Jimma University
Jimma Institute of Technology
Faculty of Electrical & Computer Engineering
Fundamentals of Electrical Engineering
(ECEg2111)
Chapter 2:Circuit Laws and Analysis Techniques
Biruk Teshome
1
Network Terminologies
 Branch: represents a single element such as a voltage source or a resistor. In
other words, a branch represents any two-terminal element. For the circuit
shown in fig.2.1, there are five branches. A 10 V voltage source, 2A current
source, 4 Ω, 5 Ω, and 3 Ω resistors.
 Node: is the point of connection between two or more circuit elements. For the
circuit shown in fig.2.1, there are four nodes. A, B, C, and D.
 Loop: A loop is simply any closed path through the circuit in which no node is
encountered more than once.. For the circuit shown in fig.2.1, there are three
loops. A-B-D-A, B-C-D-B, And A-B-C-D-A.
 Mesh: is a loop that does not enclose any other loops. For the circuit shown in
fig.2.1, there are two meshes. A-B-D-A and B-C-D-B.
 Series: two elements are said to be in series if they share a single node and if
there are no other current-carrying connections at this node. And consequently
carry the same current. In fig.2.1 , 10V & 5Ω are in series and also 3Ω & 2A.
 Parallel: two or more elements
are in parallel if they are
connected to the same two nodes
and consequently have the same
voltage across them.
Fig. 2.1
2
Kirchhoff’s Laws
Kirchhoff’s Current Law(KCL)
 Kirchhoff’s current law states that the algebraic sum of all the currents at any
node in a circuit equals zero. In other words, the sum of currents flowing into
that node is equal to the sum of currents flowing out of that node.
𝑖=0
𝑜𝑟
𝑖𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 =
𝑖𝑙𝑒𝑎𝑣𝑖𝑛𝑔
 we add the currents entering and subtract the currents
leaving. For illustration, consider Fig. 2.2. Then, we can
write:
Node a: 𝑖1 + 𝑖2 − 𝑖3 = 0
Example 2.1. Find current and voltage in the circuit shown
in Fig. 2.3.
Fig. 2.2 Partial circuit to
illustrate KCL
Solution: Applying KCL to node a, we obtain
For the 4Ω- resistor, Ohm’s law gives
Fig. 2.3
3
Kirchhoff’s Voltage Law(KVL)
 Kirchhoff’s voltage law states that the algebraic sum of all voltages around a
closed path (or loop) is zero. In other words, the summation of voltage rises is
equal to the summation of voltage drops around a closed loop.
𝑉=0
𝑜𝑟
𝑉𝑟𝑖𝑠𝑒𝑠 =
𝑉𝑑𝑟𝑜𝑝𝑠
 In applying KVL to a loop, voltages are added or subtracted depending on their
reference polarities relative to the direction of travel around the loop.
 For the circuit of Fig.
2.4, we obtain the
following equations:
Loop 1: −𝑣𝑎 + 𝑣𝑏 + 𝑣𝑐 = 0
Loop 2: −𝑣𝑐 − 𝑣𝑑 + 𝑣𝑐 = 0
Loop 2: 𝑣𝑎 − 𝑣𝑏 + 𝑣𝑑 −𝑣𝑒 = 0
Fig. 2.4 convenient
convention
Fig. 2.5 Circuit used for illustration of Kirchhoff’s
voltage law.
4
Example 2.2. Determine 𝑣0 and i in the circuit shown in Fig. 2.6.
Solution: Apply KVL around the loop. The result is
−12 + 4𝑖 + 2𝑣0 − 4 − 𝑣0 = 0
(2.2.1)
Applying Ohm’s law to the 6Ω- resistor gives
𝑣0 = −6𝑖
(2.2.2)
Substituting Eq. (2.2.2) into Eq. (2.2.1) yields
−16 + 10𝑖 − 12𝑖 = 0
And
⇒
𝐢 = −𝟖𝐀
𝒗𝟎 = 𝟒𝟖𝑽
Exercise 2.1. Find 𝑣0 and 𝑖0 in the
circuit of Fig. 2.7.
Equivalent Circuits
Fig. E2.1
Ans. 12 V, 6 A.
5
Fig. 2.6
Resistors in Series and parallel
Series Resistors and Voltage Divider Rule (VDR)
 For Fig. 2.7, If we apply KVL to the loop (moving in the clockwise direction),
we have
𝑣 = 𝑣1 + 𝑣2 = 𝑖𝑅1 + 𝑖𝑅2
= 𝑖(𝑅1 +𝑅2 )
= 𝑖𝑅𝑒𝑞
 Implying that the two resistors can be replaced
by an equivalent resistor 𝑅𝑒𝑞; that is,
𝑅𝑒𝑞 = 𝑅1 + 𝑅2
Fig. 2.7 A single-loop circuit
with two resistors in series.
 Thus, Fig. 2.7 can be replaced by the equivalent
circuit in Fig. 2.8. They are equivalent because
they
exhibit
the
same
voltage-current
relationships at the terminals a-b.
