CHAPT.4-Dc Equivalent Circuit

8/3/2019 CHAPT.4-Dc Equivalent Circuit

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At the end of the chapter, student should be able to:

1. Explain methods of analysis for resistive circuit.

2. Explain Kirchoffs Law for current and voltage

3. Analyze Thevenins Theorem

4. Analyze Nortons Theorem

5. Analyze Superposition Theorem

6. Explain Maximums Power Transfer Theorem

CHAPTER 4

OBJECTIVE


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4.1 Resistive Circuit.

a) Nodal Analysis

The term node commonly used to refer to a junction of two or more branches.

Nodal analysis will provide the nodal voltages of a network, that is, the voltage from the

various nodes (junction points) of the network to ground. Although it is not a

requirement, we will make it policy to make ground our reference node and assign it a

potential level of zero volts. All the other voltages levels will then be found with respect

to this reference level. For a network of N, we will have (N-1) nodes for which the

voltages must be determined. In other words,

The number of nodes for which the voltage must be determined using nodal analysis is 1

less than the total number of nodes.

The result of the above is (N-1) nodal voltages that need to be determined,

requirement that (N-1) independent equations be written to find the nodal voltages. In

other words,

The number of equation required to solve for all the nodal voltages of a network is 1less

than the total number of independent nodes.

Nodal analysis can be applied by a series of carefully defined steps. The examples to

follow will explain each step in detail.

1. Determine the number of nodes (junctions of two or more branches) for the

network.

2. Pick a reference node (normally the ground connection), and label each of the

INPUT-4a


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remaining nodes with a subscripted label such as V1, V2 and so on.

3. Apply kirchoffs current law at each node except the reference node. For each

application of Kirchoffs Current Law, assume that each of the unknown current

leaves the node (this removes the concern about direction; a minus sign will

appear in the solution if incorrectly chosen).

4. Solve the resulting equations for the nodal voltages.

Apply nodal analysis to the network of Figure 4.1.

Figure 4.1

As shown in Figure 4.1, this network has three nodes, that is, three points in the network

where the branches are tied together. The bottom node, connected to ground, is defined as

the reference node, and the other two are labeled as shown. Since there are two nodes

defined by subscripted voltages, two equations will be required to solve for the nodal

voltages.

Example 4.1

SOLUTION 4.1


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Figure 4.2

Figure 4.3

The current directions have been added to Figure 4.3 for the application of

Kirchoffs current law to the node V1. As required by step 3 before, those currents I1 and

I3 are defined as leaving the node.

Applying Kirchoffs current law to node V1 will result in Ii = I0

Is1 = I1 + I3

Since the voltage across R1 is the subscripted voltage V1, the current I1 is defined

by

I1 = V1 / R1

The voltage across R3 is equal to V1 V2 resulting in I3 being defined by

I3 = V1 V2R2


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The voltage across R3 is V1 V2 rather than V2 V1 because of the chosen direction for

I3. Substituting into Kirchoffs current law equation above will result

Is1 = V1 + = V1 V2

R1 R2

Is1 = V1 ( 1 + 1 ) V2 ( 1 )R1 R2 R2

For node V2, the currents are redefined as shown in Figure 2, applying Kirchoffs

current law to node V2 will result in

Ii = I0

0 = I3 + I2 + Is2

The voltage across resistor R2 is the subscripted voltage V2, so the current I2 is defined

by

I2 = V2 / R3

Due to the chosen direction for I3, the voltage V2 so the current I2 is defined by

I3 = V2 V1

R2

Substituting into Kirchoffs current law equation above will result

0 = V2 + V2 V1 + Is2R3 R2

0 = V2 ( 1 + 1 ) V1 ( 1 ) + Is2R2 R3 R2

Resulting in two equations and two unknown:

Is1 = V1 ( 1 + 1 ) V2 ( 1 )

R1 R2 R2

- Is2 = V2 ( 1 + 1 ) V1 ( 1 )

R2 R3 R2Substituting values will result in

4A = V1 ( 1 + 1 ) V2 ( 1 )

2 12 6


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- 2A = V2 ( 1 + 1 ) V1 ( 1 )

12 6 12

which, after we drop units and work through the mathematics, can be written as

0.5833V1 0.1667V2 = 4A

0.0833V1 + 0.25V2 = - 2A

V1 = 4 - 0.1667

- 2 0.250.5833 -0.1667

- 0.0833 0.25

= (4)(0.25) (-2)(-0.1667) .(0.5833)(0.25) (-0.0833)(- 0.1667)

= 1 0.1667 .

