Superposition theorem - Sathyabama University · Faculty of Electrical & Electronics Basic...

Faculty of Electrical & Electronics Basic Electrical Engineering

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Department of Electrical and Electronics Engineering

Superposition theorem

Superposition theorem states that:

In a linear circuit with several sources the voltage and current responses

in any branch is the algebraic sum of the voltage and current responses due to

each source acting independently with all other sources replaced by their internal

impedance.

OR

In any linear circuit containing multiple independent sources, the current or

voltage at any point in the network may be calculated as algebraic sum of the

individual contributions of each source acting alone

The process of using Superposition Theorem on a circuit:

To solve a circuit with the help of Superposition theorem follow the following

steps:

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

branch simply by using KCL, KVL or Ohm’s Law.

Repeat step 2 and 3 for every source the circuit has.

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.


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The superposition theorem is not applicable for the power, as power is

directly proportional to the square of the current which is not a linear function.

Steps:

1) Select any one source and short all other voltage sources and open all

current sources if internal impedance is not known. If known replace them by

their impedance.

2) Find out the current or voltage across the required element, due to the

source under consideration.

3) Repeat the above steps for all other sources.

4) Add all the individual effects produced by individual sources to obtain the

total current in or across the voltage element.

Problem 1

Find Ia in circuit shown below, where only the current source is kept in the

circuit. The 5V is zeroed out yielding a 0V source, or a short. The 9V is zeroed

out, making it a short also.

Note that the 8K resistor is shorted out (that is, 8K in parallel with 0 yields 0.


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Note that the 3mA flowing up through the 2K will split left and right at the top.

Part of it will flow through the 1K and part of it will flow through the 4K. Let's use

the label "I4" for the current flowing right through the 4K resistor. If we combine

the parallel 6K and 7K (6K||7K = 3.2K) and then add the series 4K, the total

resistance on the right is 7.2K. Now we can use a current divider to find that I4 =

[1K / (1K + 7.2K)] * 3mA = 0.37mA. Note that the 2K does not enter into this

computation because the entire 3mA flows through it. The 3mA does not split

until it gets to the junction at the top of that branch.

Now that we know I4, we can then split it again through the 6K and the 7K.

Ix = [7K/(6K+7K)]*I4 = 0.20 mA.

Problem 2

Using the superposition theorem, determine the voltage drop and current

across the resistor 3.3K as shown in figure below.

Solution:


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Step 1: Remove the 8V power supply from the original circuit, such that the

new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 2K are in parallel, therefore resultant resistance will be

1.245K.

Using voltage divider rule voltage across 1.245K will be

V1= [1.245/(1.245+4.7)]*5 = 1.047V

Step 2: Remove the 5V power supply from the original circuit such that the

new circuit becomes as the following and then measure voltage across resistor.

Here 3.3K and 4.7K are in parallel, therefore resultant resistance will be

1.938K.

Using voltage divider rule voltage across 1.938K will be

V2= [1.938/(1.938+2)]*8 = 3.9377V


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Therefore voltage drop across 3.3K resistor is V1+V2 =

1.047+3.9377=4.9847

Reciprocity Theorem

The reciprocity theorem states that in a linear passive bilateral network by

changing the voltage source from branch 1 to branch2, the current I in the branch 2

appears in branch 1.

1. Verify the reciprocity theorem for the given network.

Find the current in branch AB

RT = 2+[3 // (2+2 //2)]

= 3.5Ω

IT =V/RT =20/3.5 = 5.71A

Apply current division technique for the circuit to find current through branch AB

= 5.71 x 3/6

= 2.855A


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Current in branch AB = 2.855 x 2/4 = 1.43A

Interchanging the source to branch AB

Total resistance RT = 3.23Ω

IT = 20/3.23 = 6.19A

Apply current division technique for the circuit to find current through branch CD

= 6.19 x 2/5.2

= 2.38A

Current in branch CD = 2.38 x 3/5 = 1.43A

Hence reciprocity theorem verified.