 For N resistors in series then,
Fig. 2.8 Equivalent circuit of
the Fig.2.7 circuit.
𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
 The equivalent resistance of any number of resistors connected in series is the
sum of the individual resistances.
6
.
To determine the voltage across each resistor from the source voltage in Fig. 2.7
𝑣1 = 𝑖𝑅1 =
𝑅1
𝑅1
𝑣=
𝑣
𝑅𝑒𝑞
𝑅1 + 𝑅2
𝑣2 = 𝑖𝑅2 = 𝑖𝑅1 =
𝑅1
𝑅1
𝑣=
𝑣
𝑅𝑒𝑞
𝑅1 + 𝑅2
 In general, if we have N resistors in series with the source voltage v, the nth
resistor (𝑅𝑛 ) will have a voltage drop of
𝑅𝑛
𝑅𝑛
𝑣𝑛 =
𝑣=
𝑣
𝑅𝑒𝑞
𝑅1 + 𝑅2 + ⋯ + 𝑅𝑁
 The voltage dropped across any series resistor is proportional to the
magnitude of the resistor. The larger the resistance, the larger the voltage drop.
This is called the principle of voltage division or VDR, and the circuit in Fig.
2.7 is called a voltage divider.
Parallel Resistors and Current Divider Rule (CDR)
 For Fig. 2.9, Applying KCL at node a gives the
total current i as
𝑖 = 𝑖1 + 𝑖2 =
𝑣
𝑣
1
1
𝑣
+
=𝑣
+
=
𝑅1 𝑅2
𝑅1 𝑅2
𝑅𝑒𝑞
 Where 𝑅𝑒𝑞 is the equivalent resistance of the
resistors in parallel:
7
Fig. 2.9 Two resistors in parallel.
1
1
1
=
+
𝑅𝑒𝑞 𝑅1 𝑅2
𝑅1 𝑅2
𝑅𝑒𝑞 =
𝑅1 + 𝑅2
⇒
 Thus, the equivalent resistance of two parallel resistors is
equal to the product of their resistances divided by their
sum. This applies only to two resistors in parallel.
 We can extend the result to the general case of a circuit
with N resistors in parallel. The equivalent resistance is
Fig. 2.10 Equivalent
circuit to Fig. 2.9.
1
1
1
1
=
+
+ ⋯+
𝑅𝑒𝑞 𝑅1 𝑅2
𝑅𝑁
 Note that 𝑅𝑒𝑞 is always smaller than the resistance of the smallest resistor in
the parallel combination.
Given total current i entering node a in Fig. 2.9, We can obtain current 𝑖1 and 𝑖2
𝑖1 =
𝑖𝑅𝑒𝑞
𝑣
𝑅2
=
=
𝑖
𝑅1
𝑅1
𝑅1 + 𝑅2
𝑖2 =
𝑖𝑅𝑒𝑞
𝑣
𝑅1
=
=
𝑖
𝑅2
𝑅2
𝑅1 + 𝑅2
 Thus, in general, if we have N resistors in parallel with the source current i, the
nth resistor (𝑅𝑛 ) will have current of
𝑖𝑅𝑒𝑞
𝑖𝑛 =
=
1
𝑅𝑛
1
𝑅1
+1
𝑅2
𝑅𝑛
+⋯+1
𝑖
𝑅𝑁
8
 It shows that the total current i is shared by the resistors in inverse proportion
to their resistances. This is known as the principle of current division or CDR,
and the circuit in Fig. 2.9 is known as a current divider. Notice that the larger
current flows through the smaller resistance.
Example 2.3. Find 𝑅𝑒𝑞 for the circuit shown in Fig. 2.11.
Solution:
 The 1Ω and 5Ω resistors are in series; so their
equivalent resistance is
 The 6Ω and 3Ω resistors are in parallel, so
Fig. 2.11
 Thus the circuit in Fig. 2.11 is reduced Fig.
2.12(a). So
and
 The circuit in Fig. 2.12(a) is now replaced with
that in Fig. 2.12(b). So
Fig.2.12
9
Exercise 2.2. Calculate the equivalent resistance
a) R ab , R cb , R db & R cd in the circuit in Fig. E2.1(a), [Ans. 𝑹𝒂𝒃 = 𝟏𝟏. 𝟐𝜴]
b) R ab & R cd in Fig. E2.1(b).
c) And current I in Fig. E2.1(c).
Fig. E2.2(a)
Fig. E2.2(b)
Fig. E2.2(c)
Example 2.2. Find 𝑖𝑜 and 𝑣𝑜 in the circuit shown in Fig. 2.13. Calculate the
power dissipated in the 3Ω resistor.
Solution:
 Combine the 6Ω and 3Ω resistors, replacing them
with:
 Since 𝑣𝑜 appears across the parallel combination,
the simplification has not lost this quantity.