0.146 0.007

= 0.834 = 6 V0.139

V2 = 4 0.5833

- 2 - 0.0833 .

0.5833 -0.1667

- 0.0833 0.25

= (0.583)(-2) (4)(-0.083).0.139

= - 1.166 + 0.332 .0.139

= - 0.834 = - 6 V

0.139

Since V1 = 6V is a positive quantity, the voltage at V1 is positive quantity, the voltage at

V1 is positive with respect to ground as shown in Figure 4. The current I1 can then be

determine from


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I1 = V1 = 6V = 3A

R1 2

Since V2 is a negative quantity, the voltage across R2 has the polarity shown in Figure

4. Therefore, I2 equal to

I2 = V2 = 6V = 1A

R3 6

Figure 4.4

The voltage across R2 is now V1 V2 = 6V (-6V) = 12V and the current through R2 has

the direction shown in Figure 4, its value is

I3 = V3 = 12V = 1A

R2 12

b) Mesh analysis.

The number of mesh currents required to analyze a network will equal the number

of windows of the configuration. Referring to Figure 4.5, the defined mesh current can

initially be a little confusing because it would appear that two currents have been defined

for resistor R2. Defining the current through R2 may seem a little troublesome. Thecurrent through R2 will simply be the difference between I1 and I2 with the direction of the

larger.

The procedure for writing the equations is as follow:-

Assign a current in the clockwise direction to each independent, closet loop of the

network.


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Within each loop, insert the polarities for each resistor as determined by the

assumed direction of the loop current.

Apply kirchoffs voltage law around each closed loop in the clockwise direction.

Solve the resulting equations for the assumed loop current.

Figure 4.5

Mesh: kirchoffs voltage law

Loop 1:

- E1 + V1 + V2 = 0

The voltage across resistor R1 is determined by

V1 = I1R1

However, V3 is determined by taking the loop current of the loop of interest, and then

subtracting the other loop current if it has the opposite direction and adding it if it has the

same direction. In this example, currents I1 and I2 have opposite directions, so the voltage

across V3 is determined by

V2 = (I1-I2) R2

Substituting E2 and E3 into E1 will result in

- E1 + I1R1 + (I1-I2) R2= 0

Loop 2:

V2 + V3 + E2 = 0

E1

E2

E3

E4

E5

I

1I2


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The voltage across R2 is simply

V3 = I2R3

And for R3 the following:

V2 = (I2-I1) R2

Resulting in the following equation:

(I2-I1) R2 + I2R3 + E2 = 0

Substituting values into the two equations (E4 and E8) will result in:

- 2V + I1(2) + (I1-I2) (4)= 0

(I2-I1) (4) + I2(1)+ 6V = 0

Dropping units, rearranging, and multiplying through will result in:

2 I1 + 4 I1 - 4 I2 = 2

4 I2 - 4 I1 +I2 = - 6

Or

6 I1 - 4 I2 = 2

5 I2 - 4 I1 = - 6

I1 = 2 -4

-6 5

6 -4

-4 5

= (2)(5) (-4)(-6)(6)(5) (-4)(-4)

= 10 2430 16

= -14 = -1 A

14Substuting I1 = -1 into E4c will result:

6(-1) - 4 I2= 2

-4 I2 = 8

I2 = 8 / -4

E6

E7

E8

E8a

E4a

E4b

E4c


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I2 = -2A

The negative signs indicate that currents I1 and I2 actually have the opposite direction

from the loop that we made before. Figure 4.2 shows the actual direction for the branch

currents of the network.

Figure 1.6

Using mesh analysis, find the current through resistor R3 of Figure 4.7.

Figure 4.7

I

1

I

2

Example 4.2


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Figure 4.8

Loop I1

- E2 + V2 + V1 + E1 = 0

V2 + V1 = E2 E1

Substituting V1 = I1R1 and V2 = (I1-I2)R2

Will result in (I1-I2)R2 + I1R1 = E2 E1

I1R2 - I2R2 + I1R1 = E2 E1

I1 (R2 + R1)- I2R2 = E2 E1

Loop I2

V3 + V2 + E2 = 0

V3 + V2 = - E2

Substituting V3 = I2R3 and V2 = (I2-I1)R2Will result in I2R3 + (I2-I1)R2 = - E2

I2R3 + I2R2 - I1R2 = - E2

I2 (R3 + R2) I1R2 = - E2

Substituting the values into the two resulting equations:

Loop 1: I1 (6 + 1)- 6I2 = 10 5

SOLUTION 4.2

I

1I

2


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7 I1 - 6I2 = 5

Loop 2: I2 (2 + 6) 6I1 = - 10

8 I2 6I1 = - 10

I1 = 5 -6 = 40 60 = - 20 = 1A

-10 8 36 56 - 20

7 -6-6 8

I2 = 5 7 = - 30 (-70) = - 40 = 2A

-10 -6 - 20 - 20- 20

The only current asked for is the current through resistor R3 that is I 2. Therefore, theanswer is I2 = 2A.

