The Finals

"Thevenin's Theorem"

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load. The qualification of “linear” is identical to that found in the Superposition Theorem, where all the underlying equations must be linear (no exponents or roots). If we're dealing with passive components (such as resistors, and later, inductors and capacitors), this is true. However, there are some components (especially certain gas-discharge and semiconductor components) which are nonlinear: that is, their opposition to current changes with voltage and/or current. As such, we would call circuits containing these types of components, nonlinear circuits.

Thevenin's Theorem is especially useful in analyzing power systems and other circuits where one particular resistor in the circuit (called the “load” resistor) is subject to change, and re-calculation of the circuit is necessary with each trial value of load resistance, to determine voltage across it and current through it.

Thevenin Equivalent Circuit

             

       Any two-terminal linear network, composed of voltage sources, current sources, and resistors, can be replaced by an equivalent two-terminal 

network consisting of an independent voltage source in series with a resistor.

Thevenin Equivalent Circuit Any two-terminal linear network, composed of voltage sources, current sources, and resistors, can be replaced by an equivalent two-terminal network consisting of an independent voltage source in series with a resistor.

                                                        Steps to follow for thevenin's Theorem

                                         

• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

                                             

                                                         "Norton's  Theorem"

Norton's theorem states that a network consists of  several voltage sources, current sources and resistors with two terminals, is electrically equivalent to an ideal current source " I_NO" and a single parallel  resistor, R_NO. The theorem can be applied to both A.C and D.C cases. The Norton equivalent of a circuit consists of an ideal current source in parallel with an ideal impedance (or resistor for non-reactive circuits).

      

The Norton equivalent circuit is a current source with current "I_NO" in parallel with a resistance R_NO.To find its Norton equivalent circuit,

1. Find the Norton current "I_NO". Calculate the output current, "I_AB", when a short circuit is the load (meaning 0 resistances between A and B). This is INo.
2. Find the Norton resistance RNo. When there are no dependent sources (i.e., all current and voltage sources are independent), there are two methods of determining the Norton impedance RNo.

• Calculate the output voltage, VAB, when in open circuit condition (i.e., no load resistor — meaning infinite load resistance). RNo equals this VAB divided by INo.
  or
• Replace independent voltage sources with short circuits and independent current sources with open circuits. The total resistance across the output port is the Norton impedance RNo.
  However, when there are dependent sources the more general method must be used. This method is not shown below in the diagrams.
• Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. The quotient of this voltage divided by the 1 A current is the Norton impedance RNo. This method must be used if the circuit contains dependent sources, but it can be used in all cases even when there are no dependent sources.

Example 1:-

Consider this circuit-

To find the Norton’s equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.

When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:

With A-B Shorted :

If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.

Find the Equivalent Resistance (Rs):

10Ω Resistor in parallel with the 20Ω Resistor

Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the following Nortons equivalent circuit.

Nortons equivalent circuit.

Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.

Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:

The voltage across the terminals A and B with the load resistor connected is given as:

Then the current flowing in the 40Ω load resistor can be found as:

Steps to follow for Norton's Theorem

• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.

Learnings:

1.Norton's theorem and Thevenin's theorem are equivalent,and the equivalence leads to source transformation in electrical circuits.

2.For an electric-circuit,the equivalence is given by  V_Th= Ith x Rth ie Thevenin's volage=Norton's current x Thevenin's resistance.

3.The applications of Norton theorem is similar to that of Thevenin's theorem.The main application is nothing but the simplification of electrical circuit by introducing source transoformation.

Maximum Power Transfer Theorem

Suppose we have a voltage source or battery that's internal resistance is Ri and a load resistance RL is connected across this battery . Maximum power transfer theoremdetermines the value of resistance RL 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.

Power delivered to the load resistance,

To find the maximum power, differentiate the above expression with respect to resistance RLand equate it to zero. Thus,

Thus in this case, the maximum power will be transferred to the load when load resistance is just equal to internal resistance of the battery .
Maximum power transfer theorem can be applicable in complex network as follows-

A resistive load in a resistive network will abstract maximum power when the load resistance is equal to the resistance viewed by the load as it looks back to the network. Actually this is nothing but the resistance presented to the output terminals of the network. This is actually Thevenin equivalent resistance as we explained in Thevenin's theorem if we consider the whole network as a voltage source.

Learnings:

• The Maximum Power Transfer Theorem states that the maximum amount of power will be dissipated by a load resistance if it is equal to the Thevenin or Norton resistance of the network supplying power.
• The Maximum Power Transfer Theorem does not satisfy the goal of maximum efficiency.

                                                     Capacitors and Inductors

It is fairly simple to understand what happens when we apply a signal to a resistance. This is because resistance obeys Ohm's Law. However, the behaviour of Capacitors and Inductors is rather more complicated. Fortunately, we can use complex numbers and the concept of Reactance to help us understand and analyse what happens when we apply signals to circuits which contain capacitors and/or resistors.

Reactance of a Capacitor.

Consider first what happens when we apply sinewave voltage to a capacitor.


We can represent the voltage applied to the capacitor as a complex quantity

although only the real part of this is 'visible', of course.
The current flowing 'through' the capacitor (i.e. the rate at which we have to remove charge from one plate and put it onto the other) is proportional to how quickly we are changing the voltage at any instant. We can (as in the 'First 11' section of this guide) therefore use a form of differential calculus to analyse what is happening. Here, instead, we'll just say that the Capacitor's behaviour can be defined in terms of a Reactance, X_C, whose value is

Where C is the capacitance value (in Farads) and f is the frequency of the wave we're applying.

To work out the current we can now use this in exactly the same way as we'd use the resistance of a resistor. Hence we can say that the current will be given by

Looking at this we can see that the 'j' which made it 'imaginary' has meant that the real (i.e. 'visible') part of the current is the Sin function, when the real part of the applied voltage is the Cos function. Hence we have a current which is out of step (by 90 degrees) with the applied voltage. This is the result we'd expect from using differential calculus since we want the peaks of the current waveform to occur when the voltage is changing most swiftly (when it is passing though zero). Note also that the magnitude of the current depends on the signal frequency. This is what we'd expect as a higher frequency means the voltage will have to change more rapidly.

Reactance of an Inductor

We can define the behaviour of an inductor in a similar way. In this case it is the rate of change of the current which is proportional to the applied voltage. So when we apply a voltage V_L which is exactly the same as the V_C we used earlier we get the behaviour shown in the diagram below.


To describe this behaviour can say that the reactance of the inductor, X_L, will be

where L is the inductance value in Henries.

As a result we get a current of

Comparing this result with the capacitor we can see that in both cases an input cosine voltage waveform produces a sine wave current variation, but the signs of the currents produced are different. Because of this the current through the capacitor is said to lead the applied voltage, and that through the inductor is said to lag the applied voltage. We can see why these terms are used by looking at the waveforms in the above diagrams which show the sorts of patterns we'd see using an oscilloscope. The difference arises because the capacitor reactance contains a '1/j' whereas the inductor reactance has a 'j'. As a result, when we divide the voltage by the reactance this means we get a 'jj = -1' for the sine term when using a capacitor and a 'j/j = 1' for the sine term when using an inductor. The terms, 'lead' and 'lag' are used generally to indicate the 'sign' of the reactance of a circuit (i.e. capacitive or inductive).

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.