circuits sol ch 04 partb

8/10/2019 Circuits Sol CH 04 PartB

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Physics I/II, English 123,Statics, Dynamics, Strength, Circuits,Fluids, C++,Numerical,Structure I/II,Economy

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4.10 Thevenin and Norton Equivalents

Thevenin and Norton equivalents are circuit

simplifications techniques that focus on terminal

behavior. We can best describe a Thevenin

equivalent circuit by reference to Fig. 4.44, which

represents any circuit made up of sources

(both independent and dependent) and resistors.

The letters a and b denote the pair of terminals of interest.

Figure 4.44(b) shows the Thevenin equivalent. Thus, a Thevenin equivalent circuit

is an independent voltage source VTh in series with a resistor RTh, which replaces an

interconnection of sources and resistors. This series combination of VTh and RTh is

equivalent to the original circuit in the sense that, if we connect the same load across

the terminals a, b of each circuit, we get the same voltage and current at the terminals

of the load.

Finding a Thevenin Equivalent

1)Calculate the open-circuit voltage as inFig. 4.45 which is equal to VTh.

.


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2)Calculate the short-circuit current as inFig. 4.46.

3)Calculate the Thevenin resistance which is the

ratio of the open-circuit voltage to the short-

circuit current as in Fig. 4.47.

The Norton Equivalent

A Norton equivalent circuit consists of an independent current source in

parallel with the Norton equivalent resistance. We can derive it from a Thevenin

equivalent circuit simply by making a source transformation. Thus the Norton current

equals the short-circuit current at the terminals of interest, and the Norton resistance

is identical to the Thevenin resistance.

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Using Source Transformations (independent sources):

Sometimes we can make effective use of

source transformations to derive a Thevenin or

Norton equivalent circuit.

For example, we can derive the Thevenin and

Norton equivalents of the circuit shown in Fig 4.45

by making the series of source transformations

shown in Fig. 4.48.

This technique is most useful when the network

contains only independent sources.

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Finding the Thevenin Equivalent of a Circuit with a Dependent Source

Example 4.10:

Find the Thevenin equivalent for the

circuit containing dependent sources

shown in Fig. 4.49.

Solution:

1.Calculating VTh(Open-Circuit):

Applying KVL at left loop:

Applying KVL at right loop:

Solving, 2.

Calculating (Short-Circuit):Applying KVL for the outer right loop (with no resistor, no voltage drop)

3.Calculating :

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The Deactivation Method (for independent Sources Only)

The technique for determining RTh that

we discussed and illustrated earlier is not

always the easiest method available. Two other

methods are generally simpler to use.

The first is useful if the network contains only

independent sources. To calculate RThfor such

a network, we first deactivate all independent

sources and then calculate the resistance seen

looking into the network at the terminal pair. A

voltage source is deactivated by replacing it

with a short circuit. A current source is

deactivated by replacing it with a short circuit. For example, consider the circuit

shown in Fig. 4.52. Deactivating the independent sources simplifies the circuit to the

one shown in Fig. 4.53. The resistance seen looking into the terminals a, b is denoted

Rab, which consists of the 4 resistor in series with the parallel combinations of the

5 and 20 resistors. Thus,

4.11 The Test Method (for Independent and Dependent Sources)

If the circuit or network contains dependent and independent sources, an

alternativeprocedure for finding the Thevenin resistance RThis as follows. We first

deactivate all independent sources, and we then apply either a test voltage source or a

test current source to the Thevenin terminals a, b. The Thevenin resistance equals the

ratio of the voltage across the test sources to the current delivered by the test source.

.

Figure 4.52A circuit used to illustrate

a Thvenin equivalent


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Finding the Thevenin Equivalent Using a Test Source

Example 4.11

Find the Thevenin resistance RTh for the circuit in Fig. 4.49, using the alternative

method described.

Solution

We first deactivate the independent voltage source from the circuit and then excite

the circuit from the terminals a, b with either a test voltage source or a test current

source. If we apply a test voltage source, we will know the voltage of the dependent

voltage source and hence the controlling current i. Therefore we opt for the rest

voltage source.

The externally applied test voltage source is denoted , and the current that itdelivers to the circuit is labeled . To find the Thevenin resistance, we simply solvethe circuit for the ratio of the voltage to the current at the test source; that is,

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In general, these computations are easier than those involved in computing the

short-circuit current. Moreover, in a network containing only resistors and

independent sources, you must use the alternative method because the ratio of the

Thevenin voltage to the short-circuit current is indeterminate. That is, it is the

ratio 0/0.