4.1) Using mesh analysis, find the current, I3 through resistor R3 of Figure 4.7

Figure 4.9

4.2) Using mesh analysis, find the current, I2 through resistor R2 of Figure 4.8

ACTIVITY 4.1


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Figure 4.10

4.3) Using nodal analysis, find the voltage for V1 and V2.

Figure 4.11

4.4) Using nodal analysis find the voltage for V1 and V2.


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Figure 4.12

4.1) I3 = 2A

4.2) I2 = 2.71A

4.3) V1 = - 16.8V , V2 = 19V

4.4) V1 = 9.056V , V2 = -3.01V

4.2 Explain Kirchoffs Law for current and voltage

Kirchoffs Voltage Law

The law specifies that the algebraic sum of the potential rises and drops around a closed

path (or closed loop) is zero. In symbolic it can be written as

V = 0

INPUT-4.2

ANSWER 4.1


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where represent summation. To selecting the direction around the closed path, any

direction will work as long as you get back to the starting point but to simplify matters,

you can always try to move in a clockwise direction.

Use Kirchoffs voltage law to determine the unknown voltage of Fig. 4.3

Figure 4.13

- E1 + V1 + VR2 + E2 = 0

- 16 + V1 + 4.2 + 9 = 0

V1 = 16 9 4.2

V1 = 2.8 V

For the series circuit in Figure 4.12:

a. Determine V2 using Kirchoffs voltage law

b. Determine I

c. Find R1 and R3

KVL

Example 4.3

SOLUTION 4.3

Example 4.4


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Figure 4.14

a. - E + V3 + V2 + V1 = 0

E = V1 + V2 + V3

V2 = E V1 V3

V2 = 54V 18V 15V

V2 = 21V

b. I2 = V2 / R2 = 21V / 7 = 3A

c. R1 = V1 / I1 = 18V / 3A = 6

R3 = V3 / I3 = 15V / 3A = 5

Kirchoffs Current Law (KCL)

The algebraic sum of the currents entering and leaving a junction (or region) of a

network is zero.

The law can also be stated in the following way:

The sum of the currents entering a junction (or region) of a network must equal the

sum of the currents leaving the same junction (or region).

SOLUTION 4.4


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In equation form the above statement can be written as follows:

Ii = Io

with Ii representing the current entering, or in and Io representing the current leaving, or

out.

Figure 4.15

In Fig 4.15 for example the node can enclose an entire system or a complex network, or it

can simply provide a connection point (junction) for the displayed currents.

Ii = Io

I2 + I3 = I4 + I1

6 + 8 = 10 + 4

14 A = 14 A

Using combination of Kirchoffs current and voltage law, find the current through the R3.

I1

= 4

I2

= 6 I3

= 8

I4

= 10

Example 4.5


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Loop I1:-

-15 + 15 I1 + 25 (I1-I2) = 0

15 I1 + 25 I1- 25 I2 = 15

40 I1 25 I2 = 15

Loop I2:-

+ 25 (I2-I1) + 15 I1 -25 = 0

40 I2 25 I1 = 25

I1 = 25 40 = - 625 600 = - 1225 = - 1.256A15 -25 1600 625 975

40 -25

-25 40

I2 = 25 -25 = - 375 1000 = - 1375 = - 1.410A

15 40 1600 625 975

40 -25-25 40

The current through R3 = I2 = -1.410A

SOLUTION 4.5


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4.5) Using the kirchoffs voltage law, prove that total of the voltage drop in Figure

4.16 equal to voltage source, E.

Figure 4.16

4.6) Determine currents I3 and I5 through application of kirchoffs current law.

Figure 4.17

ACTIVITY 4.2


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4.5) VR1 = 16.2V , VR2 = 27V, VR3 = 10.8V

V total = VR1 + VR2 + VR3 = 16.2 + 27 + 10.8 = 54V

4.6) I3 = 26A

I5 = 13A

4.3 Thevenin Theorem

Thevenins theorem is quite useful when the current in one branch of a network is to be

determined or when the current in an added branch is to be calculated.

