PAGE

Sabado, Agosto 23, 2014

BLOG 9 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE SUPERPOSITION


SUPERPOSITION

 The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

to apply the superposition principle, we must keep two things in mind:

1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.

2. Dependent sources are left intact because they are controlled by circuit variables.

3 Steps to Apply Superposition 

Principle:
1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to the independent sources.


EXAMPLE:

Solution

The voltage V can be found by solving three subproblems (one circuit for each source).

In each subproblem below, just one source is kept, while the others are deactivated (zeroed out). For any superposition problem, you get as many subproblems as there are sources in the circuit. Note that even if they are the same type (like both voltage sources), they are solved separately. In this case there are three sources (1mA, 2V, 3V), so we have three subproblems.

Subproblem 1

The first subproblem is shown below, where only the current source is kept in the circuit. The 2V is zeroed out yielding a 0V source, or a short. The 3V is zeroed out, again making it a short.

The 9K ohm is shorted out now (0 in parallel with 9K is 0):

Now solve for the voltage (this is just part of the total answer). First find the current flowing right through the 5K using a current divider equation:

Now find the voltage by Ohm's law:

Subproblem 2

Now we keep the 2V source and deactivate the other two. Zeroing out the 1mA current source yields a 0 A source, or an open. Zeroing out the 3V source yields a 0 V source, or a short. We then have:
circuit image
Again, the 9K resistor is shorted out. Note that the current through the 4K resistor will be zero and thus we can eliminate this resistor as well. That leaves us with:

Let's redraw the circuit with the source on the left. Be sure to maintain the same connections and the same polarity on the voltage V.

Now apply a voltage divider (twice) to get the voltage V. Note that the voltage V is the opposite direction expected for a voltage divider, so we need a negative sign as well:

Subproblem 3

Finally, keep the 3V source and deactivate the rest. The 1mA current source becomes an open, the 2V source becomes a short:
circuit image
We can again eliminate the 4K resistor because no current can flow through it. We then have:
circuit image
Now use a voltage divider (twice), and again note the voltage is the wrong direction, so we need a negative sign:

Final Solution

To find the total V, add the answers from each subproblem:
V = 2.27 - 0.28 - 0.56 = 1.43V

LEARNING

I learn about the superposition is we need to turn off the other independent source and we need to remain one independent source. We can use the other analysis to solve the superposition like mesh analysis and nodal analysis. When we get the output voltage and current we add the two result.


Sabado, Agosto 16, 2014

BLOG 8 - ELECTRICAL CIRCUIT 1

THIS BLOG CONTAIN ABOUT THE LINEARITY PROPERTY AND SOURCE TRANSFORMATION


Linearity property in electric circuit

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 both the combination of linearity property.
The homogeneity property is that if the input is multiplied by a constant k then the output is also multiplied by the constant k. Input is called excitation and output is called response here. As an example if we consider ohm’s law. Here the law relates the input i to the output v.
Mathematically,               v= iR
If we multiply the input current  i by a constant k then the output voltage also increases correspondingly by the constant k. The equation stands,      kiR = kv
The additivity property is that the response to a sum of inputs is the sum of the responses to each input applied separately.
Using voltage-current relationship of a resistor if
                                       v1 = i1R       and   v2 = i2R
Applying (i1 + i2)gives
V = (i1 + i2)R = i1R+ i2R = v+ v2
We can say that a resistor is a linear element. Because the voltage-current relationship satisfies both the additivity and the homogeneity properties.
We can tell a circuit is linear if the circuit both the additive and the homogeneous. A linear circuit always consists of linear elements, linear independent and dependent sources.

What is linear circuit?

A circuit is linear if the output is linearly related with its input.
The relation between power and voltage is nonlinear. So this theorem cannot be applied in power.
See a circuit in figure 1. The box is linear circuit. We cannot see any independent source inside the linear circuit.
what is linear circuit
The linear circuit is excited by another outer voltage source vs. Here the voltage source vs acts as input. The circuit ends with a load resistance R. we can take the current I through R as the output.
Suppose vs = 5V and i = 1A. According to linearity property if the voltage is multiplied by 2 then the voltage vs = 10V and then the current also will be multiplied by 2 hence i = 2A.
The power relation is nonlinear. For example, if the current i1 flows through the resistor R, the power p= i12R and when current iflows through the resistor R then power p= i22R.
If the current (i1 + i2) flows through R resistor the power absorbed
   P3 = R(i1 + i2)2 = Ri12 + Ri22 + 2Ri1i2 ≠ p1 + p2
So the power relation is nonlinear. Circuit solution method superposition is based on linearity property.