10
Fig. 2.13
Example 2.3. Consider the circuit of Fig. 2.14:
A. Find the total resistance 𝑅𝑇 “seen” by the source E.
B. Calculate 𝐼𝑇 , 𝐼1 , and 𝐼2 .
C. Determine the voltages 𝑉2 and 𝑉4 .
Solution:
A. The total resistance of the circuit is
B. The current supplied by the voltage source is
Fig. 2.14
C. The voltages 𝑉2 and 𝑉4 are
now easily calculated by
using Ohm’s law:
Applying the current divider rule
11
Example 2.4. The circuit of Fig. 2.15 is referred to as a bridge circuit and is
used extensively in electronic and scientific instruments. Calculate the current I
and the voltage 𝑉𝑎𝑏 when
A. 𝑅𝑥 = 0Ω (short circuit)
B. 𝑅𝑥 = 15kΩ
C. 𝑅𝑥 = ∞ (open circuit)
Solution:
A. 𝑅𝑥 = 0Ω. The circuit is redrawn as shown in Fig. 2.16
 The voltage source “sees” a total resistance of
 Resulting in a source current of
Fig. 2.15
 The voltage 𝑉𝑎𝑏 may be determined by solving for
voltage across 𝑅1 and 𝑅2 .
Fig. 2.16
12
B. 𝑅𝑥 = 15kΩ. The circuit is redrawn as shown in Fig. 2.17
 The voltage source “sees” a total resistance of
 Resulting in a source current of
 The voltage 𝑉𝑎𝑏 may be determined by solving for
voltage across 𝑅1 and 𝑅2 .
C. 𝑅𝑥 = ∞. The circuit is redrawn as shown in Fig. 2.18
 The voltage source “sees” a total resistance of
Fig. 2.17
Fig. 2.18
 The voltage 𝑉𝑎𝑏 may be determined by
solving for voltage across 𝑅1 and 𝑅2 .
 Resulting in a source current of
13
Exercise 2.3. For the circuit in Fig. E2.3
a) show that 𝑖1 = 0 & 𝑖2 = 𝑖 in Fig. E2.3(a)
b) equivalent resistance 𝑅𝐴𝐵 in Fig. E2.3(a)
c) show that 𝑖1 = 𝑖 & 𝑖2 = 0 in Fig. E2.3(b)
d) equivalent resistance 𝑅𝐴𝐵 in Fig. E2.3(b)
A
A
B
B
(a)
(b)
Fig. E2.3
Exercise 2.4. For the circuit in Fig. E2.4 calculate
a) io and vo . b) the power dissipated in the 6Ω resistor.
c) the power developed by the current source.
Fig. E2.4
14
Wye-Delta Transformations(Y-Δ)
 Situations often arise in circuit analysis when the resistors are neither in
parallel nor in series. For example, consider the bridge circuit in Fig.2.19.
 How do we combine resistors 𝑅1 through 𝑅6 when
the resistors are neither in series nor in parallel?
 Many circuits of the type shown in Fig. 2.19 can be
simplified by using three-terminal equivalent
networks. These are the wye (Y) or tee (T) network
shown in Fig. 2.20 and the delta (Δ) or pi (𝜋)
network shown in Fig. 2.21.
Fig. 2.20 Two forms of the same network:
Y & T.
,
Fig.2.19 bridge circuit
Fig. 2.21 Two forms of the same network:
(a) Δ , (b) 𝜋.
 Our main interest here is in how to identify them when they occur as part of a
network and how to apply wye-delta transformation in the analysis of that
network.
15
 The transformation that relates the resistances 𝑅𝑎 , 𝑅𝑏 and 𝑅𝑐 to the
resistances 𝑅1 , 𝑅2 and 𝑅3 is derived as follows.
 For the two networks to be equivalent at each corresponding pair of terminals,
it is necessary that the resistance at the corresponding terminals be equal.
 Therefore, if we equate the resistances for each corresponding set of terminals,
we obtain the following equations:
Δ to Y Conversion
Y to Δ Conversion
 Solving the above set
of equations for 𝑅1 ,
𝑅2 and 𝑅3 yields
 Solving the above set
of equations for 𝑅𝑎 ,
𝑅𝑏 and 𝑅𝑐 ,we obtain
Fig. 2.22 Superposition
of Y and Δ networks as
an aid in transforming
one to the other.
16
Exercise 2.5. Find the current and power supplied by the 40 V
source in the circuit shown in Fig. E2.5.
Fig. E2.6
Fig. E2.5
Exercise 2.6. Determine V in the circuit of Fig. E2.6.
Exercise 2.7. For the circuits in Fig. E2.7, obtain the equivalent
resistance at terminals a-b..
Fig. E2.7
17
Nodal Analysis
 Nodal analysis is based on a systematic application of Kirchhoff’s current law
(KCL). It is also known as the node-voltage method.
 Nodal analysis provides a general procedure for analyzing circuits using node
voltages as the circuit variables.