It states that for the purpose of determining the current in a resistor, RL connected across

two terminals of a network which contains sources of e.m.f and resistor, the network can

be replaced by a single source of e.m.f and a series resistor, R th. This e.m.f, Eth is equal

to potential difference between the terminals of the network when the resistor, R, is

removed: the resistance of series resistor, Rth, is equal to the equivalent resistance of the

network with the resistor, R, removed.

Hence, I = E____ (RL + Rth)

Explanation:

Lets us consider the circuit shown in Figure 4.16. The following steps are required to find

the current through the load resistance, RL.

INPUT-4.3

ANSWER 4.2


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Figure 4.18

Step 1: Remove RL from the circuit terminals A and B and redraw the circuit as shown in

Figure 4.17.

Figure 4.19

Step 2: Replacing the voltage source by a short-circuit equivalent will result in the

configuration of Figure 4.18. The Thevenin resistance is then determined between

points a and b.

Figure 4.20

The thevenin resistance is then;

Rth = R2 || R1

Rth

A

B

A

B


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Step 3: Replacing the source will result in the configuration of Figure 4.20. The

important Thevenin voltage is then measured between the same two marked terminals a

and b. in this case, the thevenin voltage is the same as that across resistor R3 since points

a and b are connected directly to the ends of the R3 resistor.

Figure 4.21

Since we have series configuration, the thevenin voltage can be found by one simple

application of the voltage divider rule as follows:

Eth = R2 (E1)

R2 + R1

Step 4: The last step is to simply draw the resulting Thevenin equivalent circuit as shown

in Figure 4.20above and reattach the variable load resistor.

Figure 4.22


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The general equation for the current through the load resistor ofFigure 4.20is

IL = Eth .Rth + RL

Find the Thevenin equivalent circuit of Figure 4.21 above and find the current through RL

for values 2, 10 and 100.

Figure 4.23

Step 1:

Figure 4.24

Step 2:

Rth = R2 || R1

1 = 1 + 1 .

Rth R1 R2

1 = 1 + 1 . R th = 2

Rth 3 6

Step 3:

Eth = R2 (E1)

Example 4.6

SOLUTION 4.6


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R2 + R1

Eth = 6 (9) Eth = 6V6 + 3

Step 4:

Figure 4.25

IL = Eth .

Rth + RL

IL = 6 .

2+ RL

a) RL = 2

IL = 6 . = 1.5A2+ 2

b) RL = 10

IL = 6 . = 0.5A

2+ 10

c) RL = 100

IL = 6 . = 58.82mA

2+ 100

IL


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DC EQUIVALENT CIRCUIT AND NETWORKEE101/ CHAPTER 4 /25

4.7) Using Thevenins theorem, find the current, IL through resistor R3 of Figure 4.24

Figure 4.26

4.8) Sketch the Thevenins equivalent circuit for Figure 4.25 and find the current, IL

through resistor R4

Figure 4.27

ACTIVITY 4.3


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4.7) IL = 15.39mA

4.8)

Figure 4.28

IL = 0.937A

4.4 Nortons Theorem

Thevenins theorem clearly demonstrates that a two terminal dc network can be

replaced by a single voltage source and resistor in a series combination. Interestingly

enough, that same configuration can also be replaced by a single current source and resistor

in a parallel combination.

The current source equivalent is a result of nortons theorem, which states the

following:

INPUT-4.4

ANSWER 4.3


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Any two terminal dc networks can be replaced by an equivalent circuit consisting of a

current source and a parallel resistor as shown in Figure 4.27 above.

Figure 4.29

The Norton equivalent circuit like the Thevenin equivalent can be found by

following a series of steps that are similar in many respect to those applied to determine the

thevenin equivalent circuit. In fact, the first three steps are exactly the same, because the

thevenin and Norton resistances have the same value. The steps are as follows:

a) Remove the portion of the network across which the Norton equivalent circuit is to

be found.

b) Mark the terminals of the remaining two-terminal network

c) Find R N by first setting all sources to zero (voltage sources are replaced with short

circuits and currents sources with open circuit) and then finding the resistance between the

two marked terminals. (if a sources has an internal resistance, it must remain as part of the

network when the Norton resistance is determined.

d) Determine INby first returning all of the sources to their original positions and

finding the short-circuit current between the two marked terminals.

