Introduction

In any electric circuit, the electrical energy from the power supply is delivered to the load where it is converted into a useful work. Practically, the entire supplied power will not be present at the load due to the heating effect and other constraints in the network. Therefore, there exists a certain difference between drawing and delivering powers.

The size of the load always affects the amount of power transferred from the supply source, i.e., any change in the load resistance results a change in the power transferred to the load. Thus, the Maximum Power Transfer Theorem ensures the ideal condition to transfer the maximum power to the load. Let us see ‘how’.

Maximum Power Transfer Theorem Statement

The Maximum Power Transfer Theorem states that in a linear, bilateral DC network, Maximum Power is delivered to the load when the load resistance is equal to the internal resistance of the source.

If it is an independent voltage source, then its series resistance (internal resistance R_S) or if it is independent current source, then its parallel resistance (internal resistance R_S) must equal to the load resistance R_L to deliver maximum power to the load.

Proof of Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem ensures the value of the load resistance, at which the maximum power is transferred to the load.

Consider the below DC two terminal network (left side circuit). The condition for maximum power is determined by obtaining the expression of power absorbed by load using mesh or nodal current methods and then deriving the resulting expression with respect to load resistance R_L.

This is quite a complex procedure. But in the previous tutorials, we have seen that the complex part of the network can be replaced with a Thevenin’s equivalent as shown below.

The original two terminal circuit is replaced with a Thevenin’s equivalent circuit across the variable load resistance. The current through the load for any value of load resistance is

Form the above expression, the power delivered depends on the values of R_TH and R_L. However, as the Thevenin’s equivalent is a constant, the power delivered from this equivalent source to the load entirely depends on the load resistance R_L. To find the exact value of RL, we apply differentiation to P_L with respect to R_L and equating it to zero as shown below:

Therefore, this is the condition of matching the load where the maximum power transfer occurs when the load resistance is equal to the Thevenin’s resistance of the circuit. By substituting the R_TH = R_L in the previous equation, we get:

The maximum power delivered to the load is,

Total power transferred from source is:

Hence, the maximum power transfer theorem expresses the state at which maximum power is delivered to the load i.e., when the load resistance is equal to the Thevenin’s equivalent resistance of the circuit. Below figure shows a curve of power delivered to the load with respect to the load resistance.

Note that the power delivered is zero when the load resistance is zero as there is no voltage drop across the load during this condition. Also, the power will be maximum, when the load resistance is equal to the internal resistance of the circuit (or Thevenin’s equivalent resistance). Again, the power is zero as the load resistance reaches to infinity as there is no current flow through the load.

Power Transfer Efficiency

We must remember that this theorem states only maximum power transfer but not for maximum efficiency. If the load resistance is smaller than source resistance, power dissipated at the load is reduced while most of the power is dissipated at the source, then the efficiency becomes lower.

Consider the total power delivered from source equation (equation 2), in which the power is dissipated in the equivalent Thevenin’s resistance R_TH by the voltage source V_TH.

Therefore, the efficiency under the condition of maximum power transfer is:

 Efficiency = Output / Input × 100 

 = I_L² R_L / 2 I_L² R_L × 100 

 = 50 % 

Hence, at the condition of maximum power transfer, the efficiency is 50%, that means only half of the generated power is delivered to the load and at other conditions, a small percentage of power is delivered to the load, as indicated in efficiency verses maximum power transfer the curves below.

For some applications, it is desirable to transfer maximum power to the load than achieving high efficiency such as in amplifiers and communication circuits.

On the other hand, it is desirable to achieve higher efficiency than maximized power transfer in case of power transmission systems, where a large load resistance (much larger value than internal source resistance) is placed across the load. Even though the efficiency is high the power delivered will be less in those cases.

Maximum Power Transfer Theorem for AC Circuits

In an active network, it can be stated that the maximum power is transferred to the load when the load impedance is equal to the complex conjugate of an equivalent impedance of a given network as viewed from the load terminals.

Consider the above Thevenin’s equivalent circuit across the load terminals in which the current flowing through the circuit is given as:

The power delivered to the load,

For maximum power the derivative of the above equation must be zero, after simplification we get

Putting the above relation in equation 1, we get

This maximum power transferred,  P_MAX = V²_TH / 4 R_TH or V²_TH/ 4 R_L 

Applying Maximum Power Transfer Example to DC Circuit

Consider the below circuit to which we determine the value of the load resistance that receives the maximum power from the supply source and the maximum power under the maximum power transfer condition.

Disconnect the load resistance from the load terminals ‘a ‘and ‘b’. To represent the given circuit as Thevenin’s equivalent, we have to determine the Thevenin’s voltage V_TH and Thevenin’s equivalent resistance R_TH.

The Thevenin’s voltage or voltage across the terminals ab is  V_ab = V_a – V_b 

 V_a = V × R2 / (R1 + R2) 

 = 30 × 20 /×(20 + 15) 

 = 17.14 V 

 V_b = V × R4/ (R3 + R4) 

 = 30 × 5 /(10 + 5) 

 = 10 V 

 V_ab = 17.14 – 10 

 = 7.14 V 

 V_TH = V_ab = 7.14 Volts 

Calculate the Thevenin’s equivalent resistance R_TH by replacing sources with their internal resistances (here, let us assume that voltage source has zero internal resistance so it becomes a short circuit).

Thevenin’s equivalent resistance or resistance across the terminals ab is

 R_TH = Rab = [R1R2 / (R1 + R2)] + [R3R4 /(R3 + R4)] 

 = [(15 × 20) / (15 + 20)] + [(10 × 5) / (10+ 5)] 

 = 8.57 + 3.33 

 R_TH = 11.90 Ohms 

The Thevenin’s equivalent circuit with above calculated values by reconnecting the load resistance is shown below.

From the maximum power transfer theorem, R_L value must equal to the R_TH to deliver the maximum power to the load.

Therefore,  R_L = R_TH= 11.90 Ohms 

And the maximum power transferred under this condition is,

 P_MAX = V²_TH / 4 R_TH

 = (7.14)² / (4 × 11.90) 

 = 50.97 / 47.6 

 = 1.07 Watts 

Applying Maximum Power Transfer to AC circuit

The below AC network consists of load impedance Z_L of which both reactive and resistive parts can be varied. Hence, we have to determine the load impedance value at which the maximum power delivered from the source and also the value of the maximum power.

To find the value of load impedance, first, we find the Thevenin’s equivalent circuit across the load terminals. For finding Thevenin’s voltage, disconnect the load impedance as shown in below figure.

By voltage divider rule,

 V_TH = 20∠0 × [j6 / (4 + j6)] 

 = 20∠0 ×[6∠90 / 7.21∠56.3] 

 = 20∠0 × 0.825∠33.7 

 V_TH = 16.5∠33.7 V 

By shorting the voltage source, we calculate the Thevenin’s equivalent impedance of the circuit as shown in figure.

Therefore,

 Z_TH = (4 × j6) / (4 + j6) 

 = (4 × 6∠90) / (7.21∠56.3) 

 = 3.33∠33.7 0r 2.77 + j1.85 Ohms 

Hence, the Thevenin’s equivalent circuit across the load terminals is shown in below.

Therefore to transfer the maximum power to the load, the value of the load impedance should be

 Z_L = Z_TH

 = 2.77 – j1.85 ohms 

The maximum power delivered, P_MAX

 = V²_TH / 4 R_TH

 = (16.5)²/4(2.77) 

 = 272.25 / 11.08 

 = 24.5 W 

Practical Application of Maximum Power Transfer Theorem

Consider the practical example of a speaker with an impedance of 8 ohms. It is driven by an audio amplifier with an internal impedance of 500 ohms. The Thevenin’s equivalent circuit is also shown in figure.

According to the maximum power transfer theorem, the power is maximized at the load if the load impedance is 500 ohms (same as internal impedance). Or else internal resistance has to be changed to 8 ohms to achieve the Maximum Power Transfer condition. However, it is not possible to change either of them.