2. Verify the reciprocity theorem for the given circuit diagram.

Apply current division technique for the circuit to find current through branch 3Ω

= 10 x 2/5 = 4A


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Voltage across 3Ω = 4 x 3 = 12V

Replace the current source and find the open circuit voltage

Apply current division technique and find the current through 2Ω resistor.

=10 x 3/5 = 6

Voltage across 2Ω resistor = Voltage across AB = 6 x 2 = 12V

Thevenin’s theorems Objectives

• To understand the basic philosophy behind the Thevenin’s theorem and

its application to solve dc circuits. • Explain the advantage of Thevenin’s theorem over conventional circuit

reduction techniques in situations where load changes.

Introduction A simple circuit is considered to illustrate the concept of equivalent circuit and it

is always possible to view even a very complicated circuit in terms of much

simpler equivalent source and load circuits. Subsequently the reduction of

computational complexity that involves in solving the current through a branch for different values of load resistance ( RL ) is also discussed. In many applications,

a network may contain a variable component or element while other elements in the circuit are kept constant. If the solution for current ( I ) or voltage (V ) or power ( P ) in any


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component of network is desired, in such cases the whole circuit need to be analyzed each time with the change in component value. In order to avoid such repeated computation, it is desirable to introduce a method that will not have to be repeated for each value of variable component. Such tedious computation burden can be avoided provided the fixed part of such networks could be converted into a very simple equivalent circuit that represents either in the form

Let us consider the circuit shown in figure). Our problem is to find a

current through RL using different techniques; the following observations are

made.

Find

• Mesh current method needs 3 equations to be solved • Node voltage method requires 2 equations to be solved • Superposition method requires a complete solution through load

resistance ( RL ) by considering each independent source at a time and

replacing other sources by their internal source resistances. Suppose, if the value of RL is changed then the three (mesh current method) or

two equations (node voltage method) need to be solved again to find the new

current in RL . Similarly, in case of superposition theorem each time the load

resistance RL is changed, the entire circuit has to be analyzed all over again. Much of the tedious

mathematical work can be avoided if the fixed part of circuit or in other words, the

circuit contained inside the imaginary fence or black box with two terminals A & B

, is replaced


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The procedure for applying Thevenin’s theorem To find a current IL through the load resistance RL using Thevenin’s theorem, the

following steps are followed:


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Step-1: Disconnect the load resistance ( RL ) from the circuit, as indicated in

Step-2: Calculate the open-circuit voltage VTH at the load terminals ( A& B ) after

disconnecting the load resistance ( RL ). In general, one can apply any of the

techniques (mesh-current, node-voltage and superposition method) to compute

VTh (experimentally just measure the voltage across the load terminals using a

voltmeter). Step-3: Redraw the circuit with each practical source replaced by its internal

resistance .note, voltage sources should be short-circuited (just remove them

and replace with plain wire) and current sources should be open-circuited (just

removed).


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Step-4: Look backward into the resulting circuit from the load terminals ( A& B ) , as suggested by the eye in Calculate the resistance that would exist between the load terminals or equivalently one can think as if a voltage source is applied across the load terminals and then trace the current distribution through the circuit in order to calculate the resistance across the load terminals. The resistance RTh is described in the statement of Thevenin’s theorem. Once again, calculating this resistance may be a difficult task but one can try to use the standard circuit reduction technique or Y − or −Y transformation techniques.

Step-5: Place RTh in series with VTh to form the Thevenin’s equivalent circuit (replacing the imaginary fencing portion or fixed part of the circuit with an equivalent practical voltage source) Step-6: Reconnect the original load to the Thevenin voltage the load’s voltage, current

and power may be calculated by a simple arithmetic operation only.