Find the Norton equivalent circuit for the network applied to resistor RL in Figure 4.28

above. (the same example for Thevenin example)

Figure 4.30

Example 4.7


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Step 1: Resistor RL was removed and the resulting terminals were marked as shown in

figure 4.29

Figure 4.31Step 2: Replacing the voltage source by a short-circuit equivalent will result in the

configuration of Figure 4.32.

Figure 4.32

The result is a parallel combination of resistor R2 and R1 so that the Norton resistance is

determined by:

RN = R2 || R1 = 6 || 3 = 2

Step 4: Replacing the voltage source and placing a short circuit between the marked

terminals will result in the network of Figure 4.33 the current through that short circuit

must now be determines to find the Norton current. Since the applied short circuit is in

parallel with resistor R2 all the current entering the junction of resistor R2 and the short

circuit will take the path of least resistance, which is that of the short circuit. The current

through resistor will be zero amperes, resulting in

SOLUTION 4.7


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V2 = I2R2 = (0A) (6A) = 0V

Figure 4.33

Applying Kirchoffs voltage law around the closed loop will result in

- E + V1 + V2 = 0

V1 = E V2

V1 = 9 0

V1 = 9V

And I1 = V1 . = 9V = 3A

R1

3

Current I1 will then continue through the short circuit and define the level of IN. That is,

IN = I1 = 3A. Therefore the equivalent circuit for Figure 4.29 is:

Figure 4.34

If RL = 2, so,

IL = 2 (3) = 1.5A this answer same with the answer in Example 4.4.

4


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Transform Thevenins Equivalent circuit to Nortons Equivalent circuit

The Norton equivalent circuit can be obtained directly from the Thevenin

equivalent circuit using a simple source conversion. The conversion equation appears in

Figure 4.33 above:

Figure 4.35

4.9) Find the Norton equivalent circuit for the network applied to resistor R3 in Figure

4.36.

Figure 4.36

4.10) Convert the Norton equivalent circuit (your answer in 4.9) to a Thevenin equivalent

circuit.

ACTIVITY 4.4


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.

4.9)

Figure 4.37

4.10)

Figure 4.38

4.5 Superposition Theorem

The Superposition theorem is unquestionably one of the most powerful in this field.

It has such widespread application that people often apply it without recognizing that their

maneuvers are valid only because of this theorem.

ANSWER 4.4

INPUT-4.5


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The superposition theorem states the following:

The current through, or voltage across, any element of a network is equal to the algebraic

sum of the currents or voltages produced independently by each source.

In other words, this theorem allows us to find a solution for a current or voltage

using only one source at a time. Once we have the solution for each source, we can

combine the results to obtain the total solution.

If we are to consider the effects of each source, the other sources obviously must be

removed. Setting a voltage source zero volts is like placing a short circuit across its

terminal. Therefore,

When removing a voltage source from network schematics, replace it with a direct

connection (short circuit) of zero ohms. Any internal resistance associated with the source

must remain in the network.

Setting a current source to zero amperes is like it with an open circuit. Therefore,

When removing a current source from a network schematic, replace it by an open circuit

of infinite ohms. Any internal resistance associated with the source must remain in the

network.

The above statements are illustrated in Figure 4.39.

Figure 4.39


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Using the superposition theorem, determine the current through the 12 resistor of Figure

4.40

Figure 4.40

Considering the effects of the 54V source will require replacing the 48V source by a short-

circuit equivalent as shown inFigure 4.41

Figure 4.40

As shown in the Figure, the result is that the 12 and 4 are in parallel.

RT = R1 + R2 || R3 = 24 + 12 || 4

= 24 + 3 = 27

IS = E1 = 54V = 2A

RT 27

Using the current divider rule will result in the contribution to I2 due to the 54V

source:

12 = R3 . (IS) = 4 (2) = 0.5AR3 + R2 4 + 12

Example 4.8

SOLUTION 4.8


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If we now replace the 54V source by a short circuit equivalent, the network of

Figure 4.41 will result.

Figure 4.41

The result is a parallel connection for the 12 and 24 resistor. Therefore, the

total resistance seen by the 48V source is

RT = R3 + R2 || R1 = 4 + 12 || 24

= 4 + 8 = 12

And the source current is

IS = E2 = 48V = 4A

RT 27

Applying the current divider rule will result in

12 = R1 . (IS) = 24 (4) = 2.667 A

R1 + R2 24 + 12

It is important to realize that current I2 due to each source has a different direction.