So, it is an impedance mismatch condition and it can be overcome by using an impedance matching transformer with its impedance transformation ratio of 500:8.

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Before going in to details of the Star Connection, Delta Connection and comparing those two, let us have a very brief note on three – phase electric power.

A single phase system consists of just two conductors (wires): one is called the phase (sometimes line, live or hot), through which the current flows and the other is called neutral, which acts as a return path to complete the circuit.

In a three – phase system, we have a minimum of three conductors or wires carrying AC voltages. It is more economical to transmit power using a 3 – phase power supply when compared to a single phase power supply as a three – phase supply can transmit three times the power with just three conductors when compared to a two – conductor single – phase power supply.

Hence, most of the power generated and distributed is actually a 3 – phase power (but majority of households will receive a single phase supply). To know more about single phase and three phase, read the Difference Between Single Phase and Three Phase Power Supplies tutorial.

Further, the three – phase electric power system can be arranged in two ways. They are: Star (also called Y or Wye) and Delta (Δ).

Star Connection

In a Star Connection, the 3 phase wires are connected to a common point or star point and Neutral is taken from this common point. Due to its shape, the star connection is sometimes also called as Y or Wye connection.

If only the three phase wires are used, then it is called 3 Phase 3 Wire system. If the Neutral point is also used (which often it is), the it is called 3 Phase 4 Wire system. The following image shows a typical Star Connection.

Delta Connection

In a Delta Connection, there are only 3 wires for distribution and all the 3 wires are phases (no neutral in a Delta connection). The following image shows a typical Delta Connection.

Comparison between Star and Delta Connections

Let us understand more about these connections by using the following Comparison between Star and Delta Connections.

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Introduction

In contrast to the Thevenin’s theorem, Norton’s theorem replaces the part of the circuit with an equivalent circuit that constitute a current source and a parallel resistance. This theorem is an extension of the Thevenin’s theorem, proposed by E. L. Norton in 1926.

Similar to the Thevenin’s theorem, it is also used to calculate load variables such as load voltage, load current and load power with simple calculations over other circuit reduction techniques. Thus, this theorem is also called as the dual of the Thevenin’s theorem. In most of the cases the choice of load resistance to transfer maximum power to the load is decided by either Thevenin’s or Norton’s theorems.

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Norton’s Theorem Statement

Norton’s theorem states that , any two terminal linear network that constitute independent sources and linear resistances can be replaced with an equivalent circuit , consisting of a current source with a parallel resistor.

Magnitude of this equivalent current source is equal to the short circuit current flowing through the load terminals and the equivalent resistance is the resistance at the load terminals, when all the sources in a given circuit are replaced by their internal resistances.

In below figure a part of a network , constituting of sources (either voltage or current or both) and resistances is replaced with a current source and a parallel resistor such that current flowing through the load is same in both cases.

For an AC circuit it can be stated as , any active two terminal network consisting of independent sources and impedance can be replaced with an equivalent circuit consisting of a constant current source with a parallel impedance.

The value of the current source is equal to the current flow through the short circuited terminals of the network. And the parallel impedance is the equivalent impedance viewed from the short circuited terminals when all the sources are replaced with their internal impedance.

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Steps to Analyse the Norton’s Theorem

To find the load variables using Norton’s theorem, Norton’s equivalent parameters have to be determined. Those are Norton current or magnitude of equivalent current source and Norton resistance Rn or impedance ZN. The following steps are required to determine them.

1. Consider the given circuit and short the load terminals after disconnecting the load resistance (or impedance in case of AC circuit) from output or load terminals.

2. Determine the short circuit current IN, through the shorted terminals by applying any of the circuit reduction techniques like Mesh analysis or nodal analysis or superposition theorem. Or simply measure the load current using the ammeter experimentally.

3. Redraw the given circuit by replacing all the practical sources in the circuit with their internal voltages or simply short circuit voltages sources and open circuit the current sources . And also make sure to open or remove the short circuited terminals of the load.

4. Calculate the resistance (or impedance) that exists between the load terminals by looking from the load terminals. This resistance is equivalent Norton’s resistance RN or (impedance ZN).

5. Insert the resistance (or impedance) in parallel with a current source IN which forms a Norton’s equivalent circuit.

6. Now reconnect the load to the Norton’s equivalent circuit and calculate the current, voltage and power associated with the load as

In DC circuit,

Load current, IL = IN × [RN / (RL + RN)]

Load Voltage, VL = IL × RL

Power dissipated at the load, P = IL2 × RL

In AC circuit,

Load current, IL = IN × [ZN / (ZL + ZN)]

Load Voltage, VL = IL × ZL

Power dissipated at the load, P = IL2 × ZL

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Example for Finding Equivalent Circuit to DC Circuit

Let us consider the same DC circuit in Thevenin’s theorem example to apply the Norton’s theorem to find the current flow through branch ab i.e., through the load resistance RL = R2 = 2 ohms.

1. Disconnect the load resistance and short the load terminals a and b. Represent the current flow direction in every loop as shown in figure.

2. Apply the mesh analysis for each loop to find the current flow IN through the shorted terminals.

By applying KVL to loop 1 we get

6 – (I1 – I2) R4 = 0

Substituting I2 = -4A

I1 = 6 – 16 / 4 = – 2.5 A

By applying KVL to Loop 3 we get

– I3R1 – (I3 – I2)R3 = 0

-4I3 – 6 (I3 + 4) = 0

– 10I3 = 24

I3 = – 2.4 A

Therefore In = I1 – I3

= -2.5 + 2.4

= 0.1A which is flowing from a to b.

3. Next step is to determine the equivalent resistance RN. To compute this resistance all sources have to be replaced with their internal resistances by removing the short terminals of the load .

Then the total resistance across the terminals a and b, RN = 10 × 4 / 10 + 4

= 2.85 Ohms

4. By placing above calculated current In in parallel with resistance Rn forms a Norton’s equivalent circuit as shown in figure. To determine the load variables we reconnect the load resistance across the load terminals.

Then Load current IL = IN × [RN / (RL + RN)]

= 0.1 × [2.85/ (2 + 2.85)]

= 0.05 Amps

With the above calculated values, the original circuit is similar to the below shown figure with the representation of load current at branch ab.

For different values of the load resistance the current flow is determined as

When RL = 8 ohms

IL = 0.1 × [2.85 / (8 + 2.85)]

= 0.02 A

When RL = 12 ohms

IL = 0.1 × [2.85 / (12+ 2.85)]

= 0.01 A.

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Relation between Norton’s and Thevinin’s Theorems

By comparing the above example with example of Thevenin’s example problem, we can observe that Norton’s equivalent circuit of a linear network constitute a Norton current source IN in parallel with a Thevenin’s resistance Rth.

Therefore it is possible to perform a source transformation of Thevenin’s equivalent circuit to get the Norton’s equivalent circuit or vice-versa.

The magnitude of the voltage source (Vth) and a series resistance (Rth) from Norton’s equivalent circuit using source transformation are determined as

Vth = RN × IN and

Rth = RN

For above example

Vth = 2.85 × 0.1

= 0.28 Volts.

Therefore, we can use any of these two methods to analyse the circuit in a simple way. However the advantages of Thevenin’s theorem are also applicable to Norton’s theorem as well. By using these methods one can find the current and voltage values of different load resistance values without doing any complex calculations again and again.

Hence, the Norton’s theorem aids the designing much easier based on the application. The use of these two theorems is decided by the application where these equivalents are required, such as current follower circuits (use Norton’s equivalent) and voltage amplifiers (Thevenin’s equivalent).

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Example of Finding Equivalent Circuit to AC Circuit

Consider the below AC circuit which was already analyzed using thevenin’s theorem. In this circuit we are going to find the current through the impedance 4+ 4j ohm using Norton’s theorem.

The above circuit consists of two voltage sources which can be transformed into the current source as

Is1 = Vs1/ Rs1

= 2∠0 / 1

= 2 A

Similarly

Is2 = Vs2/ Rs2

= 4∠0/ 2

= 2 A

Then the circuit becomes

For applying Norton’s theorem, we disconnect the load impedance and short the load terminals as shown in figure. Assume the current directions as represented in figure.