Simplify Circuit Analysis by Transforming Sources in Circuits



With transformation, you can modify a complex circuit so that in the transformed circuit, the devices are all connected in series or in parallel. By transforming circuits, you can apply shortcuts such as the current divider technique and the voltage divider technique to analyze circuits.
Each device in a series circuit has the same current, and each device in a parallel circuit has the same voltage. Therefore, finding the current in each device in a circuit is easier when the devices are all connected in parallel, and finding the voltage is easier when they’re all connected in series.
Through a circuit transformation, or makeover, you can treat a complex circuit as though all its devices were arranged the same way — in parallel or in series — by appropriately changing the independent source to either a current or voltage source.
Changing the practical voltage source to an equivalent current source (or vice versa) requires the following conditions:
  • The resistors must be equal in both circuits.
  • The source transformation must be constrained by vS isR.
The constraining equation, vS isR, looks like Ohm’s law, which should help you remember what to do when transforming between independent voltage and current sources.

Convert to a parallel circuit with a current source

Transformation techniques let you convert a practical voltage source with a resistor connected in series to a current source with a resistor connected in parallel. Therefore, you can convert a relatively complex circuit to an equivalent circuit if all the devices in the external circuit are connected in parallel. You can then find the current of individual devices by applying the current divider techniques.
When switching from a voltage source to a current source, the resistors have to be equal in both circuits, and the source transformation must be constrained by vS isR. Solving the constraint equation for is allows you to algebraically convert the practical voltage source into a current source:
image0.jpg
This sample circuit shown here illustrates the conversion of a voltage source, in Circuit A, into an equivalent current source, in Circuit B. The resistors, R, are equal, and the constraint equation was applied to change the voltage source into a current source.
image1.jpg
The sample circuit below shows the conversion with some numbers plugged in. Both circuits contain the same 3-kΩ resistor, and the source voltage in Circuit A is 15 volts. With this information, you can find the source current, is, for the transformed Circuit B.
image2.jpg
Use the constraint equation to find the source current in Circuit B. Here’s what you get when you plug in the numbers:
image3.jpg

Convert to a series circuit with a voltage source

You can convert a current source connected in parallel with a resistor to a voltage source connected in series with a resistor. You use this technique to form an equivalent circuit when the external circuit has devices connected in series.
Converting a practical current source connected with a resistor in parallel to a voltage source connected with a resistor in series follows the conditions for equivalent circuits:
  • The resistors must be equal in both circuits.
  • The source transformation must be constrained by vS isR.
This circuit illustrates how to convert a current source into a voltage source.
image4.jpg
The sample circuit shown below depicts the same transformation of a current source to a voltage source with some numbers plugged in. Both circuits contain the same 3-kΩ resistor, and the current source in Circuit A is 5 mA.
image5.jpg
You can use the constraint equation to find the source voltage for Circuit B. Plugging in the numbers produces the following:
vs = is= (5 mA)(3 kΩ) = 15 V
Suppose you have a complex circuit that has a current source, a resistor connected in parallel, and an external circuit with multiple resistors connected in series. You can transform the circuit so that it has a voltage source connected with all the resistors in series.
Consider Circuit A in the sample circuit below, where the right side of Terminals A and B consists of two resistors connected in series.
image6.jpg
On the left side of Terminals A and B is a practical current source modeled as an ideal current source in parallel with a resistor.
You want all the devices to be connected in series, so you need to move R when you transform the circuit. To transform the circuit, change the current source to a voltage source and move R so that it’s connected in series rather than in parallel. When you use the constraint equation vs = isR to find the source voltage, remember that R is the resistor you moved.
Circuit B is a series circuit where all the devices share the same current. You can now find the voltage through RR1, and R2 using voltage divider techniques.

LEARNING
I learn about the linearity property is when the voltage is increasing the current also increasing and vice versa.I learn about the source transformation is when the polarity of the voltage source has a positive on top therefore the resulting current source is pointing upward and vice versa. Also when the voltage source is dependent therefore the resulting current is also dependent, same goes with independent sources.