 In nodal analysis, we are interested in finding the node voltages. Once the node
voltages have been found, it is relatively easy to find the current, voltage, and
power for each element in the circuit.
Steps to determine the node voltages:
1.
2.
3.
4.
5.
Select a node as the reference node.
Assign voltages 𝑣1 , 𝑣2 ,…, 𝑣𝑛−1 to the remaining n-1 nodes. The
voltages are referenced with respect to the reference node.
Apply KCL to each of the n-1 non-reference nodes.
Use Ohm’s law to express the branch currents in terms of node voltages.
Solve the resulting simultaneous equations to obtain the unknown node
voltages.
18
Example 2.5.
A. Use the node-voltage method of circuit analysis to find the branch currents
𝑖𝑎 , 𝑖𝑏 , and 𝑖𝑐 , in the circuit shown in Fig. 2.23.
B. Find the power associated with each source, and state whether the source is
delivering or absorbing power.
Solution:
A. the circuit has two essential nodes.
Hence we need one node-voltage
equations to describe the circuit.
At node 1, applying KCL and Ohms law gives
Fig. 2.23
Solving for 𝑣1 gives,
Hence,
Fig. 2.24 The circuit shown in Fig. 2.23
with a reference node and the unknown
node voltage 𝑣1 .
19
B. The power associated with the 50 V and 3 A sources are
 Check by the conservation of power! By calculating the sum of power
absorbed by the three resistors.
Example 2.6. Use the nodal analysis to find the power dissipated in the 5Ω
resistor in the circuit shown in Fig. 2.25.
Solution:
At node 1, applying KCL and Ohms law
At node 2, applying KCL and Ohms law
Fig. 2.25
We must express the controlling
current in terms of the node voltages,
Fig. 2.26
20
Substituting this relationship into the node 2 equation and solving for 𝑣1 and 𝑣2
gives, 𝑣1 = 16 𝑉 and 𝑣2 = 10 𝑉
Thus,
𝑃5Ω = 𝑖5Ω 2 × 5 = 𝑖ϕ 2 × 5 = 7.2 𝑊
The Node-Voltage Method: Some Special Cases
 CASE 1: If a voltage source is connected between the reference node and a
non reference node, we simply set the voltage at the non reference node
equal to the voltage of the voltage source. In Fig. 2.27, for example,
𝑣1 = 10 𝑉
Thus, our analysis is somewhat simplified by this knowledge of the voltage at
this node.
 CASE 2: If the voltage source (dependent or independent) is connected
between two non reference nodes, the two non reference nodes form a
generalized node or super node; we apply both KCL and KVL to determine
the node voltages.
21
 A supernode is formed by enclosing a
(dependent or independent) voltage
source connected between two non
reference nodes and any elements
connected in parallel with it.
 In Fig. 2.27, nodes 2 and 3 form a
supernode.
 We analyze a circuit with super-nodes
using the same steps mentioned above
except that the super-nodes are treated
differently.
Fig. 2.27 A circuit with a supernode.
Basic steps:
1. Take off all voltage sources in super- nodes and apply KCL to supernodes.
for Fig. 2.27
⇒
2. Put voltage sources back to the nodes and apply KVL to relative loops.
22
Example 2.7. For the circuit shown in Fig. 2.28, find the node voltages.
Solution:
The supernode contains the 2 V source, nodes
1 and 2, and the 10 Ω resistor.
At supernode, applying KCL and Ohms law
Applying KVL,
2 = 𝑣2 − 𝑣1
Solving the equations, 𝑣1 = −7.333 𝑉 𝑎𝑛𝑑 𝑣2 = −5.333 𝑉
Fig. 2.28
 Note that the 10Ω resistor does not make any difference because it is
connected across the supernode.
Exercise 2.8. Use the node-voltage
method to find the value of 𝑣0 in the
circuit in Fig. E2.8.
Fig. E2.8
23
Exercise 2.9. Use the node-voltage method to find the value of 𝐼4Ω in the circuit
in Fig. E2.9.
Exercise 2.10. Use the node-voltage method to find the value of 𝑣𝑥 in the
circuit in Fig. E2.10.
𝐼4Ω
Fig. E2.9.
Fig. E2.10.
24
Mesh Analysis
 Mesh analysis is based on a systematic application of Kirchhoff’s voltage law
(KVL). Mesh analysis is also known as loop analysis or the mesh-current
method. It provides another general procedure for analyzing circuits, using
mesh currents as the circuit variables. In this analysis, we are interested in
finding the mesh current. Once the mesh current have been found, it is
relatively easy to find the current, voltage, and power for each element in the
circuit.
 But, It is not quite as general as nodal analysis because it is only applicable to
a circuit that is planar. A planar circuit is one that can be drawn in a plane
with no branches crossing one another; otherwise it is nonplanar. A circuit
may have crossing branches and still be planar if it can be redrawn such that it
has no crossing branches.
Fig.29 A nonplanar circuit.
Fig.30 (a) A planar circuit with crossing branches,
(b) the same circuit redrawn with no crossing branches.