(Thevenin Method)

Method Indep. Only Dep. Only Indep + Dep.

Basic Method ---

Source Transformation --- ---

Deactivation --- ---

Test (alternative)

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Assessment Problem 4.16

Solution

To find RTh, replace the 72 V source with a short circuit:

The 5 and 20 resistors are in parallel, with

an equivalent resistance of 5 || 20 = 4 .

The equivalent 4 resistance is in series with

the 8 resistor for an equivalent resistance of

4 + 8 = 12 . The 12 equivalent resistance is

in parallel with the 12 resistor, so

Using node voltage analysis to find :The node voltage equations are:

Solving,

and

.

Find the Thvenin equivalent circuit with respect

to the terminals a, b for the circuit shown.


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Find the Norton equivalent circuit with respect


Solution

We perform a source transformation, turning the parallel combination of the 15 A

source and 8 resistor into a series combination of a 120 V source and an 8

resistor. Next, combine the 2 , 8 and 10 resistors in series to give an equivalent20 resistance. Then transform the series combination of the 120 V source and the

20 equivalent resistance into a parallel combination of a 6 A source and a 20

resistor.

Finally, combine the 20 v and 12 parallel resistors to give RN= 20||12 = 7.5 .

Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a7.5 resistor.

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A voltmeter with an internal resistance of 100 k

is used to measure the voltage in the circuitshown. What is the voltmeter reading?

Solution

Using source transformations, convert the series

combination of the -36 V source and 12 k

resistor into a parallel combination of a -3 mA

source and 12 k resistor.

Combine the two parallel current sources and the

two parallel resistors to give a -3 + 18 = 15 mA

source in parallel with a 12 k || 60 k = 10 k

resistor. Transform the 15 mA source in parallel

with the 10 k resistor into a 150 V source in

series with a 10 k resistor, and combine this

10 k resistor in series with the 15 k resistor.

The Thevenin equivalent is thus a 150 V source in series with a 25 k resistor

Using voltage division:

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Find the Thvenin equivalent circuit with respect


Solution

Calculating the open circuit voltage, which is also vTh,

Solving,

Using the test source method to calculate RTh,replace

the voltage source with a short circuit, the current

source with an open circuit:

Applying KCL equation at the middle node:

Solving,

The Thevenin equivalent is an 8 V source in series with a 1 resistor.

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Problem 4.63

Find the Thvenin equivalent with respect to the

terminals a, b for the circuit in Fig. P4.63.

Solution

Using deactivating method, we can calculate RTh

as follows:

Using voltage divider:

The final circuit of Thevenin is

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Problem 4.64



Solution

Applying node-voltage to get v1:

Using deactivation method to get :

The final circuit of Thevenin is

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Problem 4.65



Solution

Using source transformation, the circuit becomes:

Applying node-voltage method:

Solving,

and

Using deactivation method to get RTh:

The final Thevenin circuit is:

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Problem 4.66

Find the Norton equivalent with respect to the

terminals a, b in the circuit in Fig. P4.66.

Solution

The 8 mA current source and the 20 k resistor

will have no effect on the behavior of the circuit

with respect to the terminals a, b. This is because

they are in parallel with an ideal voltage source.

Which can be transformed to:

Which can be simplified to Norton equivalent:

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Problem 4.67

A voltmeter with a resistance of 100 is used tomeasure the voltage in the circuit in Fig. P4.67.

a)What is the voltmeter reading?

b)What is the percentage of error in the

voltmeter reading if the percentage of error is

defined as [(measured-actual)/actual] 100?

Solution:

a)

Using source transformation method, we can get through the following steps:

1.

Converting 30 V along with series 10 kresistor to 3 mA current source parallel to

10 kresistor, then combining 10 k|| 40 kto 8 k.

2.Converting 3 mA current source along with

parallel 8 k resistor to 24 V voltage source in

series along with 8 k resistor.

3.Converting 24 V voltage source in series

along with 12 k resistor to 2 mA current

source in parallel along with 12 k resistor,

and adding current sources together.