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Load current IL =

VTh

RTh

+

RL

VTh

Voltage across the load V = × R = I L ×R

L L L

RTh + RL

Power absorbed by the load P = I L

2 ×R

L L

Remarks: (i) One great advantage of Thevenin’s theorem over the normal circuit

reduction technique or any other technique is this: once the Thevenin equivalent circuit

has been formed, it can be reused in calculating load current ( IL ), load voltage (VL )

and load power ( PL ) for different loads using the equations (ii) Fortunately, with help of

this theorem one can find the choice of load resistance RL that results in the maximum

power transfer to the load. On the other hand, the effort necessary to solve this

problem-using node or mesh analysis methods can be quite complex and tedious from

computational point of view. Application of Thevenin’s theorem

Example: 1 For the circuit find the current through

RL = R2 = 1 Ω resistor ( Ia −b branch) using Thevenin’s theorem & hence calculate the

voltage across the current source (Vcg ).


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Solution: Step-1: Disconnect the load resistance RL and redraw the circuit Step-2: Apply any method (say node-voltage method) to calculateVTh . At node C: 2 + I1 + I 2 =0

2 +

(3

−3

Vc

) +

(0

−6V

c

) ⇒ Vc = 6 volt

Now, the currents I1 & I2 can easily be computed using the following expressions.

I1 = Va

−3

Vc = 3

−

3 6

= −1 A (note, current I1 is flowing from ‘c’ to ‘a’)

I2 = 0

−

6Vc =

−6

6 = −1 A (note, current I2 is flowing from ‘c’ to ‘g’)

Step-3: Redraw the circuit indicating the direction of currents in different branches. One

can find the Thevenin’s voltage VTh using KVL around the closed path ‘gabg’ (see


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VTh = Vag − Vbg = 3 − 2 =1volt

Step-4: Replace all sources by their internal resistances. In this problem, voltage

source has an internal resistance zero (0) (ideal voltage source) and it is short-circuited

with a wire. On the other hand, the current source has an infinite internal resistance

(ideal current source) and it is open-circuited (just remove the current source).

Thevenin’s resistance RTh of the fixed part of the circuit can be computed by looking at

the load terminals ‘a’- ‘b’

RTh = ( R1 + R3 ) & R4 = ( 3 + 4 )&2 = 1.555Ω

Step-5: Place RTh in series with VTh to form the Thevenin’s equivalent circuit (a simple

practical voltage source). Reconnect the original load resistance RL = R2 = 1 Ω to the

Thevenin’s equivalent circuit (note the polarity of ‘a’ and ‘b’ is to be considered carefully)


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IL =

VTh

= 1

= 0.39 A ( a to b)

1.555 +1

R

Th

+

RL

Step-6: The circuit is redrawn to indicate different branch currents. Referring to one can

calculate the voltage Vbg and voltage across the current source (Vcg ) using the following

equations.

Vbg = Vag − Vab = 3 − 1 × 0.39 =2.61 volt.

Ibg = 2.61

2 = 1.305 A; Icb = 1.305 − 0.39 = 0.915 A

Vcg = 4 × 0.915 + 2 ×1.305 =6.27 volt.

Example-2 For the circuit shown in fig.8.4 (a), find the current IL through 6 Ω resistor

using Thevenin’s theorem.

Solution: Step-1: Disconnect 6 Ω from the terminals ‘a’ and ‘b’ and the corresponding circuit diagram. Consider point ‘g’ as ground potential and other voltages are measured with


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respect to this point. Step-2: Apply any suitable method to find the Thevenin’s voltage (VTh ) (or potential between the terminals ‘a’ and ‘b’). KVL is applied around the closed path ‘gcag’ to compute Thevenin’s voltage. 42 − 8 I − 4 I − 30 = 0 ⇒ I =1 A Now, Vag = 30 + 4 = 34 volt ; Vbg = 2 × 3 =6 volt.

VTh = Vab = Vag − Vbg = 34 − 6 =28volt

Step-3: Thevenin’s resistance RTh can be found by replacing all sources by their internal resistances ( all voltage sources are short-circuited and current sources are just removed or open circuited)


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RTh = ( 8 & 4)+ 2 = 8

12×4

+ 2 = 14

3 =4.666 Ω

Step-4: Thevenin’s equivalent circuit is now equivalently represents the original circuit

IL =

VTh

= 28 =2.625 A

4.666 +6

RTh + RL

Example-3 The box consists of independent dc sources and resistances.