25
Steps to determine the mesh currents:
1. Assign mesh currents i1, i2, …, in to the n meshes.
2. Apply KVL to each of the n meshes.
3. Use Ohm’s law to express the voltages in terms of the mesh currents.
4. Solve the resulting n simultaneous equations to get the mesh currents.
Example 2.7.
A. Use the mesh-current method to determine the power associated with each
voltage source in the circuit shown in Fig. 2.31.
B. Calculate the voltage 𝑣0 across the 8Ω resistor.
Solution: A. Applying KVL for each mesh,
the three mesh current equations are
Mesh 1
Fig. 2.31
Mesh 2
Mesh 3
Solving the equations, The three mesh currents are
𝑖𝑎 = 5.6 𝐴, 𝑖𝑏 = 2 𝐴, 𝑎𝑛𝑑 𝑖𝑐 = −0.8 𝐴
26
Fig. 2.32
The power associated with the sources are
B. The branch current in the 8Ω resistor in the direction of the voltage drop 𝑣0
is 𝑖𝑎 − 𝑖𝑏 . Therefore
Example 2.8. Use the mesh-current method of circuit analysis to determine the
power dissipated in the 4Ω resistor in the circuit shown in Fig. 2.33.
Solution: Applying KVL for each mesh, the
three mesh current equations are
Mesh 1
Mesh 2
Mesh 3
Fig. 2.33
We must express controlling current in terms of
the mesh currents as
Substituting controlling current into the mesh
current equation three and solving for mesh
currents, 𝑖1 = 29. 6 𝐴, 𝑖2 = 26 𝐴, & 𝑖3 = 28 𝐴
27
Fig. 2.34
The current in the 4Ω resistor oriented from left to right is 𝑖3 − 𝑖2 , or 2 A. Therefore the
power dissipated is
Note: What if you had not been told to use the mesh current method? Would you have
chosen the node voltage method? It reduces the problem to finding one unknown node
voltage because of the presence of two voltage sources between essential nodes.
The Mesh-Current Method: Some Special Cases
 CASE 1: When a current source(dependent or independent) exists only in
one mesh: Consider the circuit in Fig. 2.35, for example. We set 𝑖2 = −5 𝐴
and write a mesh equation for the other mesh in the usual way; that is,
 CASE 2: When a current source exists between two meshes: Consider the
circuit in Fig. 2.36(a), for example. We create a supermesh by excluding the
current source and any elements connected in series with it, as shown in Fig.
2.36(b). Thus, As shown in Fig. 2.36(b), we create a supermesh as the
periphery of the two meshes and treat it differently. Therefore, applying KVL
to the supermesh in Fig. 3.23(b) gives
28
We apply KCL to a node in the branch where the
two meshes intersect. Applying KCL to node 0 in
Fig. 2.36(a) gives
Solving the equations,
Fig. 2.35 A circuit with a
current source.
Fig. 2.36 (a) Two meshes having a current source in common, (b) a supermesh, created by
excluding the current source.
29
Example 2.9. For the circuit in Fig. 2.37, find 𝑖1 to 𝑖4 using mesh analysis.
Solution: Note that meshes 1 and
2 form a supermesh since they have
an independent current source in
common. Also, meshes 2 and 3
form another supermesh because
they have a dependent current
source in common. The two
supermeshes intersect and form a
larger supermesh as shown.
Applying KVL to the larger
supermesh,
Fig. 2.37
For the independent current source, we apply
KCL to node P:
Applying KVL in mesh 4,
For the dependent current source, we apply KCL
to node Q:
Solving the four equations,
30
Exercise 2.11. Using mesh analysis, find 𝐼0 in the circuit of Fig. E2.11.
Exercise 2.12. Using mesh analysis, determine the current through the 6Ω
resistor in the circuit of Fig. E2.12.
Fig. E2.11
Fig. E2.12
31
Source Transformation
 It is the process of replacing a voltage source 𝑣𝑠 in series with a resistor R by a
current source 𝑖𝑠 in parallel with a resistor R, or vice versa.
 An equivalent circuit is one whose v-i characteristics are identical with the
original circuit.
𝑣𝑠 = 𝑖𝑠 𝑅
𝑜𝑟
𝑣𝑠
𝑖𝑠 =
𝑅
 The arrow of the current
source is directed toward the
positive terminal of
the
voltage source.
 The source transformation is
not possible when
R = 0
for voltage source and R = ∞
for current source.
Fig.2.38 Transformation of independent sources
Fig.2.39 Transformation of dependent sources
32
Example 2.10. For the circuit shown in Fig. 2.40, find the power associated
with the 6 V source and state whether it is absorbing or delivering the power.
Solution: we must reduce the circuit
in a way that preserves the identity of
the branch containing the 6 V source.