Consider the above figure as single node and the total current becomes 6 amperes and the total parallel combination of resistance is 0.574 ohms. This can be transformed into a voltage source for ease of finding the Norton’s current is given as

Vs = 6 ∠0 × 0.574

= 3.44∠0

Therefore, IN = VN / 0.574

= 3.44∠0 / 0.574

= 5.97∠0 A

The Norton’s equivalent impedance is equal to the circuit equivalent impedance, ZN = 0.574

Therefore the Load current across the 4 + j4 impedance is, IL = IN × [ZN / (ZL + ZN)]

= 5.97∠0× [0.574/ (4 + j4 + 0.574)]

= 3.42 / 6.07∠41.17

= 0.56 ∠-41.17 A

This value is identical to the value obtained in the case thevenin’s example of AC circuit. And hence Norton’s theorem is the dual of thevenin’s theorem. Limitations of Thevenin’s theorem are also applicable for norton’s theorem.

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Introduction

For many linear circuits, analysis is greatly simplified by the use of two circuit reduction techniques or theorems as Thevenin’s and Norton’s theorems. The Thevenin’s theorem is named after a French engineer, M. L. Thevenin’s in 1883 and Norton’s theorem after a scientist E. L. Norton.

By using these theorems a large or complex part of a network is replaced with a simple equivalent. With this equivalent circuit, we can easily make necessary calculations of current, voltage and power delivered to the load (as the original circuit delivers). This type of application ensures to select the best value of the load resistance. Let us see in detail about Thevenin’s theorem.

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Why Do We Use Thevenin’s Theorem?

In most of the applications, a network may consists of a variable load element, while other elements are constant. The best example is our household outlet which is connected to different appliances or loads. Therefore, if desired, it is necessary to calculate voltage or current or power in every element in a given circuit for every change in variable component.

This repeated procedure is somehow complicated and burdensome. Such repeated computations are avoided by introducing an equivalent circuit for a fixed part in circuit so that circuit analysis with change in load resistance becomes easy.

Consider the above simple DC circuit where the current flowing through the load resistance can be determined by using different techniques like mesh analysis or nodal analysis or superposition methods. Suppose the load resistance is changed to some other value than previous then we have to apply any one of these methods again.

This tedious method of applying reduction technique for every load change is avoided by replacing the fixed part of a circuit (inside the black box) with a practical voltage source nothing but a manifestation of Thevenin’s theorem. In practice, the Thevenin’s theorem helps to find the maximum power delivered to the speakers which are supplied from the amplifier in a transistor power amplifier.

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Thevenin’s Theorem Statement

The Thevenin’s theorem states that any linear two terminal circuit consisting of sources and resistors connected to a given load RL can be replaced by an equivalent circuit consisting a single voltage source of magnitude Vth with a series resistance Rth across the terminal of RL.

The below figure shows the Thevenin’s model of two terminal network where the current through the load is same therefore, these two circuits are equivalent to each other.

Similar to the DC circuits, this method can be applied to the AC circuits consisting of linear elements like resistors, inductors, capacitors. Like thevinin’s equivalent resistance, equivalent thevinin’s impedance is obtained by replacing all voltage sources by their internal impedances.

In AC circuits the thevenin’s theorem can be stated as any two terminal, linear bilateral circuit consisting of linear elements and active sources connected across the terminal of ZL can be replaced by a single equivalent voltage source of Vth with a single impedance Zth across the two terminals of ZL.

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Steps to Analyse the Thevenin’s Theorem

The following are the steps to simplify the circuit so that the load current is determined using Thevenin’s theorem.

1. Consider the given circuit and disconnect the load resistance RL (load impedance ZL) or branch resistance (branch impedance in AC circuit) through which current flow is to be calculated.

2. Determine the open circuit voltage Vth across the load after disconnecting RL. For finding Vth, one can apply any methods from available circuit reduction techniques like mesh analysis, nodal voltage method, superposition, etc. Or simply, we can measure the voltage at the load terminals using a voltmeter.

3. Redraw the circuit by replacing all the sources with its internal resistances (Internal impedances in case of AC circuit) and make sure that voltage sources are to be short circuited and current sources to be open circuited (for ideal sources).

4. Calculate the total resistance Rth (or Zth) that exist between the load terminals.

5. Insert this equivalent resistance Rth (or Zth) in series with voltage Vth and this circuit is referred as the Thevenin’s equivalent circuit.

6. Now reconnect the load resistance (load impedance ZL) across the load terminals and calculate the current, voltage and power of the load by simple calculations.

In DC circuit,

Load current,

IL = Vth/ (RL + Rth)

Voltage across the load,

VL= RL × Vth/ (RL + Rth)

Power dissipated in the load resistance,

PL = RL × IL2

In case of AC circuit Load current,

IL1 = Vth/ (ZL + Zth)

Voltage acorss the load,

VL= ZL × Vth/ (ZL + Zth)

Power dissipated in the load resistance, PL = ZL × IL12

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Example of Finding Equivalent Circuit to DC circuit

Consider the DC circuit shown in below. We are going to find the current through the resistance R2 = RL = 2 ohms (connected between terminals a and b) by applying Thevenin’s theorem.

1. Remove the load resistance R2 or RL and assume the closed path direction towards point C.

2. Apply nodal analysis at node C to calculate the thevenin’s voltage Vth.

By applying KCL at node C we get

4 + I1 + I2 = 0

4 + (6 – Vc)/ 4 + (0 – Vc)/ 10 = 0

Vc = 15.714 Volts

Then, the currents in each branch can be determined as

I1 = Va – Vc/ 4

= 6 – 15.714/4

= 2.0715 Amps

I2 = 0 – Vc/ 10

= – 15.714/ 10

= – 1.571 Amps

The negative symbol indicates that current is flowing from node C to their respective point (as ‘a’ and ground points for I1 and I2 respectively).

By redrawing the circuit with these currents and by applying KVL, the voltage across the terminal ab is determined as,

Vth = Va – Vb (with respect to ground terminals)

= Va – (I2× R4)

= 6 – (1.571 × 4)

= 0.28 Volts

3. Next step is to replace all the sources with their internal sources. Consider that voltage source is an ideal source so the internal resistance is zero, therefore it is short circuited and the current source is an ideal current source therefore it has infinite resistance hence it is open circuited. Then the equivalent Thevenin’s resistance circuit is shown below.

4. Next, we have to find the Thevenin’s equivalent resistance Rth by looking at the terminals a and b (load terminals).

Rth = [(R1 + R3) × R4] / [(R1 + R3) + R4] (parallel resistors)

= 10 × 4 / 10 + 4

= 2.85 Ohms

5. By placing the above calculated voltage source in series with equivalent resistance forms Thevenin’s equivalent circuit as shown in below figure.

By reconnecting the load resistance across the terminals a and b, we calculate the current flowing through the load as

IL = Vth / (Rth + RL)

= 0.28/ (2.85 + 2)

= 0.057 Amps

And below figure shows the original circuit where the current through the load resistance is indicated.

We can also find the currents through the load by changing the value of the load resistor as

When RL = 8 ohms

IL = 0.28 / (2.85 + 8)

= 0.02Amps

When RL = 12 ohms

= 0.28 / (2.85 + 12)

= 0.01 Amps

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Example of Finding Equivalent Circuit to AC circuit

Consider the below AC circuit to which we are going to find the current through the impedance 4+ 4j ohm using thevenin’s theorem.

In the above circuit, the current source of 2∠0 in parallel with 4 ohm resistor. And hence, this can be converted into a voltage source 8∠0 with a series resistance of 4 ohms as shown in figure.

After making the above change, the circuit is redrawn by disconnect the load terminals as shown in below figure.