Measurements are taken by connecting an ammeter in series with the resistor R and

the results are shown in table.


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Table

R I

10 Ω 2 A

20 Ω 1.5 A

? 0.6 A

Solution: The circuit can be replaced by an equivalent Thevenin’s voltage source .The

current flowing through the resistor R is expressed as

I = V

Th

R

+ R

Th

The following two equations are written from measurements recorded in table.

V

Th = 2 ⇒ V − 2 R =20

RTh

+ 10

Th Th

V

Th = 1.5 ⇒ V − 1.5 R =30

RTh

+ 20

Th Th

VTh = 60 volt ; RTh =20Ω

I =

V

Th =

60 = 0.6 ⇒ R =80 Ω.

R + R 20 +R

Th


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Proof of Thevenin’s Theorem The basic concept of this theorem and its proof are based on the principle of

superposition theorem.

It is assumed that the dc resistive network is excited by the independent voltage and

current sources. In general, there will be certain potential difference (Voc =VTh ) between

the terminals ‘a’ and ‘b’ when the load resistance RL is disconnected from the terminals.

Additional voltage source E (ideal) is connected in series with the load resistance RL in

such a way ( polarity of external voltage source E in opposition the open-circuit voltage

Voc across ‘a’ and ‘b’ terminals) so that the combined effect of all internal and external

sources results zero current through the load resistance RL .

According to the principle of superposition, zero current flowing through RL can be

considered as a algebraic sum (considering direction of currents through RL ) of (i)

current through RL due to the external source E only while all other internal sources are

replaced by their internal resistances (all voltage sources are short-circuited and all


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current sources are open circuited), and (ii) current through RL due to the combined

effect of all internal sources while the external source E is shorted with a wire.

For the second case, the current I2 (due to combined effect of all internal sources only)

is flowing through RL with same magnitude of I1 but in opposite direction (left to right).

Note that the resultant current I through the resistor RL is zero due to the combination of

internal and external sources

NORTON’S THEOREM

Network with voltage and current sources and only resistances can be replaced

at terminals A-B by an equivalent current source INO in parallel connection with an

equivalent resistance RNO. This equivalent current INO is the current obtained at

terminals A-B of the network with terminals A-B short circuited. This equivalent

resistance RNO is the resistance obtained at terminals A-B of the network with all its

voltage sources short circuited and all its current sources open circuited.

Simple Steps to Analyze Electric Circuit through Norton’s Theorem

1. Short the load resistor. 2. Calculate / measure the Short Circuit Current. This is the Norton Current (IN). 3. Open Current Sources, Short Voltage Sources and Open Load Resistor. 4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN). 5. Now, Redraw the circuit with measured short circuit Current (IN) in Step (2) as current Source and measured open circuit resistance (RN) in step (4) as a parallel resistance and connect the load resistor which we had removed in Step (3). 6. This is the Equivalent Norton Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed. 7. Now find the Load current flowing through and Load Voltage across Load Resistor by using the Current divider rule. IL = IN / (RN / (RN+ RL)) .

https://en.wikipedia.org/wiki/Voltage_source

https://en.wikipedia.org/wiki/Current_source

https://en.wikipedia.org/wiki/Resistance_(electricity)

https://en.wikipedia.org/wiki/Series_and_parallel_circuit

https://en.wikipedia.org/wiki/Short_circuit

https://en.wikipedia.org/wiki/Short_circuit

https://en.wiktionary.org/wiki/open_circuit


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Example:

Find RN, IN, the current flowing through and load Voltage across the load resistor in fig (1) by using Norton’s Theorem.

Step 1.

Short the 1.5Ω load resistor. Step 2.

Calculate / measure the Short Circuit Current. This is the Norton Current (IN). We have shorted the AB terminals to determine the Norton current, IN. The 6Ω and 3Ω are then in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω. So the Total Resistance of the circuit to the Source is:- 2Ω + (6Ω || 3Ω) ….. (|| = in parallel with). RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → IT= 2Ω + 2Ω = 4Ω. RT = 4Ω IT = V / RT IT = 12V / 4Ω IT = 3A.. Now we have to find ISC = IN… Apply CDR… (Current Divider Rule)… ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

ISC= IN = 2A.

