Circuit theorems linear circuit analysis

Chapter-4: Circuit Theorems
4.1 Introduction
• A major advantage of analyzing circuits using Kirchhoff’s laws is that
we can analyze a circuit without altering its original configuration.
• Disadvantage associated with this approach is that, analysis of a large,
complex circuit, involves tedious computations.
• Growth and diversity of electrical products has boosted the complexity
of modern day electrical circuits.
• To handle this complexity, engineers over the years have developed
some theorems to simplify circuit analysis.

• Such theorems include Thevenin’s and Norton’s theorems.
• In addition to circuit theorems, concepts like superposition, source
transformation, and maximum power transfer also aid circuit analysis.
• Since these theorems are applicable to linear circuits, we first discuss
the concept of circuit linearity.
4.2: Linearity Property.
• Linearity is the property of an element describing a linear relationship
between cause and effect.
• Although the property applies to many circuit elements, we shall limit
its applicability to resistors in this chapter.

• Linearity property is a combination of both the homogeneity (scaling)
property and the additivity property.
• The homogeneity property requires that if the input (also called the
excitation) is multiplied by a constant, then the output (also called the
response) is multiplied by the same constant.
• For a resistor, for example, Ohm’s law relates input i to output v, as,
• v = iR (4.1).
• If the current is increased by a constant k, then the voltage increases
correspondingly by k; that is, kiR = kv (4.2).

• The additivity property requires that the response to a sum of inputs is
the sum of the responses to each input applied separately.
• Using the voltage-current relationship of a resistor,
• If v1 = i1R and v2 = i2R, then, applying (i1 + i2) gives back
• (4.3).
• We say that a resistor is a linear element because the voltage-current
relationship satisfies both the homogeneity & the additivity properties.
• In general, a circuit is linear if it is both additive and homogeneous.
• A linear circuit consists of only linear elements, linear dependent and
independent sources.

• Simply put, a linear circuit is one whose output is linearly related (or
directly proportional) to its input.
• Throughout this book we consider only linear circuits.
• Note that as p = i2R = v2/R is a quadratic rather than a linear function,
the relationship between power and voltage (or current) is nonlinear.
• Thus, theorems covered in this chapter are not applicable to power.
• To illustrate the linearity principle, consider
the linear circuit shown in Fig. 4.1.
• It has no independent sources inside it & is
excited by voltage source vs serving as input.

• The circuit is terminated by a load R.
• We may take the current i through R as the output.
• Suppose vs = 10 V gives i = 2 A, then according to the linearity
principle, vs = 1 V will give i = 0.2 A.
• By the same token, i = 1 mA must be due to vs = 5 mV.
• Example 4.1: Find IO in circuit of Fig. 4.2, for vs = 12 V & vs = 24 V.
• Applying KVL to the two loops, yield,
• (4.1.1).
• (4.1.2).

• But vx = 2i1, Eq. (4.1.2) becomes, (4.1.3).
• Adding Eqs. (4.1.1) and (4.1.3) yields
•
• For vs = 12 V, IO = i2 = 12/76 A & for vs = 24 V, IO = i2 = 24/76 A.
• How? By inserting first vs = 12 V and then vs = 24 V in Eq. (4.1.3).
• Showing that when the source value is doubled, Io also doubles.
• Home Work: P_Problem 4.1, Problem 4.1 and 4.5.
• Example 4.2: Assume IO = 1 A and use linearity to find the actual
value of IO in the circuit of Fig. 4.4.

• Applying KCL at node 1 yield: I2 = I1 + IO = 1 +2 = 3 A.
• and I3 = V2/7 = 14/7 = 2 A.
• Applying KCL at node 2 rveal: I4 = I3 + I2 = 2 +3 = 5 A.
• Therefore; IS = I4 = 5 A.
• This shows that assuming IO = 1 A, gives IS = 5 A, thus actual source
current of 15 A will yield IO = 3 A. How? (5/1 = 15/x leads to x = 15/5 =3)
• When IO = 1 A, then,
• V1 = (3 + 5)Io = 8 V.
• And I1 = V1 ÷ 4 = 2 A.