Fig. 2.40
I
The current in the direction of the
voltage drop across the 6 V source is
𝐼=
19.2 − 6
16
= 0.825 A
Therefore the power associated with the
6 V source is
𝑃6𝑉 = 0.825 6 = 𝟒. 𝟗𝟓 𝐖 (absorbing)
33
Example 2.10. Find 𝑣𝑥 in Fig. 2.41 using source transformation.
Solution:
Step 1
Fig. 2.41
Applying KVL to the loop containing
only the 3 V voltage source, the 1Ω
resistor, and 𝑣𝑥 yields,
Solving the two equations,
Step 2
𝑖 = −4.5 𝐴 𝑎𝑛𝑑 𝑣𝑥 = 7.5 𝑉
Notice that the terminals for 𝑣𝑥 are intact.
Applying KVL around the loop in Fig.
step 2 gives
34
Exercise 2.13. Find 𝑖𝑥 in the circuit of Fig.E2.13 using source transformation.
Answer: 7.056 mA.
Fig.E2.13
Exercise 2.14. Use source transformation to find 𝑖0 in the circuit shown in
Fig.E2.14.
Answer: 7.056 mA.
Fig.E2.14
35
Superposition Theorem
 Superposition Theorem states that the voltage across (or current through) an
element in a linear circuit is the algebraic sum of the voltage across (or
currents through) that element due to each independent source acting alone.
 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. A system is said to be linear, it should satisfy
both Homogeneity & additive property. Superposition is not applicable to the
effect on power.
 Keep in mind that superposition is based on linearity. For this reason, it is not
applicable to the effect on power due to each source, because the power
absorbed by a resistor depends on the square of the voltage or current. If the
power value is needed, the current through (or voltage across) the element
must be calculated first using superposition.
Steps to apply superposition principle:
1.
Select one of the independent sources. Turn off all other independent
sources(note that dependent sources keep on).
 Turn off independent voltage source
0 V (short circuit)
 Turn off independent current source
0 A (open circuit)
36
2.
3.
4.
5.
6.
Relabel voltages and currents using suitable notation(e.g., 𝑣 ′ , 𝑣 ′′). Be sure to
relabel controlling variables of dependent sources to avoid confusion.
Analyze the simplified circuit to find the desired currents and/or voltages.
Repeat steps 1 through 3 until each independent source has been considered.
Add the partial currents and/or voltages obtained from the separate
analyses. Pay careful attention to voltage signs and current directions when
summing.
Do not add power quantities. If power quantities are required, calculate only
after partial voltages and/or currents have been summed.
Example 2.11. Use the superposition theorem to find V in the circuit of
Fig.2.42
Solution:
Since there are two sources, let
𝑣 = 𝑣 ′ + 𝑣 ′′
Where 𝑣 ′ and 𝑣 ′′ are the contributions
due to the 6 V voltage source and the 3
A current source, respectively.
37
Fig. 2.42
I. Contribution of the 6V voltage source:
 We need to turn off the current source
by replacing it with an open circuit.
Using current division rule,
8
𝑖1′′ =
3 = 2𝐴
8+4
Hence,
𝑣′
𝑣 ′′ = 4𝑖1′′ = 4 2 = 8 V
Therefore,
Using voltage division rule,
𝑣′ =
𝑣 = 𝑣 ′ + 𝑣 ′′
4
6 =2𝑉
4+8
𝒗 = 2 + 8 = 𝟏𝟎 𝐕
II. Contribution of the 3A current source:
 We need to turn off the voltage source
by replacing it with short circuit.
𝑖2′′
𝑖1′′
𝑣 ′′
38
Example 2.12. Use the superposition theorem to find 𝑖𝑥 in the circuit of
Fig.2.43
Solution:
I. Contribution of the 10V voltage source:
 We need to turn off the independent current
source by replacing it with an open circuit.
Fig. 2.43
write the single node equation,
𝑣 ′′ 𝑣 ′′ − 2𝑖𝑥′′
+
=3
2
1
Apply KVL,
−10 + 2𝑖𝑥′ + 𝑖𝑥′ + 2𝑖𝑥′ = 0 ⇒ 𝒊′𝒙 = 𝟐 𝑨
II. Contribution of the 3A current source:
 We need to turn off the independent voltage
source by replacing it with short circuit.
Relate the dependent-sourcecontrolling quantity to 𝑣 ′′ :
𝑣 ′′ = 2(−𝑖𝑥′′ )
Solving, we find
𝒊′′
𝒙 = −𝟎. 𝟔 𝑨
Thus,
𝒊𝒙 = 𝑖𝑥′ + 𝑖𝑥′′ = 𝟏. 𝟒𝐀
39
Exercise 2.15. Find I in the circuit of Fig. E.2.15 using the superposition
[Answer: 0.75 A]
principle.
Exercise 2.16. Use the principle of superposition to find 𝑣0 in the circuit shown
in Fig. E.2.16.
