Super Position Theorem

the basic concept of superposition is to analyze the circuit, one source at a time. by using the superposition in the circuit we remove all the independent sources except one and analyze the circuit for that source.

To apply the superposition theorem to calculate the current through resistor R₁ in the two loop circuit shown. the individual current supplied by each battery is calculated with the other battery replaced by a short circuit.

• the strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active. 

Example:

  

• Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect and one for the circuit with only the 7 volt battery in effect:

•  When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire.
• Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:

  

• Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:

• When superimposing these values of voltage and current, we have to be very careful to consider polarity (voltage drop) and direction (electron flow),
• Applying these superimposed voltage figures to the circuit, the end result looks something like this.

• Currents add up algebraically as well, and can either be superimposed as done with the resistor voltage drops, or simply calculated from the final voltage drops and respective resistances (I=E/R). Either way, the answers will be the same. 

Process of using Superposition

• First of all make sure the circuit is a linear circuit; or a circuit where Ohm’s law implies, because Superposition theorem is applicable only to linear circuits and responses.
• Replace all the voltage and current sources on the circuit except for one of them.While replacing a Voltage source or Current Source replace it with their internal resistance or impedance. If the Source is an Ideal source  or internal impedance is not given then replace a Voltage source with a short ; so as to maintain a 0 V potential difference between two terminals of the voltage source. And replace a Current source with an Open ; so as to maintain a 0 Amps Current between two terminals of the current source.
• Determine the branch responses or voltage drop and current on every branches simply by using KCL , KVL or Ohm’s Law.
• Repeat step 2 and 3 for every source the circuit have.
• Now algebraically add the responses due to each source on a branch to find the response on the branch due to the combined effect of all the sources.
  
  

Some Learnings:

• The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they'll do with all power sources in effect.
• To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open circuit.

Linearity 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 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 = v₁ + v₂

We can say that a resistor is a linear element. Because the voltage-current relationship satisfies both the additivity and the homogeneity properties.

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.

The Source Transformation

Source Transformation - is a circuit that transforming voltage to current and vice versa. and it is a application of the Thevenin's Theorem and the Norton Theorem.

Performing a source transformation consists of using Ohm's law to take an existing voltage source in series with a resistance, and replace it with a current source in parallel with the same resistance. Remember that Ohm's law states that a voltage on a material is equal to the material's resistance times the amount of current through it (V=IR). Since source transformations are bilateral, one can be derived from the other. 

- Specifically, source transformations are used to exploit the equivalence of a real current source and a real voltage source, such as a battery. Application of Thévenin's theorem and Norton's theorem gives the quantities associated with the equivalence. Specifically, suppose we have a real current source I, which is an ideal current source in parallel with an-impedance.

Examples:

Sample Problem:

Some Learnings:

• Before solving the unknown, we transform 1st the sources from voltage to current and vice versa.
• Source transformation can be applied if the voltage is in series with the resistor and the current is in parallel with the resistor.
•  Remember that Ohm's law states that a voltage on a material is equal to the material's resistance times the amount of current through it (V=IR). 

Another Example:

Mesh Analysis

(Week 8)

Mesh Current is a current that loops around the essential mesh and the equations are set solved in terms of them. A mesh current may not correspond to any physically flowing current, but the physical currents are easily found from them. It is usual practice to have all the mesh currents loop in the same direction. This helps prevent errors when writing out the equations. The convention is to have all the mesh currents looping in a clockwise direction.

Mesh Set-up

Each mesh produces one equation. These equations are the sum of the voltage drops in a complete loop of the mesh current. For problems more general than those including current and voltage sources, the voltage drops will be the impedance of the electronic component multiplied by the mesh current in that loop.

If a voltage source is present within the mesh loop, the voltage at the source is either added or subtracted depending on if it is a voltage drop or a voltage rise in the direction of the mesh current. For a current source that is not contained between two meshes, the mesh current will take the positive or negative value of the current source depending on if the mesh current is in the same or opposite direction of the current source. The following is the same circuit from above with the equations needed to solve for all the currents in the circuit.

Steps

1.  Assign mesh currents i₁, i₂, . . . , i_n to the n meshes.

2.  Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.