• Problem 4.2; In the circuit of Fig 4.2,
let I = 4.5 mA, R1 = 8 k, R2 = 2 k,
R3 = 6 k, R4 = 3 k and R5 = 4 k..
• Find real value of vO employing linearity & assumption that vO = 1V.
• When vO = 1 V  iR5 = iR4 = 1 V  4 k = 0.25 mA,
• Therefore, vR4 = iR4 x R4 = 0.25 mA x 3 k = 0.75 V.
• Let the node between R2 and R4 as node-1 and R1 and R2 as node-2.
• The node voltage v1 = vO + vR4 = 0.75 + 1 = 1.75 V.
• We can now compute iR3 = v1  R3 = 1.75  6 k = 0.292 mA.

• Applying KCL at node-1, iR2 = iR3 + iR5 = 0.292 + 0.25 = 0.542 mA.
• Thus vR2 = iR2 x R2 = 0.542 mA x 2 k = 1.084 V.
• Voltage vR3 = iR3 x R3 = 0.292 mA x 6 k = 1.752 V
• Therefore, the node voltage v2 = vR2 + vR3 = 1.084 + 1.752 = 2.836 V.
• We can now find iR1 = v2  R1 = 2.836  8 k = 0.355 mA.
• KCL at node-2 reveal, I = iR1 + iR2 = 0.355 + 0.542 = 0.897  0.9 mA.
• Solving 0.9  1 = 4.5  x leads to x = 4.5  0.9 =5.
• Hence real value of vO, when I = 4.5 mA is vO = 5 x 1 = 5 V.
• Home Work: P_Problem 4.2 and Problem 4.4.

• OTB Example; Using
linearity & assumption
Io = 1 mA, determine
true value of Io in the circuit shown above, when I = 6 mA?
• When Io = 1 mA  V1 = 3kIo = 3 V.
• Hence I1 = V1  6k = 0.5 mA and therefore, I2 = I1 + Io = 1.5 mA.
• Total resistance on right & left side of source is 4 k & 12 k
respectively, thus source current must split in 4  12 = 1:3 ratio.
• That is when I2 = 1.5 mA then I3 = 0.50 mA and I = I2 + I3 = 2 mA.
• Since we know that I = 6 mA, therefore, Io = 3 mA.

4.3: Superposition
• If a circuit has two or more independent sources, it can be solved by
determining the contribution of each independent source to the
variable of interest and then adding them all together.
• This approach is known as the superposition.
• The idea of superposition rests on the linearity property.
• The superposition principle states that the voltage across (or current
through) an element in a linear circuit is the algebraic sum of the
voltages across (or currents through) that element due to each
independent source acting alone.

• Superposition is not limited to circuit analysis only but is applicable in
many fields where cause and effect show a linear relationship.
• The principle of superposition helps us to analyze a linear circuit with
more than one independent source by calculating the contribution of
each independent source separately.
• To apply superposition, every independent source is considered one at
a time while all other independent sources are turned off.
• This implies that we replace every voltage source by 0 V (or a short
circuit), and every current source by 0 A (or an open circuit).

• This way we obtain a simpler and more manageable circuit.
• Dependent sources are left intact because they are controlled by
circuit variables.
• Analyzing a circuit using superposition involve more work e.g. a
circuit with n-independent sources is equivalent to solving n-circuits.
• But, superposition does reduce a complex circuit to simpler circuits.
• As superposition is based on linearity, it is not applicable to the effect
on power due to each source.
• To find power the element’s current (voltage) must be calculated first.

• Example 4.3: Use the superposition
theorem to find v in the circuit of Fig. 4.6.
• There are two sources, so let v = v1 + v2.
• Where v1 and v2. are the contributions due to the 6 V voltage source
and the 3 A current source, respectively.
• To obtain v1 set the current source to zero by
open circuiting it, as shown in Fig. 4.7(a).
• Applying KVL to resulting loop in Fig. 4.7(a)
yield, 12 i1 – 6 = 0  i1 = 0.5 A.

• Thus
• We may also use voltage division to get v1 by computing.
•
• To get v2 short circuit the voltage source to zero, as in Fig. 4.7(b).
• Employing current division
• Hence
• Finally
• Home Work: P_Problem 4.3 and Problems 4.8 plus 4.9.
• Example 4.4: Find iO in the circuit of Fig. 4.9 using superposition

• Turning off the 20 V source results in circuit of Fig. 4.10(a).
• Circuit in Fig. 4.9 contains a dependent
source, which must be left intact.
• Let , where iʹO and iʺO are due to
current and voltage sources respectively.

• Applying mesh analysis to loop 1 yield; i1 = 4 A (4.4.2).
• For loop 2: (4.4.3).
• And for loop 3: (4.4.4).
• But at node 0: (4.4.5).
• Substituting values of i1 and i3 from Eqs. (4.4.2) and (4.4.5) into Eqs.
(4.4.3) and (4.4.4) reveal two simultaneous equations, i.e.
• (4.4.6).
• (4.4.7).
• Solving them yield: iʹO = 3.059 A (4.1.8).

• To find the current iʺO turn off the 4-A current source so that the circuit
looks like as shown in Fig. 4.10(b).
• For loop 4, KVL gives: (4.4.9).
• And for loop 5: (4.4.10).
• As , substituting it in Eqs (4.4.9) & (4.4.10) reveal;
• (4.4.11).
• (4.4.12).
• Solving these for iʺO yield = – 3.529 (4.1.13).
• Next substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives back,

• Current iO = iʹO + iʺO = – 0.471 A.
• Home Work: P_Problem 4.4 & Problem 4.11.
• Example 4.5: In circuit of Fig. 4.12, use
the superposition theorem to find i.
• In this case, there are three sources and hence three circuits to solve.
• Let i = i1 + i2 + i3,where where i1 , i2 , and i3 are due to the 12 V, 24 V,
and 3-A sources respectively.
• To get i1, remove all but 12 V source, as in the circuit of Fig. 4.13(a).
• Combining (4+8) || 4 = 3, reduces the circuit into single loop.

• By Ohm’s law i1 = 12 ÷ 6 = 2 A.
• To get i2, remove all but 24 V
source as in Fig. 4.13(b).
• Applying mesh analysis;
• In mesh ia:
• OR (4.5.1).
• In mesh ib: (4.52).
• Solving Eq. (4.5.1) and Eq. (4.5.2) reveals i2 = ib = – 1 A.
• To get i3, remove all but 3 A source as shown in circuit of Fig. 4.13(c).

• Using nodal analysis;
• At node v2: (4.5.3).
• At node v1: (4.5.4).
• Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to v1 = 3 V.
• Therefore; i3 = v1 ÷3 = 1 A.
• Thus
• Home Work: P_Problem 4.5 and Problems 4.12 and 4.13

4.5: Source Transformation
• Like series-parallel combination & Y transformation, source
transformation is another tool for simplifying circuits.
• Basic to these tools is the concept of equivalence.
• What is an equivalent circuit?
• It is circuit with same i-v characteristics as the original circuit.
• This permits a one to one correspondence between a voltage source in
series with resistor and its equivalent current source in parallel with
same resistor or vice versa, as shown in Fig. 4.15.

• Either substitution is known
as a source transformation.
• Two circuits shown in Fig. 4.15 are equivalent; provided they have the
same voltage-current relation at terminals a-b.
• It is easy to show that they are indeed equivalent.
• If the sources are turned off, the equivalent resistance at terminals a-b
in both circuits is R.
• When terminals a & b are short-circuited in both circuits of Fig.4.15,
the short-circuit current flowing from a to b is isc = vs÷ R in the circuit
on the left-hand side and isc = is for the circuit on the right-hand side.

• Thus, in order for the two circuits to be equivalent source
transformation requires that (4.5).
• Source transformation also applies to dependent sources, as shown in
Fig. 4.16, as long as we can ensure that Eq. (4.5) is satisfied.
• Like the  Y transformation, a source transformation does not affect
the remaining part of the circuit.
• Keep the following points in
mind when dealing with
source transformation.

1. From Fig. 4.15 (or Fig. 4.16) note that arrow of the current source is
directed toward the positive terminal of the voltage source.
2. Note from Eq. (4.5) that source transformation is not possible when
R = 0, which is the case with an ideal voltage source.
• However, for a practical, (non-ideal) voltage source, R ≠ 0.
• Also an ideal current source with R =  cannot be replaced by a finite
voltage source. More on ideal & non-ideal sources to follow later.
• Example 4.6: Using the source transformation find the voltage vO in
the circuit of Fig. 4.17.

• Transforming the current and voltage
sources produces the circuit of Fig. 4.18(a).
• Combining 4  and 2  resistors in series and transforming the 12 V
voltage source into a current source gives us Fig. 4.18(b).

• Next combine the 3  and 6  resistors in parallel to get 2 .
• Also combine the 2 A and 4 A current sources to get a 2 A source.
• This repeatedly application of source transformations, leads to the
circuit as shown in Fig. 4.18(c).
• Applying current division in Fig. 4.18(c);
• Hence;
• Class Work: Is there another way to find vO in circuit of Fig.4.18(c)?
• As both resistors in Fig. 4.18(c) are in parallel, they have the same
voltage across them, therefore;

• Home Work: P_Problem 4.6 and
Problems 4.20 plus 4.22
• Example 4.7: Find vX in Fig. 4.20?
• Circuit in Fig. 4.20 involves a voltage-
controlled dependent current source.
• Transform this dependent current source as well as the 6 V
independent voltage source as shown in Fig. 4.21(a).

• Can we transform the 18-V voltage source?
• NO, as it is not connected in series with any resistor.
• Two 2  resistors in parallel combine to give a 1  resistor, which is
in parallel with the 3 A current source.
• Thus 3 A current source is transformed to a 3 V voltage source as
shown in Fig. 4.21(b).
• Notice that the terminals for vX are intact.
• Applying KVL around the loop in Fig. 4.21(b) gives,
• (4.7.1).

• Applying KVL to the loop containing only the 3 V voltage source, the
1  resistor, and vX yields;
• (4.7.2).
• Substituting Eq. (4.7.2). into Eq. (4.7.1), yield;
•
• Also applying KVL to loop over 4  resistor, dependent and 18 V
voltage source in circuit of Fig. 4.21(b), yields same answer i.e.
•
• Replacing i = – 4.5A in Eq. (4.7.2) yield; vx = 3 – i = 3 + 4.5 = 7.4 V.
• Home Work: P_Problem 4.7 and Problems 4.28 plus 4.29.

4.5: Thevenin’s Theorem
• In every practical circuit some particular element (called load) has
variable value while other elements have fixed parameters.
• For example, a household outlet terminal may be connected to
different appliances constituting a variable load.
• Each time the variable element is changed, the entire circuit has to be
analyzed all over again.
• To resolve this problem, Thevenin’s Theorem provides a technique by
which the fixed part of the circuit is replaced by an equivalent circuit.

• Load in circuits of Fig. 4.23 may be a
single resistor or another circuit.
• Thevenin’s circuit consists of a voltage
source VTh in series with a resistor RTh.
• Thevenin’s equivalent circuit is illustrated
in Fig. 4.23(b) to left of the load.
• Thevenin’s Theorem states that a linear two-terminal circuit, with load
connected at terminals a & b, such as one represented in Fig. 4.23(a),
can be replaced by an equivalent circuit called Thevenin’s circuit.

• VTh is the open-circuit voltage (voc) at the terminals a & b.
• RTh is the input equivalent resistance (Rin) at the terminals a & b when
the independent sources are turned off.
• In find Thevenin resistance (RTh = Rin), we need to consider two cases.
• CASE 1: If the network has no dependent sources, then turn off all
independent sources. RTh is the input resistance (Rin) of the network
looking between terminals a & b, as shown in Fig. 4.24(b).

• CASE 2: If the network has dependent sources, we turn off all
independent sources.
• As with superposition, dependent sources are not to be turned off
because they are controlled by circuit variables.
• Next apply a voltage source vO at terminals
a & b and determine resulting current, then
RTh = vo/io, as shown in Fig. 4.25(a).
• Alternatively, insert a current source iO at
terminals a-b as shown in Fig. 4.25(b) and
find the terminal voltage, again; RTh = vo/io.

• Either of the two approaches will give the same result.
• In either approach we may assume any value of vO and iO.
• Most commonly used values are vO = 1 V or iO = 1A.
• It can and often occurs that RTh takes a negative value.
• In this case, the negative resistance (v = –iR) implies that the circuit is
supplying power, which is possible in circuit with dependent sources.
• Load current IL and voltage VL in a linear circuit terminated by a load
RL, as shown in Fig. 4.26(a) can easily be determined once its
Thevenin equivalent is obtained, as shown in Fig. 4.26(b).

• From Fig. 4.26(b); (4.8a).
• And (4.8b).
• Note from Fig. 4.26(b) that the Thevenin equivalent is a
simple voltage divider, yielding VL by mere inspection.
• Example 4.8: Find the Thevenin equivalent circuit of the circuit
shown in Fig. 4.27, to the left of the terminals a-b and then find the
current through RL = 6 , 16  & 36 ?
• Remove voltage source by short circuit
and the current source by open circuit.

• Resulting circuit is shown in Fig. 4.28(a).
• Thus,
• To find VTh apply mesh analysis to the circuit in Fig. 4.28(b).
• In mesh-2; i2 = – 2 A and in mesh-1; 16 i1 – 12 i2 = 32.
• Solving it reveals i1 = 0.5 A.
• Open Circuit voltage VOC = VTh = 12 (i1 – i2) = 12 (0.5 + 2) = 30 V.
• Class Work; Apply nodal analysis at node VTh , it yields same answer.

• Solution:  multiplying by 12 & solving it,
•
• Why have we ignored 2  resister? (no current).
• Can we apply source transformation to compute VTh? Try it at home.
• Using RTh & VTh values Thevenin circuit is drawn in Fig. 4.29.
• Current IL through load RL is given by;
• Class Work: Compute IL for RL equal to;
6 , 16  and 36 ? (3, 1.5 & 0.75 A).
• Home Work: P_Problem 4.8 & Problem 4.36.

• Example 4.9; Find Thevenin equivalent of the
circuit in Fig. 4.31 at terminals a-b?
• This circuit contains a dependent source.
• To find RTh set the independent source equal to zero but leave the
dependent source alone.
• Because of the dependent source, excite the network with a voltage
source vO connected to the terminals as indicated in Fig. 4.32(a).
• Since the circuit is linear, set vO = 1 V for ease of calculation.
• Next find current iO through the terminals and then obtain RTh = vO/iO.

• Alternatively, we may insert a iO = 1 A current source, find the
corresponding voltage vO and compute RTh = vO/iO.
• Applying mesh analysis to the circuit of Fig. 4.32(a);
• For mesh-1:
• Since: hence (4.9.1).

• For mesh 2: (4.9.2).
•  Putting i1 = – 3i2, reduces Eq. 4.9.2 to, 18i2 – 6i3 = 0.
• For mesh 3: (4.9.3).
•  Simplification reduces Eq. 4.9.3 to, – 6i2 + 8i3 = – 1.
• Solving above two equations reveal; i3 = – 1/6 A.
• Hence iO = – i3 = 1/6 A
• therefore; RTh = vO/iO = 1 ÷ 1/6 = 6 .
• To compute VTh, restore current source and determine vOC as
illustrated in the circuit of Fig. 4.32(b).

• Applying mesh analysis to circuit of Fig. 4.32(b);
• For mesh 1: i1 = 5 A (4.9.4).
• For mesh 2:
•  (4.9.6).
• For mesh 3: (4.9.5).

• Class Work: From acquired simultaneous equations find current i2?
• Note that from circuit in Fig 4.32(b); vX = 4(i1 – i2)
• Answer: Mesh current i2 = 10/3 A.
• Therefore;
• Thevenin equivalent is drawn in Fig 4.33.
• Home Work: P_Problem 4.9, problems 4.47 & 4.48 (apply only Thevenin).
• Example 4.10; Determine the Thevenin equivalent of the circuit in
Fig. 4.35(a) at terminals a-b.
• It is important to note that there are no independent sources in circuit.

• Since there is no independent sources in this
circuit, the circuit must be excited externally.
• Also, when there are no independent sources,
there will be NO VTh & need find only the RTh.
• To excite the circuit insert either a 1-V voltage
source or a 1-A current source, as in both cases
we end up with same equivalent resistance.
• In last example we experimented with a
voltage source, so now we use current source.

• We start by writing the nodal equation at node a in Fig. 4.35(b)
• Assuming that iO = 1 A;
• To find node voltage vO, express ix in terms of node voltage. How?
• Current ix expressed in Ohm’s Law reveal
• Substituting this value of iX in drawn nodal equation above, yield;
•
• Class Work: Solve above equation for vO? (vO = – 4 V)
• Since vO = 1 x RTh, therefore; RTh = vO/1 = – 4 .
• Negative sign, as per passive sign rule, implies “supplying power”.

• The resistors in Fig. 4.35(a) cannot supply power (they absorb power);
it is the dependent source that supplies the power.
• This is an example of how a dependent source and resistors could be
used to simulate negative resistance.
• In this circuit we do have an active device (the dependent current
source), thus, the equivalent circuit is essentially an active circuit that
can supply power.
• The book also evaluates the result at length, which is left as self study.
• Home Work: P_Problem 4.10 and Problem 4.64.

4.6: Norton’s Theorem
• Norton’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a current source IN in
parallel with a resistor RN, as shown in Fig. 4.37.
• Such equivalent circuit at terminals a & b is called Norton circuit.
• IN in Norton circuit is short-circuit current between terminals a & b.
• RN is the equivalent resistance at terminals a & b
with all independent sources removed.
• We find RN in the same way as we find RTh.

• Thevenin and Norton resistances are equal i.e. RN = RTh (4.9).
• Short-circuit current flowing from terminal a to b in both circuits of
Fig. 4.37, in either case is IN.
• Since the two circuits are equivalent, thus, as shown in Fig. 4.38;
• IN = ISC (4.10).
• Dependent and independent sources are treated in Norton theorem the
same way as in Thevenin’s theorem.
• Relationship between Norton’s & Thevenin’s
theorems is based on Eq. (4.9), & defined as;

• IN = VTh  RTh (4.11).
• This essentially defines the source transformation that is why it is
often called Thevenin-Norton transformation.
• Since IN, VTh and RTh are interlinked (Eq. 4.11), to determine the
Thevenin or Norton equivalent circuit requires that we find:
1. The open-circuit voltage vOC across terminals a and b.
2. The short-circuit current ISC through terminals a and b.
3. The equivalent or input resistance Rin at terminals a and b when all
independent sources are turned off.

• We can calculate any two of these three using the method that takes
the least effort and then use results to get the third by Ohm’s law.
• Also since: VTh = vOC (4.12a).
• IN = ISC (4.12b).
• And; RTh = vOC  ISC (4.12c).
• Therefore open-circuit and short-circuit tests are sufficient to find any
Thevenin or Norton equivalent of a circuit which contains at least one
independent source.
• Example 4.11: Find Norton equivalent of circuit in Fig. 4.39 at a-b?

• To find RN set the independent sources
equal to zero, as shown in Fig. 4.40(a).
• Thus, RN = 5||(8 + 4 + 8) = 4 .
• Short circuiting a-b, as in Fig. 4.40(b),
renders 5  resistor redundant.
• Applying the mesh analysis;
•
• Solving it;
• Instead; we may find IN from VTh/RTh.

• Find VTh, as the open-circuit
voltage across terminals a-b in
Fig. 4.40(c).
• Using mesh analysis; i3 = 2A.
• Thus,
• And therefore, IN = VTh  RTh = 4/4 = 1 A, that is the same answer.
• Hence the Norton equivalent
circuit is as shown in Fig. 4.41

• Home Work: P_Problem 4.11.
• Example 4.12: Using Norton’s theorem,
find RN and IN of the circuit in Fig. 4.43
at terminals a-b.
• To find RN set the independent
voltage source equal to zero and
connect voltage source of vO = 1 V
to the terminals resulting in a circuit
as shown in of Fig. 4.44(a).

• As the 4  resistor results in short-circuit, hence current iX = 0 and so
will be the output of the dependent current source.
• Hence; iO = vO  R = 1  5 = 0.2 A.
• Therefore; RN = vO  iO = 1  0.2 = 5 .
• To find IN short-circuit terminals a-b and find the current as indicated
in Fig. 4.44(b).
• Note from this figure that both resistors,
the voltage source and the dependent
current source are all in parallel.

• Hence iX = 10 4 = 2.5 A.
• Applying KCL at node a; IN =
• Home Work: P_Problem 4.12.
• Self Study: sec 4.7: Derivations of Thevenin’s & Norton’s Theorems.
4.8: Maximum Power transfer Theorem
• In a circuit designed to provide power to a load, it is desirable to
maximize the power delivered to a load.
• We now address the problem of delivering the maximum power to a
load in a system with known internal losses.

• It should be noted that this will result in significant internal losses
greater than or equal to the power delivered to the load.
• The Thevenin equivalent is useful in finding the maximum power a
linear circuit can deliver to a load.
• Assume that load resistance RL is adjustable or variable.
• If the entire circuit is replaced by its Thevenin equivalent except for
the load, as shown in Fig. 4.48, the power delivered to the load is;
• (4.21).
• For a given circuit, VTh and RTh are fixed.

• By varying the load resistance the power delivered to the load varies
as sketched in Fig. 4.49.
• In Fig. 4.49 power is maximum for
some value of RL between 0 & .
• We now want to show that this
maximum power occurs when RL is
equal to RTh.
• This is known as the maximum power
theorem.

• More formally maximum power theorem states that maximum power
is transferred to the load when the load resistance equals the Thevenin
resistance as seen from the load (RL = RTh).
• To prove power transfer theorem, differentiate power p in Eq. (4.21)
with respect to RL and set the result equal to zero, which yields;
• RL = RTH (4.23).
• The maximum power transferred is obtained by substituting Eq. (4.23)
into Eq. (4.21), i.e. (4.24).
• Equation (4.24) applies only when RL = RTH.

• When RL  RTH we compute the power delivered using Eq. (4.21).
• Example 4.13: Find the value of RL
for maximum power transfer in the
circuit of Example 4.13
• Using the circuit in Fig. 4.51(a) computing RTH reveal;
•

• To find VTh, applying mesh analysis to circuit of Fig. 4.51(b) provide,
•
• Solving reveal i1 = – 2/3 A.
• Applying KVL around the outer loop, yields value for VTh, i.e.
•
• For maximum power transfer RL = RTh = 9 .
• Maximum power transferred;
• Home Work: P_Problem 4.13 plus problems 4.69 and 4.72
• Self Study: 4.10: Applications, 4.11: Summary and all review questions.

Circuit theorems linear circuit analysis

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