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Continue Problem 4.67

4.Combining (10 mA with 12 k) to (120 Vwith 12 k, adding (12 k with 3 k) to15 k, transferring (120 V with 15 k) to

(8 mA with 15 k), combining resulting

(15 k || 10 k) to 6 k, finally transferring

(8 mA and 6 k) to (48 V and 6 k).

b)

( )

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a)

Find the Thvenin equivalent with respect to

the terminals a, b for the circuit in Fig. P4.68

by finding the open-circuit voltage and the

short-circuit current.

b)

Solve for the Thvenin resistance by

removing the independent sources. Compare

your result to the Thvenin resistance found in (a)

Solution

a)

Open circuit:

Short circuit:

b)

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8.4


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Problem 4.71

Determine the Thvenin equivalent with

respect to the terminals a, b the circuit shown

in Fig. P4.71.

Solution

Open circuit:

Applying Ohms law for right loop:

Applying KVL for middle loop:

Applying KCL at middle node:

Solving,

.Short circuit:

The final Thevenin circuit is:

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Problem 4.72

Find the Thvenin equivalent with respect to

the terminals a, b for the circuit seen in

Fig.P4.72

Solution

Open circuit:

Short circuit:

Applying mesh method to get isc

Solving,

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4.73 When a voltmeter is used to measure the

voltage in Fig. P.4.73, it reads 7.5V.a. What is the resistance of the voltmeter?

b.

What is the percentage of error in thevoltage measurement?

Solution

a)

Use source transformationsto simplify the left side of the circuit, as follows:

1. Transfer (16 V in series with 4 k) to (4 mA in series with 4 k).

3. Transfer (4 mA in parallel with 2.4 k) to (9.6 V in series with 2.4 k)

4. Add (9.6 V to 0.4 V) and (2.4 k to 0.1 k)to get the following shape:

Applying KVL for left loop:

b)

( )

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Problem 4.77


terminals a, b in the circuit in Fig. P4.77.

Solution

VTh = 0, since there are no independent sources,

we must apply the test method:

Solving,

The equivalent Thevenin circuit is:

Note: is zero since there are no independent sources.

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1A


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Problem 4.78


terminals a, b for the circuit seen in Fig. P.4.78.

Solution:

= 0 since there are no independent resourcesin the circuit. Therefore, we must apply the test

methodas there are no independent resources.

Applying node-voltage method

Substituting in :

Substituting in :

1

2

3

3 2

4 1

4


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4.12 Maximum Power Transfer

Maximum power transfer can best be

described with the aid of the circuit shown in

Fig. 4.58. We assume a resistive network

containing independent and dependent

sources and a designated pair of terminals, a,

b to which a load, RLis to be connected.

The problem is to determine the value of RL

that permits maximum power delivery to RL.

The first step in this process is to recognize

that a resistive network can always be

replaced by its Thevenin equivalent.

Therefore, we redraw the circuit shown in

Fig. 4.58 as the one shown in Fig. 4.59.

Replacing the original network by its

Thevenin equivalent greatly simplifies the

task of finding RL.

Derivation of RL requires expressing the

power dissipated in RL as a function of the

three circuit parameters VTh,

RTh, and RL.

Thus,

( )

.

Fig. 4.59


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Next, we recognize that for a given circuit, VThand RThwill be fixed.

Therefore the power dissipated is a function of the single variable RL.

To find the value of RLthat maximizes the power, we use the elementary calculus.

We begin by writing an equation for the derivative of p with respect to RL:

[ ]

The derivative is zero and p is maximized when

Solving this equation yields

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Example 4.12: Calculating The Condition for Maximum Power Transfer

a)For the circuit shown, find the value of RL that

results in maximum power being transferred to RL.

b)Calculate the maximum power that can be

delivered to RL.

c)When RLis adjusted for maximum power transfer, what percentage of the power

delivered by the 360 V source reaches RL?

Solution

a)

Applying the node-voltage equation

Using Deactivation method:

c) When RLequals , the voltage is

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Find the Thvenin equivalent circuit with respect to

the terminals a, b for the circuit shown. (Hint: Define

the voltage at the leftmost node as

, and write two

nodal equations with as the right node voltage.)

Solution

Using the node voltage method

Solving with:

Using test source method to calculate the testcurrent and thus . Replace the current sourcewith a short circuit and apply the test source.

Applying KCL equation at the rightmost node:

Solving with:

Thus, the Thevenin equivalent is a 30 V source in series with a 10 resistor.

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a)Find the value or R that enables the

circuit shown to deliver maximum power

to the terminals a, b.

b)Find the maximum power delivered to R.