Sabado, Agosto 9, 2014

BLOG 7 - ELECTRICAL CIRCUIT 1

This blog contain about the Mesh analysis and Mesh current analysis.

Mesh analysis


Figure 1: Essential meshes of the planar circuit labeled 1, 2, and 3. R1, R2, R3, 1/sc, and Ls represent the impedance of the resistors, capacitor, and inductor values in the s-domain. Vs and is are the values of the voltage source and current source, respectively.
Mesh analysis (or the mesh current method) is a method that is used to solve planar circuits for the currents (and indirectly the voltages) at any place in the circuit. Planar circuits are circuits that can be drawn on a plane surface with no wires crossing each other. A more general technique, called loop analysis (with the corresponding network variables calledloop currents) can be applied to any circuit, planar or not. Mesh analysis and loop analysis both make use of Kirchhoff’s voltage law to arrive at a set of equations guaranteed to be solvable if the circuit has a solution. Mesh analysis is usually easier to use when the circuit is planar, compared to loop analysis.


Mesh Current Analysis Circuit

mesh current analysis circuit
One simple method of reducing the amount of math’s involved is to analyse the circuit using Kirchoff’s Current Law equations to determine the currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2. So Kirchoff’s second voltage law simply becomes:
  • Equation No 1 :    10 =  50I1 + 40I2
  • Equation No 2 :    20 =  40I1 + 60I2
therefore, one line of math’s calculation have been saved.

Mesh Current Analysis

A more easier method of solving the above circuit is by using Mesh Current Analysis or Loop Analysis which is also sometimes called Maxwell´s Circulating Currents method. Instead of labelling the branch currents we need to label each “closed loop” with a circulating current.
As a general rule of thumb, only label inside loops in a clockwise direction with circulating currents as the aim is to cover all the elements of the circuit at least once. Any required branch current may be found from the appropriate loop or mesh currents as before using Kirchoff´s method.
Another way of simplifying the complete set of Kirchhoff’s equations is the mesh or loop current method. Using this method, Kirchhoff’s current law is satisfied automatically, and the loop equations that we write also satisfy Kirchhoff’s voltage law. Satisfying Kirchhoff’s current law is achieved by assigning closed current loops called mesh or loop currents to each independent loop of the circuit and using these currents to express all the other quantities of the circuit. Since the loop currents are closed, the current that flows into a node must also flow out of the node; so writing node equations with these currents leads to identity.
Let us first consider the method of mesh currents.
We first note that the mesh current method is only applicable for “planar” circuits. Planar circuits have no crossing wires when drawn on a plane. Often, by redrawing a circuit which appears to be non-planar, you can determine that it is, in fact, planar. For non-planar circuits, use the loop current method described later in this chapter.
To explain the idea of mesh currents, imagine the branches of the circuit as “fishing net” and assign a mesh current to each mesh of the net. (Sometimes it is also said that a closed current loop is assigned in each “window” of the circuit.)
The schematic diagram
The “fishing net” or the graph of the circuit
The technique of representing the circuit by a simple drawing, called a graph, is quite powerful. Since Kirchhoff’s laws do not depend on the nature of the components, you can disregard the concrete components and substitute for them simple line segments, called the branches of the graph. Representing circuits by graphs allows us to use the techniques of mathematical graph theory. This helps us explore the topological nature of a circuit and determine the independent loops. Come back later to this site to read more about this topic.

