three circuits shown here are also equivalent: R_{x} = {R_{1}*R_{2})/(R_{1} + R_{2}) R_{x} = {12*12)/(12 + 12) = 144/24 = 6 Ohms R_{eq.} = {R_{3}*R_{x})/(R_{3} + R_{x}) R_{eq.} = (3*6)/(3+6) = 18/9 = 2 Ohms 

Combining Resistors in Parallel:
If we always combine resistors in parallel "twoatatime"then we can use the equation: R_{equivalent} = {R_{1}*R_{2})/(R_{1} + R_{2}) Otherwise we must use the more general form for 'n' resistors: 1/R_{equivalent} = 1/R_{1} + 1/R_{2} + .... + 1/R_{n} 
Continuing with analysis: R_{eq.} = R_{1} + R_{2} + R_{3} I_{S} = V_{S}/R_{eq.} V_{1} = I_{S} * R_{1} V_{2} = I_{S} * R_{2} V_{3} = I_{S} * R_{3} Now plug in I_{S}: V_{1} = (V_{S}/R_{eq.}) * R_{1} = (R_{1}/R_{eq.})*V_{S} V_{2} = (V_{S}/R_{eq.}) * R_{2} = (R_{2}/R_{eq.})*V_{S} V_{3} = (V_{S}/R_{eq.}) * R_{3} = (R_{3}/R_{eq.})*V_{S} In plain english what do these mathematical equations say? The voltage across a series resistor is some percentage of the total voltage, V_{S} This percentage is equal to the individual resistor's resistance divided by the equivalent resistance, R_{eq.} The voltage is being divided among the series resistors, hence this is called Voltage Division 

With the circuit above does KVL still hold true? 
Continuing with analysis: From the circuit on the left we know that I_{1} = V_{s}/R_{1} and I_{2} = V_{s}/R_{2} From the reduced circuit on the right we know that V_{s} = I_{s}*R_{eq.} Now we will plug the second equation into the first equation: I_{1} = (I_{s}*R_{eq.})/R_{1} I_{2} = (I_{s}*R_{eq.})/R_{2} From the section on parallel resistors we know: R_{eq.} = R_{1}*R_{2}/(R_{1}+R_{2}) If we plug this into the last set of equations we get: I_{1} = (I_{s}*[R_{1}*R_{2}/(R_{1}+R_{2})])/R_{1} I_{2} = (I_{s}*[R_{1}*R_{2}/(R_{1}+R_{2})])/R_{2} Note that R_{1} factors out of the first equation and R_{2} factors out the the second equation giving the following: I_{1} = I_{s}*[R_{2}/(R_{1}+R_{2})] I_{2} = I_{s}*[R_{1}/(R_{1}+R_{2})] 

In plain english what do these mathematical equations say? The current through a parallel resistor is some percentage of the total source current, I_{S} This percentage is equal to the OTHER resistor's resistance divided by the sum of the two resistors, R_{1} and R_{2} 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. 