[Answer: 24 V]
Fig. E.2.15
Fig. E.2.16
Exercise 2.17. For the circuit of Fig.E2.17, use superposition to obtain the
voltage across each current source. [Answer: 𝒗𝟏 = 𝟏𝟏. 𝟏𝟒𝟕 𝐕, 𝒗𝟐 = −𝟏. 𝟑𝟗𝟒 𝐕 ]
Fig. E.2.17
40
Thevenin’s theorem
 Thevenin’s theorem states that a linear two-terminal circuit can be replaced
by an equivalent circuit consisting of a voltage source 𝑉𝑇ℎ in series with a
resistor 𝑅𝑇ℎ , where 𝑉𝑇ℎ is the open-circuit voltage at the terminals and 𝑅𝑇ℎ is
the input or equivalent resistance at the terminals when the independent
sources are turned off.
Finding Thevenin’s Voltage (𝑉𝑇ℎ )
 If the terminals a-b are made open-circuited
(by removing the load), no current flows, so
that the open-circuit voltage across the
terminals a-b in Fig. 2.44(a) must be equal
to the voltage source 𝑉𝑇ℎ in Fig. 2.44(b),
since the two circuits are equivalent.
 Thus 𝑽𝑻𝒉 is the open-circuit voltage across
the terminals as shown in Fig. 2.45; that is,
𝑽𝑻𝒉 = 𝒗𝒐𝒄
Fig. 2.44 Replacing a linear two-terminal circuit
by its Thevenin equivalent: (a) original circuit,
(b) the Thevenin equivalent circuit.
Fig.2.45 Finding 𝑉𝑇ℎ
41
Finding Thevenin’s Resistance (𝑅𝑇ℎ )
 CASE 1: If the network has no dependent
sources, 𝑅𝑇ℎ is the input resistance at the
terminals when the independent sources are
turned off, as shown in Fig. 2.46; that is,
𝑹𝑻𝒉 = 𝑹𝒊𝒏
 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. We apply a voltage source 𝑣0
at terminals a-b as shown in Fig. 2.47(a), and
determine the resulting current 𝑖0 . Alternatively,
we may insert a current source 𝑖0 at terminals ab as shown in Fig. 247(b), and find the terminal
voltage 𝑣0 . Then,
𝒗𝟎
𝑹𝑻𝒉 =
𝒊𝟎
42
Fig. 2.46 Finding R Th when circuit
has no dependent sources.
(a)
(b)
Fig. 2.47 Finding R Th when circuit
has dependent sources.
 We may assume any value of We may assume any value of 𝑣0 and 𝑖0 For
example, we may use 𝑣0 = 1 V or 𝑖0 = 1 A , or even use unspecified values
of 𝑣0 or 𝑖0 .
 It often occurs that 𝑉𝑇ℎ takes a negative value. In this case, the negative
resistance (𝑉 = −𝑖𝑅) implies that the circuit is supplying power. This is
possible in a circuit with dependent sources.
 As mentioned earlier, a linear circuit with a variable load can be replaced by
the Thevenin equivalent, exclusive of the load. The equivalent network
behaves the same way externally as the original circuit. Consider a linear
circuit terminated by a load 𝑅𝐿 , as shown in Fig. 2.48(a).
 The current through the load and the voltage across the load are easily
determined once the Thevenin equivalent of the circuit at the load’s terminals
is obtained, as shown in Fig. 2.48(b). From Fig. 2.48(b), we obtain
𝑉𝑇ℎ
𝐼𝐿 =
𝑅𝑇ℎ + 𝑅𝐿
𝑉𝐿 = 𝑅𝐿 𝐼𝐿 =
(a)
𝑅𝐿
𝑉
𝑅𝑇ℎ + 𝑅𝐿 𝑇ℎ
(b)
Fig.2.48 A circuit with a load: (a) original circuit, (b) Thevenin equivalent.
43
Example 2.13. Find the Thevenin equivalent circuit of the circuit shown in Fig.
2.49, to the left of the terminals a-b. Then find the current through 𝑅𝐿 =
6, 16 𝑎𝑛𝑑 36 Ω.
Solution:
 Finding 𝑽𝑻𝒉
Remove the load as shown in Fig. 2.50.
Fig. 2.49
 Finding 𝑹𝑻𝒉
Turn off the independent voltage
sources as shown in Fig. 2.51
32 V voltage source ⇒ short
2 A current source ⇒ open
𝑅𝑇ℎ =
Fig. 2.50
4 × 12
+1=4Ω
16
Fig. 2.51
44
 The Thevenin’s equivalent circuit is as shown in
Fig.2.52
Thus,
Fig. 2.52
Example 2.14. Find the Thevenin equivalent of the circuit in Fig. 2.53 at
terminals a-b.
Solution:  Finding 𝑽𝑻𝒉
Applying mesh analysis for Fig. 2.53, we get
Solving the equations,
Thus,
Fig. 2.53
45
 Finding 𝑹𝑻𝒉
Applying mesh analysis to loop 1 in the
circuit of Fig. 2.54 results in
For loops 2 and 3, applying KVL produces
Solving these equations gives
Fig. 2.54
 The Thevenin equivalent is
46
Exercise 2.18. Apply Thevenin’s theorem to find 𝑣0 in the circuit of Fig.