Assume the mesh currents as shown in modified diagram and KVL equation of the meshes are

2∠0 – I1 – 2(I1– I2) – 4∠0 = 0

– 3I1 + 2I2 = 2 …….(1)

For mesh 2

4∠0 – 2(I2 – I1) – 4I2– 8∠0 = 0

2I1 – 6I2 = 4 …….(2)

By solving above two equations, we get

I2 = –1.142∠0 A

Therefore,

Vth = 8∠0 – 4× (1.142∠0)

= 3.43∠0 V

The equivalent impedance,

Zth = 1/ (1 + (1/2) + (1/4))

= 0.574 ∠0

Therefore, the current through the 2 + 2j impedance is given as,

IAB = Vth/ Zth + ZL

= 3.43∠0/ (0.574 ∠0 + 4 + 4j)

= 3.43∠0 / (6.07 ∠41.17)

= 0.56∠ – 41.17 A

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Limitations of Thevinen’s Theorem

• If the circuit consists of non linear elements, this theorem is not applicable.

• Also to the unilateral networks it is not applicable.

• There should not be magnetic coupling between the load and circuit to be replaced with the thevinen’s equivalent.

• There should not be controlled sources on the load side which care controlled from some other parts of the network.

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Introduction

Many of the electrical circuits are complex in nature and the computations required to find the unknown quantities in such circuits, using simple ohm’s law and series/parallel combination simplifying methods is not possible. Therefore, in order to simplify these circuits Kirchhoff’s laws are used.

These laws are the fundamental analytical tools that are used to find the solutions of voltages and currents in an electric circuit whether it can be AC or DC. Elements in an electric circuit are connected in numerous possible ways, thus to find the parameters in an electrical circuit these laws are very helpful.

Before going to know more about Kirchhoff’s law, we have to consider some of the terms related to electric circuits.

Node: Node or junction is a point in the circuit where two or more electrical elements are connected. This specifies a voltage level with a reference node in a circuit.

Branch: The continuous conducting path between two junctions which contains electrical element in a circuit is referred as branch.

Loop: In an electrical circuit a loop is an independent closed path in a circuit that follows the sequence of branches in such a way that it must start and ends with same node and it shouldn’t touch any other junction or node more than once.

Mesh: In an electrical circuit mesh is a loop that doesn’t contain any other loop in its interior.

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Kirchhoff’s Laws

In 1847, Gustav Robert Kirchhoff, a German physicist was developed these laws to describe the voltage and current relationship in an electric circuit. These laws are: Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL).

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Kirchhoff’s Current Law (KCL)

This is also called as the law of conservation of charge because charge or current cannot be created or destroyed at the junction or node. It states that the algebraic sum of currents at any node is zero. Thus the current entering at a node must be equal to sum of current out of the node.

In the above figure, currents I1 and I2 are entering to the node while the currents I3 and I4 are leaving from the node. By applying KCL at the node , assume that entering currents are positive and leaving currents are negative ,we can write as

I1 + I2 + (-I3) + (-I4) = 0
I1 + I2 = I3 + I4

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Example Problem of KCL

Consider the below figure where we have to determine the currents IAB and Ix by using KCL .

By applying Kirchhoff’s Current Law at point A, we get

IAB = 0.5 – 0.3

IAB = 0.2 Amps

Similarly by applying KCL at point B, we get

IAB = 0.1 + Ix

0.2 = 0.1 + Ix

Ix = 0.2 – 0.1 = 0.1 Amps

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Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law states that the algebraic sum of voltages in a closed path is equal to zero that is the sum of source voltages is equal to the sum of voltage drops in a circuit. If the current flows from higher potential to lower in an element, then we consider it as a voltage drop.

If the current flows from lower potential to higher potential, then we consider it as a voltage rise. Thus, the energy dissipated by the current must be equal to the energy given by the power supply in an electric circuit.

Consider above circuit where the direction of current flow is taken clockwise. Various voltage drops in the above circuit are V1 is positive, IR1is negative (drop in voltage), IR2 is negative (drop in voltage), V2 is negative, IR3 is negative (drop in voltage), IR4 is negative (drop in voltage), V3 is positive, IR5 is negative and V4 is negative. By applying KVL, we get

V1 + (-IR1) + (-IR2) + (-V2) + (-IR3) + (-IR4) + V3 + (-IR5) + (-V4) = 0

V1 – IR1 – IR2 – V2 – IR3 – IR4 + V3 – IR5 – V4 = 0

V1 – V2 + V3 – V4 = IR1+ IR2 +IR3 + IR4 + IR5

Hence the KVL is also known as the law of conservation of electrical energy because the sum of voltage drops (product of resistance and current) is equal to the sum of voltage sources in a closed path.

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Kirchhoff’s Voltage Law Example

1. Let us consider the single loop circuit which is shown below and assume the current flow direction as DEABCD closed path. In this circuit, by using KVL we have to find the voltage V1.

By applying KVL to this closed loop, we can write as

VED + VAE + VBA + VCB + VDC = 0

Where

Voltage of point E with respect to point D, VED = -50 V

Voltage of point D with respect to point C, VDC = -50 V

Voltage of point A with respect to point E. VAE = I * R

VAE = 500m* 200

VAE = 100 V

Similarly Voltage at point C with respect to pint B, VCB = 350m*100

VCB = 35V

Consider voltage at point A with respect to point B, VAB = V1

VBA= -V1

Then by using KVL

-50 + 100 – V1 + 35 – 50 = 0

V1 = 35 Volts

2. Consider the below typical two loop circuit where we have to find the currents I1 and I2 by applying the Kirchhoff’s laws.

There are two loops inside the circuit and consider the loop paths as shown in figure.

By applying KVL to these loops we get

For first loop,

2 (I1 + I2) + 4I1 – 28 = 0

6I1 + 2I2 = 28 ——— (1)

For second loop,

-2(I1 + I2) – 1I2 + 7 = 0
-2I1 – 3I2 = -7 ——– (2)

By solving the above 1 and 2 equations we get,

I1 = 5A and I2 = -1 A

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Example Problem on Kirchhoff’s laws

Now let us use both Kirchhoff’s current and voltage laws to find the current and voltage drops in below circuit. Similar to the above problem this circuit also contains two loops and two junctions. Consider the current direction given in the figure.

Apply Kirchhoff’s current law at both junctions, then we get

At junction 1, I = I1 + I2

At junction 2, I1 + I2 = I

Apply Kirchhoff’s voltage law to the both loops, then we get

In first loop,

1.5 V – 100 I1 = 0

I1 = 1.5 / 100

= 0.015 Amps

In second loop

100(I1- I2) – 9V – 200I2 = 0

100I1- 300I2 = 9

Substituting I1 value in the above equation then

1.5 – 300I2 = 9

– 300I2 = 7.5

I2 = -0.025

Then the current at the junction I = I1 + I2

I = 0.015 – 0.025

I = – 0.01

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Applications of Kirchhoff’s Laws

• By using these laws, we can find the unknown resistances, voltages and currents (direction as well as value).
• In the branch method, finding the currents through each branch carried by applying KCL at every junction and KVL in every loop of a circuit.
• In the loop current method, finding current through each independent loop is carried by applying KVL for each loop and counting all the currents in any element of a circuit.
• Used in nodal method of finding voltages and currents.
• These laws can be applied to analyse any circuit regardless of the composition and structure of it.

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The post Kirchhoff’s Laws: A Comprehensive Guide to Analyzing Electrical Circuits appeared first on ElectronicsHub.

For analysing the linear electric circuits that consists of two or more independent sources (Voltage or current or both), superposition theorem is extremely used (particularly for time domain circuits with elements operated at different frequencies). If a linear DC circuit has more than one independent source, we can find the current (through a resistance) and voltage (across the resistance) by using nodal or mesh analysis methods.

Alternatively , we can use the superposition theorem that adds each individual source effect on the value of the variable to be determined. This means superposition theorem considers the each source in a given circuit separately for finding the value of the variable (whether current or voltage) and finally produce the resultant variable by adding all the variables caused by each source effect. Even though it is of complex procedure, but still can be applied for any linear circuit.