The steps of mesh current analysis:
  1.  Assign a mesh current to each mesh. Although the direction is arbitrary, it is customary to use the clockwise direction.
  2. Apply Kirchhoff’s voltage law (KVL) around each mesh, in the same direction as the mesh currents. If a resistor has two or more mesh currents through it, the total current through the resistor is calculated as the algebraic sum of the mesh currents. In other words, if a current flowing through the resistor has the same direction as the mesh current of the loop, it has a positive sign, otherwise a negative sign in the sum. Voltage sources are taken into account as usual, If their direction is the same as the mesh current, their voltage is taken to be positive, otherwise negative, in the KVL equations. Usually, for current sources, only one mesh current flows through the source, and that current has the same direction as the current of the source. If this is not the case, use the more general loop current method, described later in this paragraph. There is no need to write KVL equations for loops containing mesh currents assigned to current sources.
  3. Solve the resulting loop equations for the mesh currents.
  4. Determine any requested current or voltage in the circuit using the mesh currents.
For example: :    i1 = I1 , i2 = -I2  and  I3 = I1 – I2
We now write Kirchoff’s voltage law equation in the same way as before to solve them but the advantage of this method is that it ensures that the information obtained from the circuit equations is the minimum required to solve the circuit as the information is more general and can easily be put into a matrix form.
For example, consider the circuit from the previous section.
mesh current analysis
These equations can be solved quite quickly by using a single mesh impedance matrix Z. Each element ON the principal diagonal will be “positive” and is the total impedance of each mesh. Where as, each element OFF the principal diagonal will either be “zero” or “negative” and represents the circuit element connecting all the appropriate meshes. This then gives us a matrix of:
mesh current matrix
Where:
  • [ V ]   gives the total battery voltage for loop 1 and then loop 2.
  • [ I ]     states the names of the loop currents which we are trying to find.
  • [ R ]   is called the resistance matrix.
and this gives I1 as -0.143 Amps and I2 as -0.429 Amps
As :    I3 = I1 – I2
The combined current of I3 is therefore given as :   -0.143 – (-0.429) = 0.286 Amps.


LEARNING
 Our topic was all about mesh analysis. I learned was that you need to follow the three steps to determine the mesh analysis such as  Assign a mesh current to each mesh. Although the direction is arbitrary, it is customary to use the clockwise direction. Apply Kirchhoff’s voltage law (KVL) around each mesh, in the same direction as the mesh currents. If a resistor has two or more mesh currents through it, the total current through the resistor is calculated as the algebraic sum of the mesh currents. In other words, if a current flowing through the resistor has the same direction as the mesh current of the loop, it has a positive sign, otherwise a negative sign in the sum. Voltage sources are taken into account as usual, If their direction is the same as the mesh current, their voltage is taken to be positive, otherwise negative, in the KVL equations. Usually, for current sources, only one mesh current flows through the source, and that current has the same direction as the current of the source. If this is not the case, use the more general loop current method, described later in this paragraph. There is no need to write KVL equations for loops containing mesh currents assigned to current sources. Solve the resulting loop equations for the mesh currents. Determine any requested current or voltage in the circuit using the mesh currents. in order to easy for you to answer the problem.

Sabado, Agosto 2, 2014

BLOG 6 - ELECTRICAL CIRCUIT 1

This blog contain about the WYE - DELTA TRANSFORMATION 

WYE - DELTA TRANSFORMATION
A delta-wye transformer is a type of three-phase electric power transformer design that employs delta-connected windings on its primary and wye/star connected windings on its secondary. A neutral wire can be provided on wye output side. It can be a single three-phase transformer, or built from three independent single-phase units. An equivalent term is delta-star transformer. Delta-wye transformers are common in commercial, industrial, and high-density residential locations, to supply three-phase distribution systems.

.


WYE to DELTA and DELTA to WYE CONVERSION

In many circuits, resistors are neither in series nor in parallel, so the rules for series or parallel circuits described in previous chapters cannot be applied. For these circuits, it may be necessary to convert from one circuit form to another to simplify the solution. Two typical circuit configurations that often have these difficulties are the wye (Y) and delta ( D ) circuits. They are also referred to as tee (T) and pi ( P ) circuits, respectively.

Delta and wye circuits:


And the equations for converting from delta to wye:





The equations can be presented in an alternate form based on the total resistance (Rd) of R1, R2, and R3 (as though they were placed in series):

Rd = R1+R2+R3

and:

RA = (R1*R3) / Rd

RB = (R2*R3) / Rd

RC = (R1*R2) / Rd

Wye and delta circuits:


And the equations for converting from wye to delta:

 

An alternate set of equations can be derived based on the total conductance (Gy) of RA, RB, and RC (as though they were placed in parallel):

Gy = 1/RA+1/RB+1/RC

and:

R1 = RB*RC*Gy

R2 = RA*RC*Gy

R3 = RA*RB*Gy

LEARNING 

This week we tackled about the wye - delta transformation. I learned that the Wye networks are some times called T networks and Delta networks are occasionally called P networks. Wye and Delta Networks have 3 terminal arrangements  commonly used in power systems. T and P 2 of the terminals are connects at one node. The node is a distributed node in the case of the P network.