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. 

 

Independent current sources can be turned into independent voltage sources, and vice-versa, by methods called "Source Transformations." These transformations are useful for solving circuits.

Voltage Source Transformation

We will first go over voltage source transformation, the transformation of a circuit with a voltage source to the equivalent circuit with a current source.

In order to get a visual example of this, let's take the circuit below which has a voltage source as its power source:

Using source transformation, we can change or transform this above circuit with a voltage power source and a resistor, R, in series, into the equivalent circuit with a current source with a resistor, R, in parallel, as shown below:

We transform a voltage source into a current source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula, I=V/R.

Example

Let's do an actual example to demonstrate the mathematics of ohm's law, using the circuit shown below:

Here, we have a circuit with a voltage source of 10V with a resistor in series of 2Ω.

To calculate what the equivalent current source would be, we calculate it using the formula: I=V/R, which is I= 10V/2Ω= 5A. So the equivalent circuit would be:

The new power source is now a 5A current source. The resistor value, however, as with all source transformations stays the same. The only thing that changes is it is now in parallel for a current source transformation.

Try out our calculator below. With this calculator, you can try out as many examples as you want. The calculator does source transformations and presents the new circuits with the new values.

---

Current Source Transformation

We will now go over current source transformation, the transformation of a circuit with a current source to the equivalent circuit with a voltage source.

In order to get a visual example of this, let's take the circuit below which has a current source as its power source:

Using source transformation, we can change or transform this above circuit with a current power source and a resistor, R, in parallel, into the equivalent circuit with a voltage source with a resistor, R, in series, as shown below:

We transform a current source into a voltage source by using ohm's law. A voltage source can be changed into a current source by using ohm's formula, V= IR.

Example

Let's do an actual example to demonstrate the mathematics of ohm's law, using the circuit shown below:

Here, we have a circuit with a current source of 2A with a resistor in parallel of 3Ω.

To calculate what the equivalent current source would be, we calculate it using the formula: V= IR, which is V= 2A*3Ω = 6V. So the equivalent circuit would be:

The new power source is now a 6-volt voltage source. The resistor value, however, again, as with all source transformations stays the same. The only thing that changes is it is now in series for a voltage source transformation.

Again, you can try as many examples as you would like if our calculators below, which do source transformations.

Some Learnings:

In Voltage Source

This voltage source transformation calculator transforms the above circuit with a voltage source and resistor in series into the equivalent circuit with a current source with a resistor in parallel.

The value of the current source is calculated according to ohm's law, I=V/R, current= 

voltage/resistance. 

The value of the power source will be different; however, the resistor value always remains unchanged.  

In Current Source

This current source transformation calculator transforms the above circuit with a current source and resistor in parallel into the equivalent circuit with a voltage source with a resistor in series.

The value of the voltage source is calculated according to ohm's law, V= IR, voltage= current * resistance. 

The value of the power source will be different; however, the resistor value, just as with voltage source transformation, always remains unchanged.
 

 

Superposition Theorem

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman's certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.
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. 

Another prerequisite for Superposition Theorem is that all components must be “bilateral,” meaning that they behave the same with electrons flowing either direction through them. Resistors have no polarity-specific behavior, and so the circuits we've been studying so far all meet this criterion.

The Superposition Theorem finds use in the study of alternating current (AC) circuits, and semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with DC. Because AC voltage and current equations (Ohm's Law) are linear just like DC, we can use Superposition to analyze the circuit with just the DC power source, then just the AC power source, combining the results to tell what will happen with both AC and DC sources in effect. For now, though, Superposition will suffice as a break from having to do simultaneous equations to analyze a circuit.

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 (break).

 

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. 

"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.

Let's start by drawing a general circuit consisting of a source and a load, as a block diagram:

Let's say that the source is a collection of voltage sources, current sources and resistances, while the load is a collection of resistances only. Both the source and the load can be arbitrarily complex, but we can conceptually say that the source is directly equivalent to a single voltage source and resistance (figure (a) below).

(a)	(b)

We can determine the value of the resistance R_s and the voltage source, v_s by attaching an independent source to the output of the circuit, as in figure (b) above. In this case we are using a current source, but a voltage source could also be used. By varying i and measuring v, both v_s and R_s can be found using the following equation:

There are two variables, so two values of i will be needed.We can easily see from this that if the current source is set to zero (equivalent to an open circuit), then v is equal to the voltage source, v_s. This is also called the open-circuit voltage, v_oc.
This is an important concept, because it allows us to model what is inside a unknown (linear) circuit, just by knowing what is coming out of the circuit. This concept is known as Thévenin's Theorem after French telegraph engineer Léon Charles Thévenin, and the circuit consisting of the voltage source and resistance is called the Thévenin Equivalent Circuit.

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.