Another example of Nodal analysis with voltage source

 Nodal analysis with voltage source

Example 1:

Determine Ib in the circuit in Fig. 3.58 using nodal analysis.



Let V1 be the unknown node voltage to the right of the 250-Ω resistor. Let the ground
reference be placed at the bottom of the 50-Ω resistor.



But Substituting this into the nodal equation leads to
Thus,

Example 2:

Find Vo and the power dissipated in all the resistors in the circuit of Fig. 3.60.

At the top node, KVL gives

Nodal Analysis with Voltage Source

Nodal Analysis with Voltage Source

There are 2 cases:

1st case:  If a voltage source is connected between the reference
node and a non reference node, we simply set the voltage at the non reference
node equal to the voltage of the voltage source.

2nd case: If the voltage source (dependent or independent) is connected between two non reference nodes, the two non reference nodes form a generalized node or super node; we apply both KCL and KVL to determine the node voltages.

To know about super node is a formed by enclosing a (dependent or independent)
voltage source connected between two non reference nodes and any
elements connected in parallel with it.

Example:

Question:
     Solve for the current through the 5 ohm resistor and the current throughthe 18V source using Node-Voltage Analysis.

We don't know the current through the 12V nor through the 18V, and we cannot use Ohm's law to find it, since they are not resistors. So add unknown current labels to the circuit, as shown:

Now we can write the KCL equations:
iy + (V3 - V2)/5 - ix = 0

Current leaving V4

-iy + (V4 - V1)/2 = 0

Current leaving V5

ix + (V5 - 0)/1 = 0

Extra Equations

We now have five equations, but we have seven unknowns (the 5 unknown voltages plus 2 more unknown currents). We can get two extra equations -- just look at the voltage sources that caused the problems in the first place, where we had to add ix and iy. We'll write the voltage relationships between the node voltages:

V4 - V3 = 12V
V3 - V5 = 18V


Now we have 7 equations and 7 unknowns. To solve these equations, let's simplify it down to fewer equations. Solve for V4 and solve for V5 from those last two we added:

V4 = V3 + 12V
V5 = V3 - 18V


Substitute these two equations into the previous five, so that we eliminate V4 and V5. We now have 5 equations and 5 unknowns:

(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - ix = 0
-iy + ([V3 + 12] - V1)/2 = 0
ix + ([V3 - 18] - 0)/1 = 0


Continue using substitution (Gaussian elimination) to narrow it down. Next, solve for ix in the last equation:
ix = 18 - V3 and substitute back to get 4 equations and 4 unknowns:

(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - [18 - V3] = 0
-iy + ([v3 + 12] - V1)/2 = 0

Next, solve for iy in the last equation:
iy = ([v3 + 12] - V1)/2 and substitute back to get 3 equations and 3 unknowns:

(V1 - [V3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
([V3 + 12] - V1)/2 + (V3 - V2)/5 - [18 - V3] = 0

Now multiply each equation by a constant to clear out fractions. That is multiply by the Least Common
Denominator (multiply first by 6, second by 60, third by 10).

3 * (V1 - [V3 + 12]) + 2 * (V1 - V2) + 3 * (V1 - 0) = 0
12 * (V2 - V3) + 15 * (V2 - 0) + 20 * (V2 - V1) = 0
5 * ([V3 + 12] - V1) + 2 * (V3 - V2) - 10 * [18 - V3] = 0

Next, multiply through

3*V1 - 3*V3 - 36 + 2*V1 - 2*V2 + 3*V1 = 0
12*V2 - 12*V3 + 15*V2 + 20*V2 - 20*V1 = 0
5*V3 + 60 - 5*V1 + 2*V3 - 2*V2 -180 + 10*V3 = 0

Now gather like terms or use matrix (basket) to solve:
8*V1 -2*V2 -3*V3 = 36
-20*V1 +47*V2 = 12*V3 = 0
-5*V1 -2*V2 + 17*V3 = 120

Nodal Analysis With Current Source

Nodal Analysis With Current Source

      In nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables. In nodal analysis, we are interested in finding the node voltages. Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps.

Steps to Determine Node Voltages:

1. Select a node as the reference node. Assign voltages v2, p , vn -1
to the remaining n -1 nodes. The voltages are
referenced with respect to the reference node.
2. Apply KCL to each of the n-1 nonreference nodes. Use
Ohm’s law to express the branch currents in terms of node
voltages.
3. Solve the resulting simultaneous equations to obtain the
unknown node voltages.


The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential.

Example:

Determine the node voltages in the following circuit.

---

Summing the currents INTO node A gives

1m - V_A / 4K + (V_B - V_A) / 8K = 0

Collecting terms and simplifying gives

3 V_A - V_B = 8   --------eq 1

Node B:

Summing the currents into node B gives

(V_A - V_B) / 8K - V_B / 16K - 2m = 0

Collecting terms and simplifying gives

2 V_A - 3 V_B = 32  --------eq 2 

Solving   eq1 and  eq2 just use metrix

V_A = - 1.143
V_B = - 11.43

Wye-Delta Transformation

        In this situation how we combine the resistor R1 through R6 if the resistor may be in series or parallel.


But we can simplify the circuit using of the tree-terminal equivalent circuits the Wye ,Tee and Delta.
We discuss this tree-terminals is can be convert to the other tree-terminal equivalent circuit with the same value or equivalent resistance.

The formula:

Pursue of Basic Law: Series and Parallel with Example

  Basic Law: Series and Parallel

In series resistor same current but not in voltage.



V1=iR1 & V2=iR2
Using KVL
     V-V1-V2=0
     V=i(R1+R2)
     V/(R1+R2)=v/Req
     or V=i(R1+R2)=iReq
     Req=R1+R2
    Because the voltage is not equal so we use the voltage division:
        V1=iR1   and V2=iR2
         i=V/(R1+R2 )

     V1=VR1/(R1+R2)
     V2=VR2/(R1+R2)
In parallel resistor same voltage but not in current

















V=i1R1=i2R2
     i2+i2
     V/R1+R2
     V/R1+V/R2
    R(I/R1+ I/R2)
    V/Req
     V=iReq
    I/Req=I/R1+1/R2
    Req=R1R2/(R1+R2)

     The current is not the same so to solve the current we use the current division:
      V=i1R1=i2R2
      V=iReq=iR1R2/(R1+R2)
      and i1=V/R1  and   i2=V/R2

          i1=iR2/(R1+R2)
          i2=iR1/(R1+R2)

Conductance(G)

Series Coductance 
         1/Geq = G1+G2+...Gn
Parallel Conductance
          Geq =G1+G2+....Gn




Example 1:







Use KVL
    -15+(1+5+2)I+Vx=0
      Vx=51
             -15+81+101=0,     I= 5/6
              Vx=51=25/6=4.17V