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CpE 311 Blog

Mid -Term

Blog Entry No. 1            

    
     Charge,Current,Voltage,Power and Energy

              Circuit Element
              Kirchoff Law

            

Blog Entry No. 2

              Basic Laws: Series and Parallel and Introduction to Experiment

Blog Entry No. 3

             Pursue of Basic Law: Series and Parallel with Example

             Wye-Delta Transformation

Blog Entry No. 4

             Nodal Analysis With Current Source

            

Blog Entry No.5

            Nodal Analysis with Voltage Source

            

Blog Entry No. 6

            Another example of Nodal analysis with voltage source

Final

Blog Entry No.8

        Mesh Analysis

        Circuit Theorem

 

Blog Entry No.9

        Thevenin and Norton Theorem

 

Blog Entry No.10

         Maximum power transfer

 

Blog Entry No.11

         First Order Circuits: Source Free RC/RL Circuit
         Capacitor and Inductor

Submitted by:

     Charlie Chavez & Alvin Tutor

Capacitor and Inductor

Capacitor and Inductor

Capacitor

         Capacitor consists of two or more parallel conductive (metal) plates which are not connected or touching each other, but are electrically separated either by air or by some form of a good insulating material such as waxed paper, mica, ceramic, plastic or some form of a liquid gel as used in electrolytic capacitors. The insulating layer between a capacitors plates is commonly called the Dielectric.

Inductor

         An inductor is a passive electronic component that storesenergy in the form of a magnetic field. In its simplest form, an inductor consistsof a wire loop or coil. The inductance is directly proportional to the number ofturns in the coil. Inductance also depends on the radius of the coil and on the type of material around which the coil is wound.

First Order Circuits: Source Free RC/RL Circuit

First Order Circuits: Source Free RC/RL Circuit

    A first-order circuit can only contain one energy storage element (a capacitor or an inductor). The circuit will also contain resistance.
So there are two types of first-order circuits:
RC circuit
RL circuit

RC

Formula for V voltage response of the RC circuit is an exponential decay of the initial voltage. Vo at time t=0

 

RC Time constant

RL

Formula for I current response of the RL circuit is an exponential decay of the initial current. Io at time to=0

RL Time constant

Maximum power transfer

Maximum power transfer

   Maximum power transfer theorem determines the value of resistance R_L for which, the maximum power will be transferred from source to it. Actually the maximum power, drawn from the source, depends upon the value of the load resistance. There may be some confusion let us clear it. ]

Thevenin and Norton Theorem

Thevenin and Norton Theorem

Thevenin's Theorem

      Thevenin’s theorem for linear electrical networks states that any combination of voltage sources, current sources and resistors with two terminals is electrically equivalent to a single voltage source and a single series resistor.It mean that it is possible to simplify any electrical circuit, no matter how complex it is. For Example:

Method 1

Step 1:  Find the open circuit voltage that will be  which is

Step 2: Take out the voltage source and make a short circuit in source connection

Method 2:  Use source transformation

Voltage source transformation

2 parallel resistances

Current source transformation

2 serial resistances

Norton's Theorem

       Norton’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. In Norton theorem, we just replace the circuit connected to a particular branch by equivalent current source . In this theorem, the circuit network is reduced into a single constant current source in which, the equivalent internal resistance is connected in parallel with it. Every voltage source can be converted into equivalent current source .It mean that this  theorem is just alternative of Thevenin theorem.

Mesh Analysis

Mesh Analysis

       Another way of simplifying the complete set of Kirchhoff’s equations is the mesh or loop current method. Note: Mesh current method is only applicable for “planar” circuits.
       There are 3 steps to consider in mesh analysis:
1.) Assign a mesh current to each mesh.
2.) Apply Kirchhoff’s voltage law (KVL) around each mesh, in the same direction as the mesh current so loop every mesh in the circuit .For example:

 

There are five mesh in this circuit.

3.) Solve the resulting loop equations for the mesh currents.

Example:

 

    Solve for the current through the 5 ohm resistor and the current through the 18V source using Mesh-Current Analysis.

Now write KVL equations for each loop.
KVL for i1:
-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0
then gather terms:
10i1 - 5i2 - 4 i3 -18V = 0
Note that the i1 term is positive, and all other current terms are negative (because they are all clockwise, all other panes will contribute a negative term). Let's do the other two panes with terms gathered up directly (write the total resistance of the loop multiplied by the mesh current that goes through that total resistance):
KVL for i2:
-5i1 + 10i2 - 3i3 - 12 = 0
KVL for i3:
-4i1 -3i2 +9i3 = 0
Now solve the three equations in three unknowns:
i1 = 7.02A
i2 = 6.28A
i3 = 5.21A     The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current through the 18V is i1, or 7.02A.

 

 

Circuit Theorem

Circuit Theorem

Linear Property

        Linear property is the linear relationship between cause and effect of an element. This property gives linear and nonlinear circuit definition. The property can be applied in various circuit elements.

The homogeneity (scaling) property and the additivity property are combination of linearity property.

Superposition  Property

      Suppose a branch of an electrical circuit is connected to numbers of voltage and current source s. As we can consider electrical current as electrical quantity, it can be easily assumed that total current flows through the branch is nothing but the summation of all individual currents, contributed by the each individual voltage or current source .

Source Transformation

     The source transformation of a circuit is the transformation of a power source from a voltage source to a current source, or a current source to a voltage source. In other words, we transform the power source from either voltage to current, or current to voltage.

      In voltage source transformation  just divide the voltage source to the resistor parallel or series for example:

       The other one is current source transformation just multiply the current source to the resistor parallel or series for example:

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: