CHAPTER 4 - CIRCUIT THEOREMS

List of topics for this chapter :Linearity PropertyPrinciple of SuperpositionSource TransformationsThevenin's TheoremNorton's TheoremMaximum Power TransferVerifying Circuit Theorems with PSpice

LINEARITY PROPERTY

Linearity is the condition in which the change in value of one quantity is directly proportional tothat of another quantity. A linear circuit is one whose output is linearly related (or directlyproportional) to its input.

Problem 4.1 If all the independent sources are multiplied by a value, K, and all thecurrents and voltages of the circuit increase by the same value, then a circuit is linear. Show thatthe circuit in Figure 4.1 is linear by solving for xI and xV .

Figure 4.1

Use nodal analysis to find 1v and 2v .

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ

+−

K10 V

Ix

K2 A5 ΩΩΩΩ

+ −−−−Vx

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ

+−

K10 V

v1 v2Ix

K2 A5 ΩΩΩΩ

+ −−−−Vx

At node 1 : At node 2 :

05

vv

10

0v

10

10Kv 2111 =−

+−

+−

2K5

0v

5

vv 212 =−

+−

Simplifying,0)vv)(2(v10Kv 2111 =−++− 10Kvvv 212 =+−

10Kv2v4 21 =− 10Kv2v- 21 =+

The system of simultaneous equations is

=

10K

10K

v

v

21-

2-4

2

1

=

=

++

=

−=

K)325(

K)320(

K)650(

K)640(

40K10K

20K20K

6

1

10K

10K

41

22

28

1v

v

2

1

Clearly, K)35-(K)325(K)320(vvV 21x =−=−=and K)31-()51K]()35-[(5VI xx ===

The voltage source and current source are multiples of K. The voltage xV and the current xI aremultiples of K. Therefore, the circuit is linear.

Problem 4.2 [4.3](a) In the circuit in Figure 4.1, calculate ov and oi when V1vs = .(b) Find ov and oi when V10vs = .(c) What are ov and oi when each of the 1-Ω resistors is replaced by a 10-Ω resistor

and V10vs = ?

Figure 4.1

1 ΩΩΩΩ

1 ΩΩΩΩ

+−

io+

vo

−−−−

vs 1 ΩΩΩΩ

1 ΩΩΩΩ

1 ΩΩΩΩ

First, transform the Y (or T) subcircuit to its ∆ (or Π) equivalent.

R4

3

R4

R3R3||R

2

== R2

3R

4

3R

4

3=+

2vv so = , independent of Rand

Rvi oo =

(a) When V1vs = and R = 1 Ω,=ov V5.0 =oi A5.0

(b) When V10vs = and R = 1 Ω,=ov V5 =oi A5

(c) When V10vs = and Ω= 10R ,=ov V5 =oi A5.0

3R

3R

+−

io+

vo

−−−−

vs R

R

3R

(a)

3R+−1 V1.5R

(b)

PRINCIPLE OF SUPERPOSITION

The superposition principle states that the voltage across (or current through) an element in alinear circuit is the algebraic sum of the voltages across (or currents through) that element due toeach independent source acting alone.

Problem 4.3 Solve for xI in Figure 4.1 using superposition.

Figure 4.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.Using the principle of superposition, we will need to find the desired current when the currentsource is turned off, or set equal to zero. This implies the replacement of the current sourcewith an open circuit. Let this value be equal to Ix′. We will also need to find the desiredcurrent when the voltage source is turned off. This implies the replacement of the voltagesource with a short circuit. Let this value be equal to Ix″. Then, the desired current is the sumof these two currents, i.e.,

Ix = Ix′ + Ix″

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.This circuit, having two independent sources, can be analyzed using nodal or mesh analysis,as seen in Chapter 3. The problem statement requires the use of superposition. The principleof superposition can be very useful, especially when there are several independent sources.More work may be required to find the desired voltage or current, but the analysis isperformed using simpler circuits.

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ

+−

30 V

Ix

–6 A5 ΩΩΩΩ

+ −−−−Vx

ATTEMPT a problem solution.Setting the current source to zero, the circuit becomes

Use nodal analysis to find 1v and 2v .At node 1 : At node 2 :

05

vv

10

0v

10

30v 2111 =−+−+− 05

0v

5

vv 212 =−+−

Simplifying,0)vv)(2(v30v 2111 =−++− 0vvv 212 =+−

30v2v4 21 =− 0v2v- 21 =+15vv2 21 =−

The system of simultaneous equations is

=

0

15

v

v

21-

1-2

2

1

=

=

−=

5

10

15

30

3

1

0

15

21

12

14

1v

v

2

1

Clearly, 5510vv'V 21x =−=−= volts.

and 15

5

5

'V'I xx === amp.

Setting the voltage source to zero, the circuit becomes

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ

+−

30 V

v1 v2Ix'

5 ΩΩΩΩ

+ −−−−Vx'

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ v1 v2Ix"

–6 A5 ΩΩΩΩ

+ −−−−Vx"

Use nodal analysis to find 1v and 2v .At node 1 : At node 2 :

05

vv

10

0v

10

0v 2111 =−+−+− 6-5

0v

5

vv 212 =−+−

Simplifying,0)vv)(2(vv 2111 =−++ 30-vvv 212 =+−

0v2v4 21 =− 30-v2v- 21 =+0vv2 21 =−

The system of simultaneous equations is

=

30-

0

v

v

21-

1-2

2

1

=

=

−=

20-

10-

60-

30-

3

1

30-

0

21

12

14

1v

v

2

1

Clearly, 10-20)(10-vv"V 21x =−=−= volts.

and 25

10

5

"V"I xx === amps.

The sum of the currents, 'I x and "Ix , due to the two independent sources acting alone is the

current, xI , of the circuit due to both sources.

"I'II xxx +=321Ix =+= amps.

EVALUATE the solution and check for accuracy.Find xI using both sources and nodal analysis. The circuit is as follows.

At node 1 : At node 2 :

05

vv

10

0v

10

30v 2111 =−+−+− 6-5

0v

5

vv 212 =−+−

10 ΩΩΩΩ

5 ΩΩΩΩ10 ΩΩΩΩ

+−

30 V

v1 v2Ix

–6 A5 ΩΩΩΩ

+ −−−−Vx

Simplifying,0)vv)(2(v30v 2111 =−++− 30-vvv 212 =+−

30v2v4 21 =− 30-v2v- 21 =+15vv2 21 =−

The system of simultaneous equations is

=

30-

15

v

v

21-

1-2

2

1

=

−−

=

−=

15-

0

6015

3030

3

1

30-

15

21

12

14

1v

v

2

1

Clearly, 15)15-(0vvV 21x =−=−= volts.

and 35

15

5

VI xx === amps.

This answer is the same as the answer obtained using the principle of superposition.Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to “ALTERNATIVE solutions” and continue through the process again.This problem has been solved satisfactorily.

=xI A3

Problem 4.4 [4.11] Apply the superposition principle to find ov in the circuit ofFigure 4.1.

Figure 4.1

4 ΩΩΩΩ

+−

+

vo

−−−−

3 ΩΩΩΩ

2 ΩΩΩΩ

6 ΩΩΩΩ

20 V

2 A

1 A

Let 3o2o1oo vvvv ++= , where 1ov , 2ov , and 3ov are due to the 20-V, 1-A, and 2-A sources,respectively. For 1ov , consider the circuit below.

Ω=+ 3)24(||6 V10)20)(21(v 1o ==

For 2ov , consider the circuit below.

Ω= 26||3 V1)4(224

2vo2 =

++=

For 3ov , consider the circuit below.

4 ΩΩΩΩ

+−

+

vo1

−−−−

3 ΩΩΩΩ

2 ΩΩΩΩ

6 ΩΩΩΩ

20 V

4 ΩΩΩΩ

+

vo2

−−−−

3 ΩΩΩΩ

2 ΩΩΩΩ

6 ΩΩΩΩ

1 A

4 ΩΩΩΩ

+

vo2

−−−−

3 ΩΩΩΩ

2 ΩΩΩΩ

6 ΩΩΩΩ

4 V

− +

4 ΩΩΩΩ

+

vo3

−−−−

3 ΩΩΩΩ

2 ΩΩΩΩ

6 ΩΩΩΩ

2 A

3 ΩΩΩΩ

3 ΩΩΩΩ

2 A

−−−− +vo3

Ω=+ 3)24(||6 3-)3)(1-(v 3o ==

Thus,=−+= 3110vo V8

Problem 4.5 A linear circuit has the following independent sources: 1V , 2V , and SI

along with a current, RI , through a resistor. It cannot be assumed that these are the onlyelements (resistors and dependent sources) in the circuit. Complete the following table.

Circuit number V1 (volts) V2 (volts) IS (amps) IR (amps)Condition #1 10 0 0 2Condition #2 0 5 0 –1Condition #3 0 0 3 1Condition #4 10 5 3Condition #5 10 –20 3Condition #6 6

This problem is based on linearity and the principle of superposition.

From the Conditions #1, #2, and #3, we know 2IR = amps when the only contributing independent source is 10V1 = volts. -1IR = amp when the only contributing independent source is 5V2 = volts. 1IR = amp when the only contributing independent source is 3IS = amps.

Condition #4 :

=+−=

+

+

= 112)1(3

3)1-(

5

5)2(

10

10IR A2

Condition #5 :

=++=

+

+

= 142)1(3

3)-1(

5

20-)2(

10

10IR A7

Condition #6 : The following are only three of an infinite set of solutions.

If only one source is not equal to zero

then ?V1 = =2V 0 =SI 0 6IR = )2(10V

6 1

= =1V 30

or =1V 0 ?V2 = =SI 0 6IR = )-1(5V

6 2

= =2V 30-

or =1V 0 =2V 0 ?IS = 6IR = )1(3I

6 S

= =SI 18

SOURCE TRANSFORMATIONS

A source transformation is the process of replacing a voltage source SV in series with a resistor R

with a current source SI in parallel with a resistor R or vice versa.

Problem 4.6 Given that the following circuits are linear, prove that the two circuits areequivalent, where SSS IVR = .

Two circuits are said to be equivalent if they have the same voltage-current relationship at theirterminals. Begin by finding xV and xI in the following circuit.

Clearly,

S

xSx R

VVI

−=

or

xSSx IRVV −=

RRS

+−

+−

VS V

R

+−

IS VRS

RRS

+−

+−

VS V

Vx

Ix

Using KVL,0VI)RR(V- xSS =+++

or

RR

VVI

S

Sx +

−=

Now, find xV and xI in this circuit.

Using KCL,

S

xxS R

VII +=

or

xSSSx IRIRV −=

We know that S

SS I

VR = . So,

xSSx IRVV −= .

Using mesh analysis,)II(RVIR SxSx −++VIRI)RR( SSxS −=+

RR

VIRI

S

SSx +

−=

We know that S

SS I

VR = . So,

RR

VVI

S

Sx +

−= .

In both circuits, xSSx IRVV −= and RRVV

IS

Sx +

−= .

Therefore, the two circuits are equivalent.

R

+−

IS VRS

Vx

Ix

Problem 4.7 [4.23] Given the circuit in Figure 4.1, use source transformation to find

oi .

Figure 4.1

Transforming only the current sources leads to Fig. (a). Continuing with source transformationsfinally produces the circuit in Fig. (d).

12 V

4 ΩΩΩΩ

4 A

5 ΩΩΩΩ10 ΩΩΩΩ

2 ΩΩΩΩ3 ΩΩΩΩ

+−

io

2 A

10 V

12 V

4 ΩΩΩΩ

10 ΩΩΩΩ

2 ΩΩΩΩ3 ΩΩΩΩ

+−

io 5 ΩΩΩΩ+ −+ −

12 V

(a)

12 V

4 ΩΩΩΩ

10 ΩΩΩΩ

10 ΩΩΩΩ

+−

io

(b)

22 V+−

12 V

4 ΩΩΩΩ

10 ΩΩΩΩ+−

io

2.2 A10 ΩΩΩΩ

(c)

Applying KVL to the loop in Fig. (d),

==→=++ 91i011i912- oo mA11.111

Problem 4.8 Using source transformations, solve for oI in Figure 4.1.

Figure 4.1

=oI A2

THEVENIN'S THEOREM

Thevenin's theorem states that a linear two-terminal circuit can be replaced by an equivalentcircuit consisting of a voltage source ThV in series with a resistor ThR , where ThV is the open-

circuit voltage at the terminals and ThR is the input or equivalent resistance at the terminals

when the independent sources are zero. An alternative way to find ThR is to find the short-

circuit current between the terminals, scI ; then,

ocTh VV =

sc

ocTh I

VR =

10 ΩΩΩΩ10 ΩΩΩΩ

15 ΩΩΩΩ

10 ΩΩΩΩ

8 ΩΩΩΩ

30 ΩΩΩΩ

20 ΩΩΩΩ

+−

100 V

Io

4 A

12 V

4 ΩΩΩΩ

+−

io

11 V

(d)

+−

5 ΩΩΩΩ

io

RL

RTh

+−

VTh

Note that finding the equivalent resistance between the terminals using this approach is a validmethod only if there are no dependent sources. Calculating ThR using ocV and scI is a validmethod with independent and dependent sources.

Finally, if 0IV scoc == , then the circuit must be excited with a voltage source or a currentsource at the terminals, as shown below, to find ThR .

Find oI if given a voltage source oV = 1 V. Find oV if given a current source oI = 1 A.

Now, =ThR 1/Io or RTh = Vo/1.

Problem 4.9 Solve for the Thevenin equivalent circuit as seen by the 10-Ω resistor withthe current oI flowing through it in Figure 4.1.

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.To find the Thevenin equivalent circuit, we need to find the open-circuit voltage ( ocV ) across

the terminals of the 10-Ω resistor with the current oI flowing through it. We also need to

find the short-circuit current ( scI ) through these terminals. Then,

ocTh VV = and scocTh IVR =

Because the circuit has no dependent sources, there is an alternative way to find the Theveninresistance. ThR is the input resistance at the terminals when the independent sources are setequal to zero. This method will not be used in the initial attempt to find a solution. It isbetter to find the short-circuit current and calculate the Thevenin resistance because it willwork for any circuit. However, we can find the input resistance at the terminals to check ourinitial solution.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.To find the open-circuit voltage, the obvious techniques are nodal analysis, mesh analysis, orsource transformations along with KVL. Because we want to find a voltage, let's eliminatemesh analysis as a promising technique. It is clear that nodal analysis produces a set of threeequations and three unknowns. Simplifying the circuit using source transformations yields asingle loop from which we can easily find the open-circuit voltage.

Vo

+ −Io

Io

+Vo

−−−−

To find the short-circuit current, a similar argument can be made. So, let's eliminate nodalanalysis because we are looking for a current rather than a voltage. Mesh analysis gives a setof equations to be solved. Lastly, source transformations allows the circuit to be reduced to asingle loop from which we can easily find the short-circuit current.

With the open-circuit voltage and the short-circuit current at the terminals, the Theveninequivalent circuit is found as stated above.

ATTEMPT a problem solution.Begin by finding the open-circuit voltage as seen by the element.

After a couple of source transformations, as seen in Problem 4.8, the circuit becomes

Using KVL, find ocV .

020I20I2060- =+++1I40I40 =→= amp

where

120

20V

20

V60I ococ =−=−=

20VV60 ococ −=−40Voc = volts

Now, find the short-circuit current through the element.

10 ΩΩΩΩ10 ΩΩΩΩ

15 ΩΩΩΩ8 ΩΩΩΩ

30 ΩΩΩΩ

20 ΩΩΩΩ

+−

100 V 4 AVoc

+

−−−−

Voc

+

−−−−

20 ΩΩΩΩ20 ΩΩΩΩ

+−

60 V 20 V+−

I

10 ΩΩΩΩ10 ΩΩΩΩ

15 ΩΩΩΩ8 ΩΩΩΩ

30 ΩΩΩΩ

20 ΩΩΩΩ

+−

100 V 4 AIsc

Again using source transformations, the circuit can be reduced as shown below.

Clearly, 410

40Isc == amps.

Then, 410

40R Th == ohms.

Therefore, the Thevenin equivalent circuit is

EVALUATE the solution and check for accuracy.To check the open-circuit voltage, perform nodal analysis using the circuit below.

For node 1, 08

vv

30

v

20

100v 2111 =−

++−

For node 2, 015

vv

8

vv 3212 =−

+−

For node 3, 0410

v

10

v

15

vv 3323 =−++−

10 ΩΩΩΩ

+−

40 V Isc

10 ΩΩΩΩ

10 ΩΩΩΩ

+−

40 V

10 ΩΩΩΩ10 ΩΩΩΩ

15 ΩΩΩΩ8 ΩΩΩΩ

30 ΩΩΩΩ

20 ΩΩΩΩ

+−

100 V 4 AVoc

+

−−−−

v1 v3v2

Simplifying these equations and putting them into matrix form results in

=

120

0

1200

v

v

v

82-0

8-2315-

030-50

3

2

1

which yields 48v1 = volts 40v2 = volts 25v3 = volts

Hence, 40vVV 2ocTh === volts

Since there are only independent sources, an alternative way to find ThR is to set the sourcesto zero and calculate the resistance of the modified network. After replacing the voltagesource with a short circuit and the current source with an open circuit, the network becomes

Combine the resistors to the left of the open circuit. (20 Ω || 30 Ω) + 8 Ω = 20 ΩCombine the resistors to the right of the open circuit. (10 Ω || 10 Ω) + 15 Ω = 20 Ω

An equivalent resistance of 10 Ω matches the value that was calculated using ocV and scI .

Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to “ALTERNATIVE solutions” and continue through the process again.This problem has been solved satisfactorily. The Thevenin equivalent circuit is as follows.

10 ΩΩΩΩ10 ΩΩΩΩ

15 ΩΩΩΩ8 ΩΩΩΩ

30 ΩΩΩΩ

20 ΩΩΩΩ

Req

20 ΩΩΩΩ20 ΩΩΩΩReq

10 ΩΩΩΩReq

10 ΩΩΩΩ

10 ΩΩΩΩ

+−

40 V

Problem 4.10 Find the Thevenin equivalent as seen by R in Figure 4.1.

Figure 4.1

First, find the open-circuit voltage, where ocTh VV = .

Using nodal analysis,At node 1 : At node 2 :

030

0v

20

25v 11 =−+− 35

0v2 =−

75v2v3 11 =+ 15v2 =15v1 =

Clearly, 01515vvV 21oc =−=−= volts.

Now, find the short-circuit current.

Clearly, 3020sc III −= .Perform KVL using the left loop.

0I30I2025- 3020 =++

5 ΩΩΩΩ

R

30 ΩΩΩΩ

20 ΩΩΩΩ

+−

25 V 3 A

Voc+ −−−−

5 ΩΩΩΩ30 ΩΩΩΩ

20 ΩΩΩΩ

+−

25 V 3 A

v1 v2

5 ΩΩΩΩ30 ΩΩΩΩ

20 ΩΩΩΩ

+−

25 V 3 A

I30 I5

I20 Isc

Begin by finding 20I in the following modified circuit.

0I7

30I2025- x20 =++ where 3II 20x +=

175)3I)(30(I140 2020 =++85I170 20 =

5.0I20 = amps

Now,0I30)5.0)(20(25- 30 =++

5.03015I30 == amps

Finally,05.05.0III 3020sc =−=−= amps.

Since 0/0 is undefined, we need to excite the circuit with a 1-V voltage source at the terminals ofR in order to find RTh.

Use mesh analysis to find oI . Then, ooTh IVR = .

For loop 1 : 0)ii)(30(i2025- 211 =−++For loop 2 : 0)ii)(5(V)ii)(30( 32o12 =−++− where 1Vo = voltFor loop 3 : 3-i3 = This is the constraint equation.

Simplifying,25i30i50 21 =−16-i35i30- 21 =+

30/7 ΩΩΩΩ

20 ΩΩΩΩ

+−

25 V 3 A

Ix

I20

5 ΩΩΩΩ30 ΩΩΩΩ

20 ΩΩΩΩ

+−

25 V 3 A

Vo

+ −

i1 i2

Io

i3

In matrix form,

=

16-

25

i

i

3530-

30-50

2

1

=

0588.0-

4647.0

i

i

2

1

Now, 0588.0i-I 2o == and 170588.01

I

VR

o

oTh === ohms.

Therefore, the Thevenin equivalent circuit is as follows.

Problem 4.11 [4.33] For the circuit in Figure 4.1, find the Thevenin equivalentbetween terminals a and b.

Figure 4.1

R

17 ΩΩΩΩ

+−

0 V

30 V

10 ΩΩΩΩ

20 ΩΩΩΩ10 ΩΩΩΩ

+−

a b

5 A 10 ΩΩΩΩ 10 ΩΩΩΩ

20 ΩΩΩΩ

−+ 20 V

To find ThR , consider the circuit in Fig. (a).

where Ω= 1020||20 .

Now, transform the wye subnetwork to a delta as shown in Fig. (b).

where Ω= 5.730||10 .

Hence, ==+== 15||30)5.75.7(||30RR abTh ΩΩΩΩ10

To find ThV , we transform the 20-V and the 5-V sources to obtain the circuit shown in Fig. (c).

10 ΩΩΩΩ

20 ΩΩΩΩ10 ΩΩΩΩa b

10 ΩΩΩΩ 10 ΩΩΩΩ

20 ΩΩΩΩ

(a)

10 ΩΩΩΩ

30 ΩΩΩΩa b

30 ΩΩΩΩ 10 ΩΩΩΩ30 ΩΩΩΩ

(b)

i1

30 V

10 ΩΩΩΩ

10 ΩΩΩΩ10 ΩΩΩΩ

+−

a b

10 ΩΩΩΩ

10 ΩΩΩΩ

50 V +−

10 V

− ++ –

i2

(c)

For loop 1,0i10i305030- 21 =−++

21 ii32- −= (1)

For loop 2,0i10i301050- 12 =−+−

21 i3i -6 += (2)

Solving (1) and (2),A0i1 = and A2i2 =

Applying KVL to the output loop,0i1030i10V- 21ab =−+−

V10Vab =

== abTh VV V10

Problem 4.12 Find the Thevenin equivalent as seen by R in Figure 4.1.

Figure 4.1

Therefore, the Thevenin equivalent circuit is as follows.

R

20 ΩΩΩΩ

+−

–20 V

20 ΩΩΩΩ –3 Ix

20 ΩΩΩΩ

+−

10 V

Ix

R 10 ΩΩΩΩ

NORTON'S THEOREM

Norton's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuitconsisting of a current source NI in parallel with a resistor NR , where NI is the short-circuit

current at the terminals and NR is the input or equivalent resistance at the terminals when theindependent sources are turned off.

Problem 4.13 Find the Norton equivalent of the circuit in Figure 4.1.

In Problem 4.9, 40Voc = volts, 4Isc = amps, and 10R Th = ohms.Recall that ThN RR = ; so 10R N = ohms.

Therefore, the Norton equivalent circuitis as shown.

Problem 4.14 [4.41] Given the circuit in Figure 4.1, obtain the Norton equivalent asviewed from terminals:

(a) a-b (b) c-d

Figure 4.1

(a) From the circuit in Fig. (a),

10 ΩΩΩΩ4 A 10 ΩΩΩΩ

3 ΩΩΩΩ

4 ΩΩΩΩ6 ΩΩΩΩ

+−

120 V 6 A

a b

d

c

2 ΩΩΩΩ

3 ΩΩΩΩ

4 ΩΩΩΩ6 ΩΩΩΩ

2 ΩΩΩΩ

Rth

(a)

==+= 4||4)3||62(||4R N ΩΩΩΩ2

For NI or ThV , consider the circuit in Fig. (b).

After some source transformations, the circuit becomes that shown in Fig. (c).

Applying KVL to the circuit in Fig. (c),

27i012i840- =→=++Hence,

V14i4VTh ==and === 214RVI NThN A7

(b) To get RN, consider the circuit in Fig. (d).

3 ΩΩΩΩ

4 ΩΩΩΩ6 ΩΩΩΩ

+−

120 V 6 A

+ −−−−

2 ΩΩΩΩ

VTh

(b)

4 ΩΩΩΩ2 ΩΩΩΩ

+−

40 V

+ −−−−

12 V

VTh

(c)

+−

2 ΩΩΩΩ

i

3 ΩΩΩΩ

6 ΩΩΩΩ

RN

(d)

2 ΩΩΩΩ

4 ΩΩΩΩ

==+= 6||2)3||6(4(||2R N ΩΩΩΩ5.1

To get IN, the circuit in Fig. (c) applies except that it needs slight modification asin Fig. (e).

27)242/()1240(i ++−= and 19i212VTh =+==== 5.119RVI NThN A667.12

Problem 4.15 Find the Norton equivalent of the circuit in Figure 4.1.

Figure 4.1

Therefore, the Norton equivalent circuit is as follows.

MAXIMUM POWER TRANSFER

Maximum power is transferred to the load when the load resistance equals the Theveninresistance as seen by the load ( ThL RR = ).