3.  Solve the resulting n simultaneous equations to get the mesh currents.

Supermesh occurs when a current source is contained between two essential meshes. The circuit is first treated as if the current source is not there. This leads to one equation that incorporates two mesh currents. Once this equation is formed, an equation is needed that relates the two mesh currents with the current source. This will be an equation where the current source is equal to one of the mesh currents minus the other. The following is a simple example of dealing with a supermesh.

 Dependent Source is a current source or voltage source that depends on the voltage or current of another element in the circuit. When a dependent source is contained within an essential mesh, the dependent source should be treated like an independent source. After the mesh equation is formed, a dependent source equation is needed. This equation is generally called a constraint equation. This is an equation that relates the dependent source’s variable to the voltage or current that the source depends on in the circuit. The following is a simple example of a dependent source.

Note:

•   The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh currents.
•   A supermesh has no current of its own.
•  A supermesh requires the application of both KVL and KCL.

By solving the Supermesh

1. Identify the supermesh
2. write a mesh equation for the super mesh.
3. Write mesh equations for any addition meshes
4. Write down any equations from KCL
5. Solve

matrix

Learnings:

• Circuit with mesh currents labeled as i₁, i₂, and i₃. The arrows show the direction of the mesh current.
• Supermesh occurs because the current source is in between the essential meshes.
• Circuit with dependent source. i_x is the current upon which the dependent source depends
•  In matrix you must get first the equations in the problem to get the value of the delta.  

Its All About Wye-Delta and Delta-Wye

Basic  Y- Delta Transformation

The transformation is used to establish equivalence for networks with three terminals. Where three elements terminate at a common node and none are sources, the node is eliminated by transforming the impedances. For equivalence, the impedance between any pair of terminals must be the same for both networks. The equations given here are valid for complex as well as real impedances.

figure1:

Figure 2 shows Wye (Y) and Tee (T) networks and figure 3 shows delta (∆) and pie (Π) network.

These networks are the equivalent of large network. I will discuss here how to transform wye to delta and delta to wye.

Delta to wye conversion

It is easy to work with wye network. If we get delta network, we convert it to wye to work easily. To obtain the equivalent resistance in the wye network from delta network we compare the two networks and we confirm that they are same. Now we will convert figure 3 (a) delta network to figure 2 (a) wye network.

From figure 3 (a) for terminals 1 and 2 we get,

R₁₂ (∆) = R_b || (R_a + R_b)

From figure 2 (a) for terminals 1 and 2 we get,

R₁₂(Y) = R₁ + R₃

Setting wye and delta equal,

R₁₂(Y) = R₁₂ (∆) we get,

Equations (v), (vi) and (vii) are the equivalent resistances for transforming delta to wye conversion. We do not need to memorize these equations. Now we create an extra node shown in figure 4 and follow the conversion rule,

Now we will solve a problem how to convert delta to wye network. It will give clear concept.

Problem: convert the delta network to wye network.

From equation (v), (vi) and (vii) we can find the equivalent resistance of wye network,

The equivalent Y network configuration is shown in figure 6.

Wye to delta conversion

For conversion to wye network to delta network adding equations (v), (vi) and (vii) we get,

 R₁R₂ + R₂R₃ + R₃R₁ = R_aR_bR_c(R_a +R_b + R_c)/(R_a + R_b + R_c)²

                                = R_aR_bR_c/(R_a + R_b + R_c) —————— (ix)

Dividing equation (ix) by each of the equations (v), (vi) and (vii) we get,

R_a = R₁R₂ + R₂R₃ + R₃R₁/ R₁

R_b = R₁R₂ + R₂R₃ + R₃R₁/ R₂

R_c = R₁R₂ + R₂R₃ + R₃R₁/ R₃

For Y to delta conversion the rule is followed below,

Learnings from this topic:

• Each resistor in the delta network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor.
• Each resistor in the Y network is the product of the resistors in the two adjacent Del branches, divided by the sum of the three Del resistors.

"Week Five"

Nodal analysis 
    
    In this method, we set up and solve a system of equations in which the unknowns are the voltages at the principal nodes of the circuit. From these nodal voltages the currents in the various branches of the circuit are easily determined.

In analyzing a circuit using Kirchhoff's circuit laws, one can either do nodal analysis using Kirchhoff's current law (KCL) or mesh analysis using Kirchhoff's voltage law (KVL). Nodal analysis writes an equation at each electrical node, requiring that the branch currents incident at a node must sum to zero. The branch currents are written in terms of the circuit node voltages. As a consequence, each branch constitutive relation must give current as a function of voltage; an admittance representation. For instance, for a resistor, I_branch = V_branch * G, where G (=1/R) is the admittance (conductance) of the resistor.

The steps in the nodal analysis method are:

• Count the number of principal nodes or junctions in the circuit. Call this number n. (Aprincipal node or junction is a point where 3 or more branches join. We will indicate them in a circuit diagram with a red dot. Note that if a branch contains no voltage sources or loads then that entire branch can be considered to be one node.)
• Number the nodes N₁, N₂, . . . , N_n and draw them on the circuit diagram. Call the voltages at these nodes V₁, V₂, . . . , V_n, respectively.
• Choose one of the nodes to be the reference node or ground and assign it a voltage of zero.
• For each node except the reference node write down Kirchoff's Current Law in the form"the algebraic sum of the currents flowing out of a node equals zero". (By algebraic sum we mean that a current flowing into a node is to be considered a negative current flowing out of the node.)
  
  For example, for the node to the right KCL yields the equation:

  I_a + I_b + I_c = 0

  
  
  Express the current in each branch in terms of the nodal voltages at each end of the branch using Ohm's Law (I = V / R). Here are some examples:
  
  
  
  The current downward out of node 1 depends on the voltage difference V1 - V3 and the resistance in the branch.
  
  
  
  
  In this case the voltage difference across the resistance is V1 - V2 minus the voltage across the voltage source. Thus the downward current is as shown.
  
  
  
  
  In this case the voltage difference across the resistance must be 100 volts greater than the difference V1 - V2. Thus the downward current is as shown.

• 
  The result, after simplification, is a system of m linear equations in the m unknown nodal voltages (where m is one less than the number of nodes; m = n - 1). The equations are of this form:

  

  where G₁₁, G₁₂, . . . , G_mm and I₁, I₂, . . . , I_m are constants.
  
  Alternatively, the system of equations can be gotten (already in simplified form) by using the inspection method.
• Solve the system of equations for the m node voltages V₁, V₂, . . . , V_m using Gaussian elimination or some other method.

Example 1: Use nodal analysis to find the voltage at each node of this circuit. 
Solution:

• Note that the "pair of nodes" at the bottom is actually 1 extended node. Thus the number of nodes is 3.
  
  
• We will number the nodes as shown to the right.
• We will choose node 2 as the reference node and assign it a voltage of zero.
• Write down Kirchoff's Current Law for each node. Call V₁ the voltage at node 1, V₃ the voltage at node 3, and remember that V₂ = 0. The result is the following system of equations:

  

  The first equation results from KCL applied at node 1 and the second equation results from KCL applied at node 3. Collecting terms this becomes:

  

  This form for the system of equations could have been gotten immediately by using the inspection method.
• Solving the system of equations using Gaussian elimination or some other method gives the following voltages:

  V₁=68.2 volts and V₃=27.3 volts

Basic case

Current through resistor R₁: (V₁ - V_S) / R₁The only unknown voltage in this circuit is V₁. There are three connections to this node and consequently three currents to consider. The direction of the currents in calculations is chosen to be away from the node.

1. Current through resistor R₂: V₁ / R₂
2. Current through current source I_S: -I_S

With Kirchhoff's current law, we get:

This equation can be solved in respect to V₁:

Super Node

In this circuit, we initially have two unknown voltages, V₁ and V₂. The voltage at V₃ is already known to be V_B because the other terminal of the voltage source is at ground potential.

The current going through voltage source V_A cannot be directly calculated. Therefore we can not write the current equations for either V₁ or V₂. However, we know that the same current leaving node V₂ must enter node V₁. Even though the nodes can not be individually solved, we know that the combined current of these two nodes is zero. This combining of the two nodes is called the supernode technique, and it requires one additional equation: V₁ = V₂ + V_A.

The complete set of equations for this circuit is:

By substituting V₁ to the first equation and solving in respect to V2, we get: