A little doubt about the nodal voltage method.

The current reference direction is optional, I am estimating, try to choose the actual current direction, so as not to answer the negative value sometimes troublesome, if I choose, i3 up.

Resistors always consume energy, and the voltage drop of the resistor is in the same direction as the current, and the va potential is less than 18V, according to Ohm's law.

i1 = （18 - va ）/ 3。

The nodal voltage method is an equation based on the algebraic sum of the currents flowing into the node, and the basic rules are as follows:

The sum of the self-conductance multiplied by the node voltage, minus the mutual conductance multiplied by the voltage of the adjacent node, is equal to the algebraic sum of the supply currents flowing into the node.

Self-conductance: As long as one end of the resistor is on a node, the reciprocal of the resistance is the conductance.

Mutual conductance: A resistor is connected between two nodes.

The internal resistance of the current source is infinite and the conductance is zero.

A controlled source is simply a parameter controlled by an excitation source, and its power supply properties remain unchanged.

First of all, since the underside of the circuit is grounded, in general, the voltage of VA is greater than 0.

In addition, you can assume the direction of the current by yourself, and according to the current circuit drawing, the VA voltage should be greater than 0.

You can also draw it the other way around, and the voltage is what you wrote at that time, but the fuzhe negative value is found, which means that the direction of the current you drew is reversed.

Nodal voltage is a circuit calculation method for solving objects. The node voltage is inAfter selecting any node as a reference point for the circuit (this point is usually numbered "0") and making its potential zero, the potential of the remaining nodes for that reference point

Compared with these two circuit analysis methods, when the number of nodes in the circuit is less and the number of branches is more, the node voltage method is simple because the number of equations in the column is less.

Especially when there is an ideal voltage source directly connected between two nodes, the number of equations listed can be further reduced by applying the nodal voltage method flexibly. Taking the node voltage in the circuit as an unknown quantity, the independent node current equation is written according to KCL, and then the method of solving the node voltage is simultaneously solved.

Let's put it this way, if there are three nodes of ABC in the question, now suppose that node A is selected as the reference node in the book, and UB=10V, UC=20V, which means that the potential difference between BA is 10V, 20V between CAs, and -10V between BC, that is, the potential of point B is 10 higher than that of A, and 10 less than C, then now if you choose B as the reference node (i.e., assume that the potential of B is 0), Then the potential of point A should be -10V, and the potential of point C should be 10V, so the results are naturally different for different reference nodes.

i is the current between node 1 and 2, and the current on node 1 has 4a flowing in from the left, flowing down un1 3, and flowing out of i to the right, so 4=un1 3 + i, and the equation for node 1 is un1 3=4-i....1)；

Node 2 is processed the same, i and 2a inflow, un2 6 out, un2 6=i+2 .2)；

un1-un2=6 .3)；

1)+(2) un1/3+un2/6=6，2un1+un2=36...4)；

3)+(4) 3un1=42，un1=14v，un2=8v；i=4 - un1 3=-2 3 a, the power emitted by the voltage source = -6i = 4w.

<> solution: Let the lowest end be the common node O, and the potential of the node is U1 and U2. According to the nodal voltage method:

Node 1: u1 1+(u1-u2) (6+2)=;

Node 2: (u2-u1) (2+6)+u2 12+(u2-30) 4=0.

Solve the system of equations: u1 = 6, u2 = 18.

um2=2×（u1-u2）/（6+2）=2×（6-18）/8=-3（v）。

So: v=um2+u2=-3+18=15(v).

Take the lower node as the reference node, the upper left current source output node voltage U1, the upper right three-resistor common node voltage U2, 6 and 2 resistors are directly connected in series to 8 resistors, and the column writes the node voltage equation:

Node 1: (1+1 8)u1-(1 8)u2=node 2: -(1 8)u1+[(1 8)+(1 4)+(1 12)]u2=30 4

Then apply KCl to the 6 and 2 resistor series branches (or directly use the law of voltage division):

u1-u)/6=(u-u2)/2

Solution: u=15v

A is the reference point of the circle, (1 50 x 3) UB = (25 + 25 + 100) Xun early difference 50, 3 UB = 150, UB = 50V; i1=-(50-25)/50=，i2=-(50-100)/50=1a，i3=-(50-25)/50=。acres of skin.

Use the superposition theorem to find the current i

Use the nodal voltage method to find the current i

When 6V acts alone, i'= 2a, when 12v acts alone, i"=-2a, when 3a acts alone, i'"=-1a；i=-1a。

Nodal voltage is a circuit calculation method for solving objects. The node voltage is used as a reference point for a node, usually "0".

If u4 is set to zero potential, then according to the node voltage theorem, there is:

u1=5v(1/3+1/1+1/4)u2-(1/1)u1-(1/4)u3=4+2u/4

1/3+1/4)u3-(1/4)u2= -2u/4-3i