R–1/4 A 20 ΩΩΩΩ

10 ΩΩΩΩ

R

20 ΩΩΩΩ

20 ΩΩΩΩ

+−

30 V 2 A

2 ΩΩΩΩ

+−

i

(e)

+

VTh

−−−−12 V

Problem 4.16 Given the circuit in Figure 4.1, complete the following table.

Figure 4.1

VS (volts) RS (ohms) RL (ohms) Power of RL (watts)Condition #1 20 10Condition #2 V 10Condition #3 20 10Condition #4 20

Let p be the power of the load resistor, LR . Then, L

2

LS

S RRR

Vp

+

= .

We want to find the maximum power transferred to the load. The maximum power is transferredto the load when the load resistance is equivalent to the Thevenin resistance as seen by the load.

In this case, SL RR = and L

2S

R4

Vp =

Condition #1 : == SL RR ΩΩΩΩ10 === 40400

)10)(4(

)20(p

2

watts10

Condition #2 : == SL RR ΩΩΩΩ10 == )10)(4(V

p2

watts40V 2

Condition #3 : )10(10R

20p

2

S

+

=

Clearly, for maximum power, =SR ΩΩΩΩ0 . {with RL fixed, making Rs as small as possiblemaximizes the power to RL}

Then, ==

= )10)(2()10(10

20p 2

2

watts40 .

Condition #4 : For maximum power, =SR ΩΩΩΩ0 and =LR ΩΩΩΩ++++0 .

RS

+−

VS RL

Then, == + )0)(4(20

p2

watts−−−−∞∞∞∞ .

Problem 4.17 [4.59](a) For the circuit in Figure 4.1, obtain the Thevenin equivalent at terminals a-b.(b) Calculate the current in Ω= 8R L .(c) Find LR for maximum power deliverable to LR .(d) Determine that maximum power.

Figure 4.1

(a) ThR and ThV are calculated using the circuits in Fig. (a) and (b) respectively.

From Fig. (a), =++= 642R Th ΩΩΩΩ12

2ΩΩΩΩ

4 ΩΩΩΩ

4 A

a

b

2 A

20 V

+ −

RL

6 ΩΩΩΩ

Rth2 ΩΩΩΩ

6 ΩΩΩΩ4 ΩΩΩΩ

(a)

8 V

6 ΩΩΩΩ4 ΩΩΩΩ

(b)

2 ΩΩΩΩ

+−

− +

12 V

+ −

+

VTh

−−−−

20 V

From Fig. (b),020812V- Th =+++

=ThV V40

(b) =+

=+

=812

40

RR

Vi

Th

Th A2

(c) For maximum power transfer,== ThL RR ΩΩΩΩ12

(d) ===)12)(4(

)40(

R4

Vp

2

Th

2Th watts33.33

Problem 4.18 Using Figure 4.1, what values of SR and LR result in maximum powerdelivered to the load? What is the power absorbed by the load?

Figure 4.1

=SR ΩΩΩΩ0

=LR ΩΩΩΩ20

=RLp watts125

20 ΩΩΩΩ

RS

+−

100 V RL

20 ΩΩΩΩ

VERIFYING CIRCUIT THEOREMS WITH PSPICE

Problem 4.19 Solve Problem 4.1 using PSpice.

A voltage source in PSpice needs a value to simulate rather than performing a symbolicsimulation. So, let 1K = . Then, 35-Vx = volts and 31-Ix = amps. Clearly, xV is thevoltage across R3 and xI is the current flowing through R3 from left to right. Thus, for 1K ==== ,

==−=−= 667.1-333.8667.6vvV 21x V35-and 31-5VI xx == amps or =xI mA333.333- .

These answers agree with the answers obtained in Problem 4.1 given that 1K = .

Problem 4.20 Solve Problem 4.3 using PSpice.

Clearly, xV is the voltage across R3 and xI is the current flowing through R3 from left to right.

Thus, =−×=−= -15)()1025(vvV 12-21x V15 and == 5VI xx A3 .

This answer agrees with the answer obtained in Problem 4.3.

SearchHelpEWB Help PageWe want your feedbacke-Text Main MenuTextbook Table of ContentsProblem Solving WorkbookWeb LinksTextbook WebsiteOLC Student Center WebsiteMcGraw-Hill Website

PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsLinearity PropertyPrinciple of SuperpositionSource TransformationsThevenin’s TheoremNorton’s TheoremMaximum Power Transfer Verifying Circuit Theorems with PSpice

Chapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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