E2.18.
Exercise 2.19. Determine 𝑅𝑇ℎ and 𝑉𝑇ℎ at terminals 1-2 of the circuits in Fig.
E2.19.
Fig. E2.19
Fig. E2.18
 For a network having a dependent source but no independent
source.
 𝑉𝑇ℎ = 0
𝑣
 𝑅𝑇ℎ = 𝑖 0 𝑎𝑠 𝑖𝑛 𝑐𝑎𝑠𝑒 2.
0
Exercise 2.20. Find the Thevenin equivalent for
the network of Fig. E2.20. 𝑨𝒏𝒔𝒘𝒆𝒓: 𝑹𝑻𝒉 = 𝟐𝟎𝜴
Fig. E2.20
47
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 , where 𝐼𝑁 is the short-circuit current through the terminals and 𝑅𝑁 is the
input or equivalent resistance at the terminals when the independent sources
are turned off.
Finding Norton’s Current (𝐼𝑁 )
To find the Norton current 𝐼𝑁 we determine the
short-circuit current flowing from terminal a to b as
shown in Fig. 2.56.
𝑰𝑵 = 𝒊𝒔𝒄
Fig. 2.55(a) Original circuit,
(b) Norton equivalent circuit.
Fig. 2.56 Finding 𝐼𝑁
48
Finding Norton’s Resistance (𝑅𝑁 )
 We find 𝑅𝑁 in the same way we find 𝑅𝑇ℎ .In fact, from what we know about
source transformation, the Thevenin and Norton resistances are equal; that is,
𝑹𝑵 = 𝑹𝑻𝒉
 Observe the close relationship between Norton’s and Thevenin’s theorems.
The Thevenin’s and Norton equivalent circuits are related by a source
transformation.
𝑽𝑻𝒉
𝑰𝑵 =
𝑹𝑻𝒉
Example 2.14. Find the Norton equivalent circuit of the circuit in Fig. 2.57 at
terminals a-b.
Solution:  Finding 𝑰𝑵
Fig. 2.57
Applying mesh analysis,
⇒
49
 Finding 𝑹𝑵
 Alternatively, we may determine 𝐼𝑁 = 𝑉𝑇ℎ 𝑅𝑇ℎ
Thus,
Norton equivalent circuit
Applying mesh analysis,
50
Maximum Power Transfer Theorem
 If the entire circuit is replaced by its Thevenin equivalent except for the load,
the power delivered to the load is:
2
P  i 2 RL
 VTh 
 
 RL
 RTh  RL 
 Maximum power is transferred to the
load when the load resistance equals
the Thevenin resistance as seen from
the load ( 𝑹𝑳 = 𝑹𝑻𝒉 ).
Fig. 2.58 The circuit used for
maximum power
2
RL  RTH

V
Pmax  Th
4 RL
 For a given circuit, 𝑉𝑇ℎ 𝑎𝑛𝑑 𝑅𝑇ℎ are
fixed. By varying the load resistance
𝑅𝐿 , the power delivered to the load
varies as sketched in Fig. 2.59.
Fig. 2.59 Power delivered to the
load as a function of RL.
51
 To prove the maximum power transfer theorem, we differentiate p with
respect to 𝑅𝐿 and set the result equal to zero. Then substitute the result in the
power equation.
2
P  i 2 RL
 VTh 
 
 RL
 RTh  RL 
2
dp
2  ( RTh  RL )  2 RL ( RTh  RL ) 
 VTh 

4
dRL
(
R

R
)
Th
L


 ( RTh  RL  2 RL ) 
V 
0
3
 ( RTh  RL )

0  ( RTh  RL  2 RL )  ( RTh  RL )
2
Th
RL  RTh
VTh2
pmax 
4 RTh
52
Example 2.15. Find the value of 𝑅𝐿 for maximum power transfer in the circuit
of Fig. 2.60. Find the maximum power.
Solution:  Finding 𝑽𝑻𝒉
Applying mesh analysis
Solving for 𝑖1 , 𝑖1 = −2/3 A
KVL around the outer loop
Fig. 2.60
 Finding 𝑷𝒎𝒂𝒙
 Finding 𝑹𝑻𝒉
53
Exercise 2.21. The variable resistor 𝑅0 in the circuit in Fig. E2.21 is adjusted
until it absorbs maximum power from the circuit.
A. Find the value of 𝑅0 .
B. Find the maximum power.
C. Find the percentage of the total power developed in the circuit that is
delivered to 𝑅0 .
Exercise 2.22. For the circuit shown in Fig. E2.22 determine v0 , for load
resistor RL = 10
A. Using nodal analysis
B. Using mesh analysis
C. Using source transformation theorem
D. Using superposition theorem
E. Using Thevenin theorem
Fig. E2.21
Fig. E2.22
54