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Superposition Theorem Statement

The superposition theorem states that in any linear bilateral network that consisting of two or more independent sources, current through (or voltage across) an element is the algebraic sum of the currents through (voltages across) that element caused by each independent source acting alone with all other sources are replaced by their internal resistances. We know that as long as the linearity exists between the source and contribution, the total contribution due to various sources acting simultaneously is equal to the algebraic sum of individual contributions due to individual source acting at a time.

Therefore, if the circuit consists of N independent sources, we have to analyse N circuits, each will produce a result with respect to each individual source. And finally these individual results must be added to get the whole analysis of the circuit. Therefore, this require more work however, this theorem will be very useful in analysing the various parts of a complex circuit.

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Steps to Analyse Superposition Theorem

1. Consider the various independent sources in a given circuit.

2. Select and retain one of the independent sources and replace all other sources with their internal resistances or else replace the current sources with open circuits and voltage sources with short circuits.

3. To avoid confusion re-label the voltage and current notations suitably.

4. Find out the desired voltage/currents due to the one source acting alone using various circuit reduction techniques.

5. Repeat the steps 2 to 4 for each independent source in the given circuit.

6. Algebraically add all the voltages/currents that are obtained from each individual source (Consider the voltage signs and current directions while adding).

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Example :

1. Let us consider the below simple DC circuit to apply the superposition theorem such that we will obtain the voltage across the resistance 10 Ohms (load terminals). Consider that in a given circuit there are two independent sources as voltage and current sources as shown in figure.

2. First, we retain one source at a time that means , only voltage source is acting in the circuit and the current source is replaced with internal resistance (infinite) so it becomes open circuited as shown in figure.

Consider V_L1 is the voltage across the load terminals with voltage source acting alone, then

V_L1 = Vs × R_L / (R_L + R₁)

= 20 × 10 / (10 + 20)

= 6. 66 Volts

3. Retain the current source alone and replace the voltage source with its internal resistance (zero) so it becomes a short circuited as shown in figure.

Consider that V_L2 is the voltage across the load terminals when current source acting alone. Then
V_L2 =  I_L × R_L

I_L = I × R₁ / (R₁ + R_L)

= 1 × 20 / (20 +30)
= 0.4 Amps

V_L2 = 0.4 × 10

= 4 Volts

Therefore, according the superposition theorem, the voltage across the load is the sum of V_L1 and V_L2

V_L = V_L1 + V_L2

= 6.66 + 4

= 10.66 Volts

Example 2:

Consider the below circuit to which we are going to determine the current I through the 4 ohm resistor using superposition theorem.

Consider I1, I2 and I3 are the currents due to sources 12v, 20V and 4A sources respectively. Then, based superposition theorem I = I1 + I2 + I3. So let’s determine these currents with each individual source.

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Only with 12V Voltage source:

Consider the below circuit where only 12V source is retained in the circuit and other sources are replaced by their internal resistances.

By combining the resistance 6 ohm with 10 ohm we get 16 ohm resistance which is parallel with 6 ohm resistance. Then this combination produce, 16 × 6 / (16 + 6) = 4.36 ohm. Therefore the equivalent circuit will be as shown in figure.

Then the current through 4 ohms resistance,

I₁  = 12 / 8.36

= 1.43 A

Only with 20 V Voltage Source:

Retain only 20 V voltage source and replace other sources with their internal resistance, then the circuit becomes as shown below.

Apply the mesh analysis to the loop a, we get

22Ia – 6I_b + 20 = 0

22Ia – 6I_b = -20 ……………….(1)

For Loop b, we get

10I_b – 6Ia = 0

Ia = 10I_b/6

Substituting Ib in equation 1

22 (10I_b/6) – 6I_b = -20

I_b = – 0.65

Therefore, I₂ = Ib = -0.65

Only with 4A Current Source

Consider the below circuit where only current source is retained and other sources are replaced with their internal resistances.

By applying nodal analysis at node 2 we get,

4 = (V₂/10) + (V₂ – V₁)/6 ………………..(2)

At node1,

(V₁/6) + (V₁/4) = (V₂ – V₁)/ 6

V₂ = 3.496 V₁

Substituting V2 in equation 2, we get

V₁ = 0.766 Volts.

Therefore I₃ = V₁/4

= 0.766/4

= 0.19 Amps.

Thus, as per the superposition theorem, I = I₁ + I₂ + I₃

= 1.43  – 0.65 +  0.19

= 0.97 Amps.

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Superposition Example Using AC Circuit:

Consider the below AC circuit, to which we are going to determine the value of current in the 4 ohm resistor using superposition theorem.

Case 1: Only with 20∠0 Voltage Source

By retaining the voltage source alone in the circuit, the current flow through the circuit is determined as

I1 = 20∠0/ (4 + j4)

= 20∠0 / (5.65∠45)

= 3.53∠- 45 or 2.49 –j2.49 A

Case 2: Only with 4∠90 Current Source

By retaining the current source alone in the circuit, the current I2 through the circuit is determined as

By current division method, I2 = 4∠90 × 4j/ (4 + j4)

= 4∠90 × 4∠90 (5.65∠45)

= 4∠90 × 0.707∠45

= 2.828∠135 or -1.99 + j1.99 A

The resultant current through the resistor 4 ohms is I = I1 + I2

= 3.53∠- 45 + 2.828∠135

= 0.785∠45 or 0.56 + j0.56 A

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Limitations of Superposition Theorem

1. For power calculations superposition theorem cannot be used as this theorem works based on the linearity. Because the power equation is not linear as it is the product of voltage and current or square of the current or square of the voltage. Thus the power consumed by the element in a given circuit with superposition theorem is not possible.

2. If the choice of the load is variable or the load resistance changes frequently, then it is required to perform every source contribution of current or voltage and their sum for every change in load resistance. So this very complex procedure for analysing complex circuits.

3. This theorem applicable for only linear circuits and for non linear circuits (Having transistors and diodes) we can not apply.

4. This theorem is applicable only if the circuit has more than one source.

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The main concept behind the nodal analysis is that , in a given circuit if the node voltages are known, then we can immediately determine all branch currents associated with the circuit. As we know that , for finding node voltages we use KCL. In this technique, node voltages are considered as variables in the circuit , instead of element voltages , which results in reduction of the number of equations to simplify the circuit. In nodal analysis method , with availability of all the nodes, one node is considered as a reference node (zero potential) and it is represented as ground terminal. To other remaining unknown nodes , voltages are assigned , with respect to the referenced node voltage.

To the each node in a given circuit, we apply the KCL except for the referenced node. Suppose if the given circuit has N nodes, then we get N-1 simultaneous equations to find the N-1 unknown node voltages.

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Steps to Analyse Nodal Analysis Technique

1) Check the possibility to transform voltage sources in the given circuit to the current sources and transform them.

2) Identify the nodes present in the given circuit and assign one node as reference node and with respect to this ground or reference node , label other nodes as unknown node voltages.

3) Assign the current direction in each branch in the given circuit (it is an arbitrary decision).

4) Apply KCL to N-1 nodes and write nodal equations by expressing the branch currents as node assigned voltages.

5) Solve the simultaneous equations of nodes to find the node voltages and finally branch currents. The number of node equations is equal to the number of nodes minus one (as one node is referenced).

Consider below DC circuit, in which branch currents are to be determined  using the nodal analysis.

As a first step in nodal analysis, we have to select the reference node which is to be connected to the zero or ground potential as indicated below.