Solution

Applying node-voltage to get VTh:

Solving,

Creating a short circuit between nodes a and b and use the

mesh current method to get

Solving,

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Continue Assessment Problem 4.21

Thus,

a)

For maximum power transfer, b)

The Thevenin voltage,

, is divided equally between the Thevenin

resistance and the load resistance, so


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Assume that the circuit in Assessment problem 4.21

is delivering maximum power to the load resistor R.

a)How much power is the 100 V source

delivering to the network?

b)Repeat (a) for the dependent voltage source.

c)

What percentage of the total power generated

by these two sources is delivered to the load

resistor R?

Solution

According to Assessment 4.21, R = 3 :

Using mesh current method:

Solving,

a) b)

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Problem 4.79

The variable resistor (Ro) in the circuit in

Fig.P4.79 is adjusted until the power dissipated in

the resistor is 1.5 W. Find the values of Ro that

satisfy this condition.

Solution

We need to find the Thevenin equivalent with respect to .1)Transfer ( 100 V in series with 50 ) to ( 2 A in parallel with 50 ).

2)

50 || 200 = 40 .

3)Transfer ( 2A in parallel with 40 ) to ( 80 V in series with 40 ).

4)40 + 60 = 100 .

Applying KVL for the left loop:

Using the test-source methodto find the Thevenin

resistance gives

Using the node voltage method:

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Solving,

Solving for :

( ) .

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Problem 4.80

The variable resistor (RL) in the circuit in

Fig.P4.80 is adjusted for maximum power

transfer to RL.

a)Find the numerical value of RL.

b)Find the maximum power transferred to RL.

Solution

a)

Finding the Thevenin equivalent:

Open circuit voltage:


The dependent source constraint equation is:

Solving,

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Continue Problem 4.80:

Short-circuit current:


The dependent source constraint equation is:

Solving,

b)

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Problem 4.81

The variable resistor in the circuit in Fig. P4.81 is

adjusted for maximum power transfer to Ro.

a)

Find the value of Ro.b)Find the maximum power that can be delivered to Ro.

Solution

a)

Finding Thevenin equivalent:

Open Circuit Voltage:

1)

Convert ( 9 mA in parallel with 2 k )

to ( 18 V in series with 2 k )

2)2 k + 4 k = 6 k

Applying node voltage method:

Solving,

Using deactivation method to calculate :

= b)

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Problem 4.86

The variable resistor (Ro) in the circuit in

Fig.P4.86 is adjusted for maximum power

transfer to Ro.

a)Find the value of Ro.

b)Find the maximum power that can be delivered to Ro.

Solution

a)

Finding the Thevenin equivalent:( Open Circuit Voltage ):


Solving,

( Short Circuit Current ):


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Solving,

For maximum power transfer,

b)

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4.13 Superposition

Whenever we have more than one independent source, we can study the effect

of one-by-one source and add these effects together to result in the same result of the

overall system.

We demonstrate the superposition principle by

using it to find the branch currents in the circuit

shown in Fig. 4.62. We begin by finding the

branch currents resulting from the 120 V

voltage source.

To find the component of the branch currents

resulting from the current source, we deactivate

the ideal voltage source and solve the circuit

shown in Fig. 4.64.

.

Figure 4.63The circuit shown in Fig. 4.62

with the current source deactivated.


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The currents

and

in Fig. 4.62 are:

For circuits containing both independent and dependent sources, you must recognize

that the dependent sources are never deactivated.

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Example 4.13

Use the principle of superposition to find in thecircuit shown in Fig. 4.66

Solution

We begin by finding the component of resultingfrom the 10 V source. Fig. 4.67 shows the circuit.

With the 5 A source deactivated, must equal

. Hence,

must b zero, the branch

containing the two dependent sources is open, and

When the 10 V source is deactivated, the circuit

reduces to the one shown in Fig. 4.68. And applying

the node-voltage equations yield:

Solving with:

The value of is the sum of and , or 24 V

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Problem 4.91

a)Use the principle of superposition to find the

voltage in the circuit of Fig. P4.91.b)

Find the power dissipated in the 20 resistor.

Solution

a)

75 V source acting alone:

6 A source acting alone:

Our circuit reduces to

b)

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Problem 4.93

Use the principle of superposition to find the current io

in the circuit in Fig. P4.93.

Solution



8 A current source acting alone:

Transferring to resistors:

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Problem 4.94

Use the principle of superposition to find in thecircuit in Fig. P4.94.

Solution:

Voltage source acting alone:

Using node voltage method:

(

)

Current source acting alone:

Using node voltage method:

circuits sol ch 04 partb

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