The net current will therefore be the difference of the two and the direction of the larger as

follows:

Figure 4.42


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I2 = I2 I2 = 2.667 A 0.5 A = 2.167 A

4.11) Find the voltage across the 2 of Figure 4.43 using the superposition theorem.

Figure 4.43

4.12) Using superposition, find current I1 for the network of Figure 4.44

Figure 4.44

4.11) VR1 = 6V

4.12) I1 = 3A

ACTIVITY 4.5

ANSWER 4.5


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4.5 Maximums Power Transfer Theorem

To explain the theory of Maximum Power Transfer, it if often important to know

the following:

What load should be applied to a system to ensure that the load is receiving maximum

power from the system?

And, conversely:

For a particular load, what condition should be imposed on the source to ensure that it

will deliver the maximum power available?

Even if a load cannot be set at the value that would result in maximum power

transfer, it is often helpful to have some idea of the value that will draw maximum power

so that you can compare it to the load at hand. The maximum power transfer states that:

A load will receive maximum power from a network when its resistance is exactly equal to

the Thevenin resistance of the network applied to the load. That is,

RL = RTh

In other words, for the Thevenin equivalent circuit, when the load is set equal to the

Thevenin resistance, the load will receive maximum power from the network.

With RL = RTh, the maximum power delivered to the load is set equal to thethevenin resistance, the load will receive maximum power delivered to the load can be

determined by first finding the current:

IL = Eth . = Eth . = Eth .

Rth + RL Rth + Rth 2Rth

Then substitute into the power equation:

PL = I2LRL = ( Eth / 2Rth )

2 x (Rth) = Eth2Rth / 4Rth

2

And

PL max = Eth2 / 4Rth

INPUT-4.6


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4.5.1 Power Versus Load Resistance Graphs.

The power to the load versus the range of resistor value is provided in Figure 4.45

note in particular that for values of load resistance less than the Thevenin value, the

change is dramatic as it approaches the peak value. However, for values greater than the

Thevenin value, the drop is a great deal more gradual. This is important because it tells

you the following:

If the load applied is less than the thevenin resistance, the power to the load will drop off

rapidly as its gets smaller. However, if the applied load is greater than the Thevenin

resistance, the power to the laod will not drop off as rapidly as its increases.

Figure 4.45

Although all of the above discussion centers on the power to the load, it is

important to remember the following:

The total power delivered by a supply such as Eth is absorbeb by both the Thevenin

equivalent resistance and the load resistance. Any power delivered by the source that does

not get to the load is lost to the Thevenin resistance.

The dc operating efficiency is defined as the ratio of the power delivered to the load

(PL) to the power delivered by the source (PS) that is


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% = PL x 100%

Ps

For the situation where RL = Rth

% = IL2 RL x 100% = RL x 100% = Rth x 100%IL

2 RT RT Rth + Rth

= Rth . x 100% = 1 . x 100%

2Rth 2

A dc generator, battery, and laboratory supply are connected to resistive load RL in Figure

4.46

a) For each, determine the value of RL for maximum power transfer.

b) Under maximum power condition, what are the powers to the load for each

configuration?

c) What is the efficiency of operation for each supply?

d) If a load of 1k were applied to the laboratory supply, what would the power

delivered to the load be? What is the level of efficiency? Compared your answer

with the previous answer in b)

a) dc generator b) battery c) laboratory supply

Figure 4.46

a) For maximum power transfer:

Dc generator RL = Rth = R1 = 2.5 Battery R L = Rth = Rint = 0.05

SOLUTION 4.9

Example 4.9


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Laboratory supply RL = Rth = Rint = 20

b) Load Power

Dc generator PLMAX = Eth2 = E2 = (120V)2 = 1.44kW

4Rth 4R1 4(2.5)

Battery PLMAX = Eth2 = E2 = (12V)2 = 720W

4Rth 4Rint 4(0.05)

Laboratory supply PLMAX = Eth2 = E2 = (40V)2 = 20W

4Rth 4Rint 4(20)

c) They are all operating under a 50% efficiency level because RL = Rth

d) The power to the load is determined as follows

IL = E = 40V = 40V = 39.22mA

Rint + RL 20 + 1000 1020

And PL = IL2RL = (39.22mA)

2 (1000) = 1.538W

The power level is significantly less than the 20W achieved in part b).The efficiency level

is

% = PL x 100% = 1.538W x 100% = 1.538W x 100%

PS EIS (40V)(39.22mA)

= 98.02%

which is markedly higher than achieved under maximum power conditions.

CHAPT.4-Dc Equivalent Circuit

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