Secondly, we apply Kirchoff’s Current Law (KCL) to each node in the circuit except the reference node. By applying KCL at node 1 we get,

Is1 – Is3 – I4 – I2 = 0

Is1 – Is3 – {(V1 – V2)/R4} – {(V1- V3)/R2} = 0

Is1 – Is3 = V1 { (1/R2) + (1/R4)} – V2 (1/R4) – V3 (1/R2)

Is1- Is3 = G11 V1 – G12 V2 – G13 V3 ………………………..(1)

where , G1i is the sum of total conductance at the first node. (As we know that 1/ R = G)

By applying KCL at node 2 we get

I4 – Is2 – I3 = 0

{(V1 – V2)/R4} – Is2 – {(V2 – V3)/R3} = 0

– Is2 = – V1 (1/R4) + V2 { (1/R3) + (1/R4)} – V3 (1/R3)

– Is2 = – G21 V1 – G22 V2 – G23 V3 ……………………………(2)

Applying KCL at node 3 we get

Is3 + I2 + I3 – I1 = 0

Is3 + {(V1 – V3)/R2} – {(V2 – V3)/R3}– V3 (1/R1) = 0

Is3 = – V1 (1/R2) – V2 (1/R3) + V3 { (1/R1) + (1/R2) + (1/R3)}

Is3 = – G31 V1 – G32 V2 + G33 V3…………………………..(3)

Likewise, we can write the KCL equations for i th node. And hence

∑ Iii is equal to the algebraic sum of all the currents connected at the i th node where i = 1, 2, 3……N and N = n-1 (n is the total number of nodes present in the circuit).

Gii = The sum of conductance connected to the i th node.

Gij = The sum of conductance connected between i and j nodes.

By solving the above three equations we get the branch voltages at respective nodes and thereby we can calculate branch currents.

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Example

Determine the node voltages   and  currents in each branch using nodal analysis method in the given circuit.

The given circuit contains a voltage source. This can be transformed to current source or can be analysed directly without any transformation. Now let us calculate the nodal voltages without any transformation.

As a first step in nodal analysis, we have to choose and label the nodes present in the given circuit. By choosing the bottom node as reference node, we have two another nodes in the given circuit. So these nodes are labelled as V1 and V2 as shown in below figure. And also current directions in each branch are represented.

By applying KCL at node 1, we get

5 = I3 + I10

5 = (V1/10) + (V1 – V2/3)

13V1 – 10V2 = 150 ………(1)

By applying KCL at node 2, we get

I3 = I5 + I1

(V1 – V2/3) = (V2/5) + (V2 – 10/1)

5V1 – 23V2 = -150 ……..(2)

By solving above two equations, we get

V1 = 19.85 Volts and V2 = 10.9 Volts

The currents in each branch is given as

I10 = V1/10

= 19.85/10 = 1.985

I3 = V1 – V2/3

= 19.85 – 10.9/3

= 2.98 A

I5 = V2/5

= 10.9/5

= 2.18 A

I1 = V2 – 10

= 10.9 – 10

= 0.9 A

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Concept of Supernode

Sometimes it becomes difficult to apply nodal analysis when any voltage source is present in between two branches in a circuit. One way to overcome this problem is by applying a super node technique. In super node technique, voltage source is connected between two adjacent nodes is shorted to reduce the two nodes to form a single super node.

Consider the above example in which a voltage source is connected between the 2 and 3 nodes. The calculations become more difficult if we analyse the circuit with voltage source. The analysis of this circuit becomes easier if we create a super node by shorting 2 and 3 nodes.

By applying kirchoff’s current law at the node 1 we get,

I = (V1/R1) + ((V1-V2)/R2) ……(1)

The super node technique can be applied to the given circuit by shorting the 2 and 3 nodes and by applying KCL we get

((V2-V1)/R2) + (V2/R3) + ((V3-Vy)/R4) + (V3/R5) = 0

And also voltage in the voltage source is given as

Vx = V2 – V3

From the above three equations, we can easily find out the three unknown voltages in the circuit.

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Example of Super node

Consider the below circuit and  find the three unknown node voltages V1, V2 and V3 by using super node technique.

At node 1, source is connected to the reference node and hence V1 becomes 5v

V1 = 5 V

A super node is formed by enclosing the nodes 2 and 3. By applying KCL at this super node we get

i1 = i2 + i3

(V1 – V2)/5 = (V2 / 10) + (V3/20)……………..(1)

And also KVL at super node gives,

V2 – V3 = 10…………… (2)

By solving above equations, we get V2 = 4.29V and V3 = -5.71 V

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The post Nodal Analysis appeared first on ElectronicsHub.

Many circuits consist of three terminal networks such as Wye (Y) or star or tee (T) and delta or pi networks. These networks are part of a large network or occur by themselves. The application areas of these networks include three phase networks, matching networks and electrical filters, etc. These networks are simplified using another useful technique called star-delta transformations.

Star and Delta Networks

In star connection, components are connected in such a way that one end of all the resistors or components are connected to a common point. By the arrangement of three resistors , this star network looks like a alphabet Y hence , this network is also called as Wye or Y network. The equivalent of this star connection can be redrawn as T network (as a four terminal network) as shown in below figure. Most of the electrical circuits constitute this T form network.

In a delta connection, end point of each component or coil is connected to the start point of another component or coil. It is a series connection of three components that are connected to form a triangle. The name indicates that connection look like an alphabet delta (Δ). The equivalent delta network can be redrawn , to look like a symbol Pi (or four terminal network ) as shown in figure. So this network can also be referred as Pi network.

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Delta to Star Transformation

The conversion from star- delta or delta-star can be achieved , when the similar pairs of terminals have the same impedance. This transformation produces a equivalent network by eliminating the node.

Let us discuss the conversion of delta to star. Consider that Rab, Rbc and Rca are the three series resistances in the delta network and Ra and Rb and Rc are the three resistances in a star network.

This equivalent star network produced after transformation and delta network will have same resistance , when measured between the similar pairs of terminals .

Consider the above figure , where the equivalent resistance between terminals a and c is

Ra+ Rc = Rca ||(Rab + Rbc )

Ra+ Rc = Rca * (Rab + Rbc)/(Rab + Rbc + Rca) ……………………(1)

The equivalent resistance between terminals c and b is

Rb+ Rc= Rbc || (Rab + Rca)

Rb+ Rc = Rbc * (Rab + Rca)/(Rab + Rbc + Rca) …………………….(2)

And between terminals b and a is

Rb+ Ra = Rab ||(Rca + Rbc )

Rb+ Ra = Rab * (Rca + Rbc)/(Rab + Rbc + Rca) ……………………..(3)

By combining above 1, 2 and 3 equations we get

Ra + Rb + Rc = (RabRbc + RbcRca + RcaRab)/(Rab + Rbc + Rca) ……………(4)

By subtracting the equation 2 from equation 4 we get

Ra = (Rab Rca)/(Rab + Rbc + Rca)

Subtracting equation 1 from equation 4 we get

Rb = (Rab Rbc)/(Rab + Rbc + Rca)

And subtracting equation 3 from equation 4 we get

Rc = (Rbc Rca)/(Rab + Rbc + Rca)

These Ra, Rb and Rc are the three resistance values in star network converted from delta equivalent circuit.

Ra = (Rab Rca)/(Rab + Rbc + Rca)

Rb = (Rab Rbc)/(Rab + Rbc + Rca)

Rc = (Rbc Rca)/(Rab + Rbc + Rca)

By observing the above three equations we can say that for a given terminal the equivalent resistance in the star network is equal to the product of two resistances (in the delta) connected to the same terminal divided by the sum of total resistances in the delta network.

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Example:

Consider the below figure to transform delta to star or Wye circuit where the values Ra = 20 ohms, R2 = 30 ohms and R3 = 50 ohms.