Step 3. Open Current Sources, Short Voltage Sources and Open Load Resistor Step 4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN) We have Reduced the 12V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure (4) We can see that 3Ω resistor is in series with a parallel combination of 6Ω resistor and 2Ω resistor. i.e.: 3Ω + (6Ω || 2Ω) ….. (|| = in parallel with) RN = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)] RN = 3Ω + 1.5Ω RN = 4.5Ω


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Step 5. Connect the RN in Parallel with Current Source INand re-connect the load resistor. This is shown in fig (6) i.e. Norton Equivalent circuit with load resistor.

Step 6. Now apply the last step i.e. calculate the load current through and Load voltage across load resistor by Ohm’s Law as shown in fig 7. Load Current through Load Resistor… IL = IN x [RN / (RN+ RL)] = 2A x (4.5Ω /4.5Ω +1.5kΩ) → = 1.5A IL = 1. 5A And Load Voltage across Load Resistor… VL = IL x RL VL = 1.5A x 1.5Ω VL= 2.25V Problem 2

First we remove the 10Ω resistor and short circuit the terminals A&B.

The current flowing through the short circuited terminals is called the Norton’s curren IN. To find the IN we apply nodal equation for point C (V – 30)/5 + V/10 + V/3 = 0 V = 180/19 Ohms law to 3Ω resistance I = V/R IN = (180/19) / 3

http://electricaltechnology.org/2013/10/ohms-law-with-simple-explanation.html


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IN = 60/19 This is the Norton’s current. Now we are about to find the Norton’s resistance RN = (10//5) + 3 RN = 19/3Ω

Maximum Power Transfer Theorem

In an electric circuit, the load receives electric energy via the supply sources and converts that energy into a useful form. The maximum allowable power receives by the load is always limited either by the heating effect (in case of resistive load) or by the other power conversion taking place in the load. The Thevenin and Norton models imply that the internal circuits within the source will necessarily dissipate some of power generated by the source. A logical question will arise in mind, how much power can be transferred to the load from the source under the most practical conditions? In other words, what is the value of load resistance that will absorbs the maximum power from the source? This is an important issue in many practical problems and it is discussed with a suitable example.

F

Fig:1

Let us consider an electric network the problem is to find the choice of the

resistance RL so that the network delivers maximum power to the load or in other words

what value of load resistance RL will absorb the maximum amount of power from the

network. This problem can be solved using nodal or mesh current RL analysis to obtain

an expression for the power absorbed by, then the derivative of this expression with

respect to RL will establish the condition under what circumstances the maximum power

transfer occurs. The effort required for such an approach can be quite tedious and

complex. Fortunately, the network can be represented by an equivalent Thevenin’s

voltage source.


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Fig:2 Circuit for maximum power transfer

In fig.2 a variable load resistance RL is connected to an equivalent Thevenin circuit of

original circuit(fig.8.6(a)). The current for any value of load resistance is

𝐼𝑙 =𝑉𝑇ℎ

𝑅𝑇ℎ + 𝑅𝐿

Then, the power delivered to the load is

PL = I L 2

VTh

2

× RL = ×RL

RTh + RL

The load power depends on both RTh and RL; however, RTh is constant for the equivalent Thevenin network. So power delivered by the equivalent Thevenin network to the load resistor is entirely depends on the value of RL. To find the value of RL that absorbs a maximum power from the Thevenin circuit, we differentiate PL with respect to RL.

