Node Equations:

Kirchhoff's current law, as previously begin, says in which at any junction point in a circuit the current arriving is equal to the current leaving.  Let we consider five currents entering and leaving a junction display as P in the Figure.  This junction is also known a node.

Suppose that all currents entering the node are positive, and all currents that will leave the node are negative.  Thus, I₁, I₃, and I₄ are positive, and I₂ and I₅ are negative. Kirchhoff's Law also states that the sum of all the currents meeting at the node is zero.  For Figure, Equation (2-19) represents this law mathematically.

I₁ + I₂ + I₃ + I₄ + I₅ = 0                                   (2-19)

Figure: Node Point

Through solving node equations, we could calculate the unknown node voltages. For each node within a circuit we will assign a letter or number.  Inside the Figure which is given below, A, B, C, and N are nodes, and N and C are principal nodes.  Principal nodes are those nodes along with three or more connections.  Node C will be our chosen reference node. V_AC is the voltage among Nodes A and C; V_{BC is} the voltage among Nodes B and C; and V_NC is the voltage among Nodes N and C.   We have already determined that all node voltages have a reference node; thus, we could substitute V_A for V_AC, V_B for V_BC and V_N for V_NC.

Figure: Circuit for Node Analysis

Suppose that loop currents I₁ and I₂ leave Node N, and that I₃ enters Node N as like in the figure. From Kirchhoff's current law:

∑I=0

I₁ + I₂+ I₃= 0

I₃ = I₁ + I₂                                                                        (2-20)

By using Ohm's Law and solving for the current by every resistor we acquire the following.

 I =     V_R/R   where V_R is the voltage across resistor, R.

I₃ =     V_N/R₂

I₁ =     V_A - V_N/R₁

I₂ =     V_B - V_N/R₃

Substitute these equations for I₁, I₂, and I₃ into Kirchhoff's current equation (2-20) yields the following.

V_N/R₂ = V_A - V_N/R₁ + V_B - V_N/R₃

The circuit display in Figure could be solved for voltages and currents through using the node-voltage analysis.

Figure: Nodes - Voltage Analysis

First step, suppose direction of current flow display. Mark nodes A, B, C, and N, and mark the polarity across every resistor.

Second step, using Kirchhoff's current law at Node N, solve for V_N.

I₃ = I₁ + I₂

V_N/R₂ = V_A - V_N/R₁ + V_B - V_N/R₃

V_N/6 = 60 - V_N/8 + 20 - V_N/4

Clear the fraction so which we have a general denominator:

4 V_N = 3 (60    V_N) + 6 (20     V_N)

4 V_N = 180 -    3 V_N + 120 - 6 V_N

13V_N = 300

V_N = 23.077

Third step, find out all voltage drops and currents.

V₁ =    V_A +    V_N =    60 - 23.077 =  36.923 Volts

V₂ =     V_N =    23.077 Volts

V₃ =    V_B -     V_N =    20 - 23.077 =  -3.077 Volts

The negative value for V₃ shows that the current flow by R₃ is opposite that that was supposed and that the polarity across R₃ is reversed.

I₁ = V₁/R₁ =36.923 V/ 8? =4.65 amps

I₂ = V₃/R₃ =-3.077 V/ 4? =-0.769 amps

I₃ = V₂/R₂ =23.077 V/ 6? =3.846 amps

The negative value for I₃ displays in which the current flow by R₃ is opposite that which was supposed.