 For delta to star conversions equivalent resistance equations (for this problem) are

Ra = (R1 R2)/(R1 + R2 + R3)

Rb = (R2 R3)/(R1 + R2 + R3)

Rc = (R1 R3)/(R1 + R2 + R3)

Therefore the total resistance, Rt = (R1 + R2 + R3)

= 20 + 30 + 50

= 100 ohms

Ra = (R1 R2)/(R1 + R2 + R3)

= (20 X 30) /100

= 6 ohms

similarly Rb = (R2 R3)/(R1 + R2 + R3)

= (30 X 50) /100

= 15 ohms

and Rc = (R1 R3)/(R1 + R2 + R3)

= (50 X 20)/ 100

= 10 ohms

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Star to Delta Transformation

Use the same representation of resistances for the star as Ra, Rb and RC and for delta as Rab, Rbc and Rca. Consider the star network of resistors shown below, where current through Ra resistor is given as

Ia = (Va – Vn) / Ra ………..(1)

By applying KCL at node N in star network then we get

(Va – Vn) /Ra + (Vb – Vn) /Rb + (Vc – Vn) /Rc

Vn [(1/Ra) + (1/Rb) + (1/Rc)] = (Va/Ra) + (Vb/Rb) + (Vc/Rc)

Vn = [(Va/Ra) + (Vb/Rb) + (Vc/Rc)] / [(1/Ra) + (1/Rb) + (1/Rc)] ……(2)

In delta network current at point A is

Ia = (Vab /Rab) + (Vac / Rac) ……(3)

From equations 1 and 3 we get

(Va – Vn) / Ra = (Vab /Rab) + (Vac / Rac) ……………………..(4)

Substituting Vn value from equation 2 in the equation 4 and by simplification we get

Rab = Ra + Rb + ((RaRb)/Rc)

Rac = Ra + Rc + ((RaRc)/Rb)

Similarly Ib in star network is

Ib = (Vb – Vn) / Rb ………..(5)

In delta network

Ib = (Vbc /Rbc) + (Vba / Rba) …………………(6)

By equating 5 and 6 equations

(Vb – Vn) / Rb = (Vbc /Rbc) + (Vba / Rba) ………………..(7)

By substituting equation 2 in the equation 7 and after simplification we get

Rbc = Rb + Rc + ((RbRc)/Ra)

Hence the equations required to transform delta network to equivalent star or wye network are

Rab = Ra + Rb + ((RaRb)/Rc) = ( RaRb + RbRc + RbRc )/Rc

Rbc = Rb + Rc + ((RbRc)/Ra) = ( RaRb + RbRc + RbRc )/Ra

Rac = Ra + Rc + ((RaRc)/Rb) = ( RaRb + RbRc + RbRc )/Rb

By observing the above three equations we can say that between the given two terminals the equivalent delta resistance is equal to the sum of two resistances (in the star) connected to those terminals plus the product of the same two resistances divided by remained or third star resistance.

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Example:

Consider the below figure to transform star or Wye to the delta circuit where the resistance values in star network are given as R1= 10 ohms, R2= 5 ohms and R3 = 20 ohms.

For star or wye to delta conversion, the equivalent resistance equations (for this problem) are

R12 = R1 + R2 + ((R1R2)/R3)

R23 = R2 + R3 + ((R2R3)/R1)

R31 = R1 + R3 + ((R1R3)/R2)

By simplifying the above equations we get the common numerator term as

R1R2 + R2R3 + R1R3

= 10 X 5 + 10 X 20 + 20 X 5

= 350 ohms

Then R12 = 350/ R3

= 350/20

= 17.5 ohms

R23 = 350/ R1

= 350/ 10

= 35 ohms

R31= 350/R2

= 350 /5

= 70 ohms

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The post Star Delta Transformations appeared first on ElectronicsHub.

Mesh Analysis

Mesh is a loop that doesn’t consists of any other loop inside it. Mesh analysis technique, uses  mesh currents as variables , instead of currents in the elements to analyse the circuit. Therefore, this method absolutely reduces the number of equations to be solved . Mesh analysis applies the Kirchhoff’s Voltage Law (KVL) to determine the unknown currents in a given circuit. Mesh analysis is also called as mesh-current method or loop analysis. After finding the mesh currents using KVL, voltages anywhere in a given circuit can be determined by using Ohms law.

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Steps to Analyse the mesh analysis technique

1) Check whether there is a possibility to transform all current sources in the given circuit to voltage sources.

2) Assign the current directions to each mesh in a given circuit and follow the same direction for each mesh.

3) Apply KVL to each mesh and simplify the KVL equations.

4) Solve the simultaneous equations of various meshes to get the mesh currents and these equations       are exactly equal to the number of meshes present in the network. 

   Consider the below DC circuit to apply the mesh current analysis, such that currents in different meshes can be found. In the below figure there are three meshes present as ACDA, CBDC and ABCA but the path ABDA is not a mesh. As a first step, the current through each mesh is assigned with the same direction as shown in figure.

Secondly, for each mesh we have to apply KVL. By applying KVL around the first loop or mesh we get

V1 − V3 − R2 ( I1 − I 3 ) − R4 ( I1 − I 2 ) = 0

V1 − V3 = I1 ( R2 + R4 ) − I2R4 − I3R2 ………………(1)

Similarly , by applying KVL around second mesh we get,

−V2 − R3 ( I 2 − I 3 ) − R4 ( I 2 − I1 ) = 0

− V2 = − I1R4 + I 2 ( R3 + R4 ) − I 3 R3 ………………………(2)

And by applying KVL around third mesh or loop we get,

V3 − R1I 3 − R3( I 3 − I 2 ) − R2( I 3 − I1 ) = 0

V3 = − I1R2 − I2R3 + I3(R1 + R2 + R3) ………………………(3)

Therefore, by solving the above three equations we can obtain the mesh currents for each mesh in the given circuit.

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Example problems on mesh analysis:

Example 1:

Consider the below example in which we find the voltage across the 12A current source using mesh analysis. In the given circuit all the sources are current sources.

Step 1: In the circuit there is a possibility to change the current source to a voltage source on right hand side source with parallel resistance. The current source is converted into a voltage source by placing the same value of resistor in series with a voltage source and the voltage in that source is determined as

Vs = Is Rs

= 4× 4 = 16 Volts

Step 2: Assign the branch currents as I1 and I2 to the respective branches or loops and represent the direction of currents as shown below.

Step 3: Apply the KVL to each mesh in the given circuit

Mesh -1:

Vx − 6 × (I1 − I 2) − 18 = 0

Substituting I1 = 12 A

Vx + 6I2 = 90…………………… (1)

Mesh – 2:

18 − 6 × ( I 2 − I1 ) − 4 × I 2 − 16 = 0

2 – 10 × I2 + 6(12) = 0

I2 = 74/ 10

= 7.4 Amps

Substituting in equation 1 we get

Vx = 90 – 44.4

= 45.6 Volts

Example 2:

Consider the below circuit where we determine the voltage across the current source and a branch current Iac. Assign the directions as shown below and note that current is assigned opposite to the source current in second loop.
By applying KVL to the first mesh we get

V1 − R2 ( I1 − I 3 ) − R4 ( I1 − I 2 ) = 0

4 – 2 I1 − 2I3 − 4I1 − 4I2 = 0

-6I1 − 2I3 = 4 ……………(1)

By applying KVL to the second mesh we get

−Vc − R4( I 2 − I1 ) − R3 ( I 2 − I 3 ) = 0

– Vc = 4I2 − 4I1 + 2I2 − 2I3 = 0

– Vc = – 4I1 + 6I2 – 2I3

But I2 = -2 A, then

– Vc = – 4I1 – 12 – 2 I3 ………………….(2)

By applying KVL to the third mesh we get

− R1 I 3 − R3 ( I 3 − I 2 ) − R2 ( I 3 − I1 ) = 0

−4 I3 − 2I3 + 2I2 − 2I3 + 2I1 = 0

− 8I3 − 4 + 2I1 = 0 (by substituting I2 = -2 A)

2I1 − 8I3 = 4 …………………(3)

By solving 1 and 3 equations we get I3 = -0.615 and I1 = 4.46

Therefore, the voltage Vc = 4 (4.46) + 12 + 2(-0.615)

Vc = 28.61 V

And the branch current Iac = I1- I3

Iac = 5.075 amps

Likewise we can find every branch current using the mesh analysis.

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Super Mesh Analysis

As we seen in the example 2, it contains current source in one of its branches. And before the application of the mesh analysis to that circuit, we assumed the unknown voltage across the current source and then mesh analysis is applied. This is a quite difficult approach and this can be overcome by applying supermesh technique.

A super mesh is formed when two adjacent meshes share a common current source and none of these (adjacent) meshes contains a current source in the outer loop. Consider the below circuit in which super mesh is formed by the loop around the current source.

The current source is common to the meshes 1 and 2 and hence it must be analysed independently. To achieve this, assume the branch that contains current source is open circuited and create a new mesh called super mesh.