𝑑𝑃(𝑅)

𝑑𝑅𝐿= 𝑉𝑇ℎ

2(𝑅 + 𝑅)2 − 2𝑅 ∗ (𝑅 + 𝑅)

(𝑅𝑇ℎ = 𝑅𝐿)4= 0

(𝑅𝑇ℎ + 𝑅𝐿) − 2𝑅𝐿 = 0

Therefore, 𝑅𝐿 = 𝑅𝑇ℎ For maximum power dissipation in the load, the condition given below must be satisfied

𝑑2𝑃(𝑅)

𝑑𝑅𝐿2 =

−𝑉𝑇ℎ2

8𝑅𝑇ℎ< 0

This result is known as “Matching the load” or maximum power transfer occurs when the

load resistance RL matches the Thevenin’s resistance RTh of a given systems. Also,

notice that under the condition of maximum power transfer, the load voltage is, by voltage

division, one-half of the Thevenin voltage. The expression for maximum power

dissipated to the load resistance is given by


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Efficiency =IL

2RL

2Il2RL

∗ 100 = 50%

For a given circuit, VTh and RTh are fixed. By varying the load resistance RL , the power

delivered to the load varies as shown in fig

Remarks: The Thevenin equivalent circuit is useful in finding the maximum power that a

linear circuit can deliver to a load. Problem: For the circuit shown in fig.4 find the value of RL that absorbs maximum

power from the circuit and the corresponding power under this condition.

Solution: Load resistance RL is disconnected from the terminals ‘a’ and ‘b’ and the

corresponding circuit diagram.


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The above circuit is equivalently represented by a Thevenin circuit and the corresponding Thevenin voltage VTh and Thevenin resistance RTh are calculated by following the steps given below: Now applying ‘Super position theorem’, one can find VTh (voltage across the ‘a’ and ‘b’ terminals. Note any method (node or mesh analysis) can be applied to find VTh. Considering only 20v source only

From the above circuit the current through ‘b-c’ branch =20

20= 1𝐴 (𝑓𝑟𝑜𝑚 𝑎 𝑡𝑜 𝑏).whereas

the voltage across the ‘b-a’ branch vba =1 ×10 =10 volt . (’b’ is higher potential than ‘a’).

∴vab = − 10volt. Considering only 10v source only


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Note: No current is flowing through ‘cb’-branch. Vab = 5v (‘a’ is higher potential than ‘b’). Consider only 2 A current source only

To compute RTh: Replace all voltage and current sources by their internal resistance of the circuit.

RTh = Rab = ((5+5) || 10) + (10 ||10) = 5 + 5 = 10 Ω

Thevenin equivalent circuit is drawn below:

The choice of RL that absorbs maximum power from the circuit is equal to the value of Thevenin resistance RTh

RL = RTh = 10Ω

Under this condition, the maximum power dissipated to RL is

𝑃𝑚𝑎𝑥 =1

4

𝑉𝑇ℎ2

𝑅𝑇ℎ=

1

4

25

10= 0.625𝑤𝑎𝑡𝑡𝑠

Problem: Find the value of RL for the given network below that the power is maximum?

And also find the Max Power through load-resistance RL by using maximum power

transfer theorem?


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Solution:

For the above network, we are going to find-out the value of unknown resistance called “RL”. In previous post, I already show that when power is maximum through load-resistance is equals to the equivalent resistance between two ends of load-resistance after removing. So, for finding load-resistance RL. We have to find-out the equivalent resistance like that for this circuit:-


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30


Now, for finding Maximum Power through load-resistance we have to find-out the

value of Vo.c. Here, Vo.c is known as voltage between open circuits. So, steps are:-

For this circuit using Mesh-analysis. We get Applying Kvl in loop 1st:- 6-6I1-8I1+8I2=0 -14I1+8I2=-6 ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (1) Again, Applying Kvl in loop 2nd:- -8I2-5I2-12I2+8I1=0


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8I1-25I2=0 ∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙∙ (2) On solving, eqn (1) & eqn (2), we get I1 = 0.524 A I2 = 0.167 A Now, from the circuit Vo.c is VA-5I2- VB = 0 Vo.c/ VAB= 5I2 = 5X0.167 = 0.835v So, the maximum power through the RL is given by:-

𝑃𝑚𝑎𝑥 =𝑉𝑜.𝑐

2

4𝑅𝐿

𝑃𝑚𝑎𝑥 =0.8352

4 ∗ 3.77= 0.046 𝑤𝑎𝑡𝑡𝑠

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