Writing KVL to the super mesh we get

V = I1R1 + (I2 – I3) R3

= I1R1 + I2R3 – I3R3

Applying KVL to the Mesh 3 we get

(I3 – I2) R3 + I3R4 = 0

And the difference between the two mesh currents gives the current from the current source. Here the current source direction is in the loop current direction I1. Hence I1 is more than I2, then

I = I1 – I2

Thus, by using these three mesh equations we can easily find the three unknown currents in the network.

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Example on  supermesh analysis

Consider the below example in which we have to find the current through the 10 ohms resistor.

By applying the KVL to the mesh 1 we get

1I1 + 10 (I1 – I2) = 2

11I1 – 10 I2 = 2 …………………………. (1)

The meshes 2 and 3 consist of 4A current source and hence form a super mesh. The current from 4A current source is in the direction of I3 and thus the super mesh current is given as

I = I3 – I2

I3 – I2 = 4…………………………. (2)

By applying KVL to the outer loop of the super mesh we get,

– 10 (I2 – I1) – 5I2 – 15I3 = 0

10I1 – 15I2 – 15I3 = 0…………………….. (3)

By solving 1, 2 and 3 equations, we get

I1 = –2.35 A

I2 = –2.78 A

I3 = 1.22 A

Hence the current through the 10 ohms resistor is I1 – I2

= –2.35 + 2.78 A

= 0.43 A

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Practical Voltage Source

An ideal voltage source means a device that offer a terminal voltage which doesn’t depend on the current flow through it. But such ideal sources never exist practically. Suppose a battery would produce 12V when no load is connected or no current flows through it, then the battery produce a less voltage than 12V when a load current flows through it. These sources are called as practical voltage sources. However, as long as small power or currents drawn from the load, these practical sources represent the ideal voltage sources.

                                                                   V-I characteristics of practical voltage source

Therefore, the real model of device (practical voltage source) is obtained by an ideal voltage source with a series resistor. This series resistor resembles the voltage drop in the device when the current flows through it. The series resistor is called as the internal resistance of the voltage source. This does not mean that we can find such arrangement inside every practical source device but merely represents the reduction in terminal voltage on an account for an increase of load current flow.

                    Practical voltage source with internal resistance

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Practical Current Source

Similar to an ideal voltage source, an ideal current source also never exists as there is no device or source that delivers a constant current independent of connected load resistance or the voltage across the load terminals. However, if the load voltage is small, load current in practical current source and current from ideal current source are equal. Therefore, a real model or practical current source is obtained from an ideal current source in parallel with resistance (or internal resistance). This resembles , the current changes in a practical current source with change in voltage of the load (or change in resistance of the load).

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Concept of Source Transformations

Consider both practical voltage and current sources with load resistance of RL. Let us see how the circuit behaves for resistance change at the load.

• If the load resistance, RL = 0 in practical voltage source circuit, then the load acts as a short circuit and hence the short circuit current flows through the load. So the VL is zero (VL = IL * RL) and the IL would be

IL = Vs/Rs

• Similarly for RL=0 in practical current source circuit, load also behaves as short circuit as it prefers the current flow through non-resistance path. This load current is equal to the source current Is which is equal to the value of Vs/Rs in practical voltage source circuit.

Therefore, Is = Vs/Rs when RL = 0…………(1)

• If the load resistance RL is infinity, both circuits behaves as an open circuit. Therefore load current is zero in both circuits. And the voltage drop across the resistance Rint in practical current source circuit is Is*Rint

Vint = Is * Rint

• And the voltage across the Rs in a practical voltage source circuit is equal to the Vs which is equal to the Is* Rint of the practical current source circuit.

Vs = Is * Rint when RL is infinity ………………(2)

Hence from equations 1 and 2, we get

Vs = Rs * Is

Vs = Rint * Is

By observing above two equations, if the internal resistance of the two sources is same then the two sources are electrically equivalent. These two sources are equivalent , they can produce the same values of IL and VL when connected to same load resistance. Hence, these equivalent sources can produce identical values of short circuited current and open circuit voltage when zero load resistance and infinity resistances respectively. Therefore, by interchanging the internal resistors we can transform their properties from current source to a voltage source and vice-versa.

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Conversion of Voltage Source to Current Source

From the above discussion a voltage source can be converted or transformed into a current source by interchanging a series resistor to parallel as shown in figure.

 Steps:

• Find the internal resistance of the voltage source and keep this resistor in parallel with a current source.
• Determine the current flow provided by current source by applying ohms law.

In the above figure, a voltage source with a resistance Rs is transformed into an equivalent current source with a parallel resistor Rs. And this current value is obtained by applying the simple ohms law as

Is = Vs/Rs.

Example:

Consider the below voltage source circuit with a voltage of 20 V and a internal resistance of 5 ohms. This circuit is transformed into the current source by placing a resistor of the same value with a current source. This current source value can be determined by,

Is    = Vs/Rs

= 20/ 5

= 4 amps

The equivalent current source with a current of 4A and parallel resistor of 5 ohms is shown below.

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Conversion of Current Source to Voltage Source

 The current source transformed into a voltage source by interchanging parallel resistor in series. Let us see how it could work.

Steps:

• Find the parallel resistance of the constant current source and place in series with a voltage source.
• Determine the open circuit voltage value of the voltage source by applying ohms law.

 In the above figure, a current source is converted into a voltage source by placing resistance Rs in series with a voltage source and the value of the voltage source is calculated as,

Vs = Is *Rs

 Example:

Consider the below example for current source transformation, where current source is of 10A with a parallel resistance of 3 ohms. To calculate the value of voltage in voltage source apply the simple ohms law, then,

Vs = Is * Rs

Vs = 10 * 3

= 30 Volts.

Therefore the equivalent voltage source of this transformation consists a voltage source 30 V with a series resistance 3 ohms.

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Example of Source Transformations

•  Consider the below example where we have to find the voltage Vo by applying the source transformation. This circuit consists of both current and voltage sources. Let us see how we can apply source transformation to simplify the circuit below.

•  In the circuit there are two areas where we can apply the source transformation since current source has a parallel resistor and voltage source has a series resistor as shown in figure. So these configurations are necessary requirements to apply the source transformation.

•  First, consider the current source with a parallel resistance of 4 ohms. This current source can be transformed into a voltage source by 4 ohms series with a voltage source and voltage source value is determined as

Vs = Is * R

= 3* 4

= 12 Volts

Consider the direction of the current as it downwards so the voltage terminals in voltage source are also changes as shown in figure.

•  Place the above voltage source with a series resistance in the circuit, then we get below figure.

•  Resistors 4 ohms and 2 ohms are in series , hence the total series resistance will be 6 ohms as shown below.

•  Again the voltage source of 12V with series resistor 6 ohms can be transformed into a current source. Therefore consider to transform it.

•  This 12V voltage source with 6 ohm resistor combination can be converted into the current source by placing 6 ohms resistor in parallel with a current source. And the value of current in current source can be determined as

Is    = Vs/R

= 12/6

= 2Amps

The direction of current flow is represented in below figure.

•   Insert the above current source in the main circuit, then we get

•  On the right hand side, there is a voltage source with a 3 ohms resistor so this can be transformed into a current source by placing a 3 ohm resistor in parallel with a current source and this current source value is calculated as

Is = Vs/Rs

= 12/ 3

= 4 Amps

The direction of current in current source is shown in figure.

•  Insert the above current source in simplified circuit, then we get a final circuit as

 From the above simplified circuit the current sources are appeared to be opposite to each other. The node current through the circuit will be

Is = I1 – I2

= 4-2

= 2 amps

 By applying the divider rule, the current through the resistor 8 ohms is

 Io = Is * (1/Ro/ ((1/Ro) + (1/R1) + (1/R2))

= 2 * (1/8/((1/8) + (1/6) + (1/3) )

= 0.4 Amps

Therefore, the voltage across the resistor 8 ohms is

Vo = Io * Ro

= 0.4 * 8

= 3.2 Volts

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